Primitive. The name Eps_i is a term of type (set → prop) → set.
Axiom. (Eps_i_ax) We take the following as an axiom:
∀P : set → prop, ∀x : set, P x → P (Eps_i P)
In Proofgold the corresponding term root is c3f0de... and proposition id is 756d6e...
Definition. We define True to be ∀p : prop, p → p of type prop.
In Proofgold the corresponding term root is f81b39... and object id is 94e6a7...
Definition. We define False to be ∀p : prop, p of type prop.
In Proofgold the corresponding term root is 1db705... and object id is 266ad5...
Definition. We define not to be λA : prop ⇒ A → False of type prop → prop.
In Proofgold the corresponding term root is f30435... and object id is 0dc3c7...
Notation. We use ¬ as a prefix operator with priority 700 corresponding to applying term not.
Definition. We define and to be λA B : prop ⇒ ∀p : prop, (A → B → p) → p of type prop → prop → prop.
In Proofgold the corresponding term root is 2ba7d0... and object id is 37da83...
Notation. We use ∧ as an infix operator with priority 780 and which associates to the left corresponding to applying term and.
Definition. We define or to be λA B : prop ⇒ ∀p : prop, (A → p) → (B → p) → p of type prop → prop → prop.
In Proofgold the corresponding term root is 957746... and object id is 86ae64...
Notation. We use ∨ as an infix operator with priority 785 and which associates to the left corresponding to applying term or.
Definition. We define iff to be λA B : prop ⇒ and (A → B) (B → A) of type prop → prop → prop.
In Proofgold the corresponding term root is 98aaaf... and object id is 6588e5...
Notation. We use ↔ as an infix operator with priority 805 and no associativity corresponding to applying term iff.
Beginning of Section Eq
Variable A : SType
Definition. We define eq to be λx y : A ⇒ ∀Q : A → A → prop, Q x y → Q y x of type A → A → prop.
Definition. We define neq to be λx y : A ⇒ ¬ eq x y of type A → A → prop.
End of Section Eq
Notation. We use = as an infix operator with priority 502 and no associativity corresponding to applying term eq.
Notation. We use ≠ as an infix operator with priority 502 and no associativity corresponding to applying term neq.
Beginning of Section FE
Variable A B : SType
Axiom. (func_ext) We take the following as an axiom:
∀f g : A → B, (∀x : A, f x = g x) → f = g
End of Section FE
Beginning of Section Ex
Variable A : SType
Definition. We define ex to be λQ : A → prop ⇒ ∀P : prop, (∀x : A, Q x → P) → P of type (A → prop) → prop.
End of Section Ex
Notation. We use ∃ x...y [possibly with ascriptions] , B as a binder notation corresponding to a term constructed using ex.
Axiom. (prop_ext) We take the following as an axiom:
∀p q : prop, iff p q → p = q
In Proofgold the corresponding term root is 00ac81... and proposition id is 1d1e56...
Primitive. The name In is a term of type set → set → prop.
Notation. We use ∈ as an infix operator with priority 500 and no associativity corresponding to applying term In. Furthermore, we may write ∀ x ∈ A, B to mean ∀ x : set, x ∈ A → B.
Definition. We define Subq to be λA B ⇒ ∀x ∈ A, x ∈ B of type set → set → prop.
In Proofgold the corresponding term root is 81c0ef... and object id is 317007...
Notation. We use ⊆ as an infix operator with priority 500 and no associativity corresponding to applying term Subq. Furthermore, we may write ∀ x ⊆ A, B to mean ∀ x : set, x ⊆ A → B.
Axiom. (set_ext) We take the following as an axiom:
∀X Y : set, X ⊆ Y → Y ⊆ X → X = Y
In Proofgold the corresponding term root is b3b698... and proposition id is 21238a...
Axiom. (In_ind) We take the following as an axiom:
∀P : set → prop, (∀X : set, (∀x ∈ X, P x) → P X) → ∀X : set, P X
In Proofgold the corresponding term root is 045211... and proposition id is 20faa5...
Notation. We use ∃ x...y [possibly with ascriptions] , B as a binder notation corresponding to a term constructed using ex and handling ∈ or ⊆ ascriptions using and.
Primitive. The name Empty is a term of type set.
Axiom. (EmptyAx) We take the following as an axiom:
¬ ∃x : set, x ∈ Empty
In Proofgold the corresponding term root is 4da931... and proposition id is c7b0d7...
Primitive. The name ⋃ is a term of type set → set.
Axiom. (UnionEq) We take the following as an axiom:
∀X x, x ∈ ⋃ X ↔ ∃Y, x ∈ Y ∧ Y ∈ X
In Proofgold the corresponding term root is bf4c26... and proposition id is 40660f...
Primitive. The name 𝒫 is a term of type set → set.
Axiom. (PowerEq) We take the following as an axiom:
∀X Y : set, Y ∈ 𝒫 X ↔ Y ⊆ X
In Proofgold the corresponding term root is 746e5a... and proposition id is d67d71...
Primitive. The name Repl is a term of type set → (set → set) → set.
Notation. {B| x ∈ A} is notation for Repl A (λ x . B).
Axiom. (ReplEq) We take the following as an axiom:
∀A : set, ∀F : set → set, ∀y : set, y ∈ {F x|x ∈ A} ↔ ∃x ∈ A, y = F x
In Proofgold the corresponding term root is 4f2357... and proposition id is 471122...
Definition. We define TransSet to be λU : set ⇒ ∀x ∈ U, x ⊆ U of type set → prop.
In Proofgold the corresponding term root is 538bb7... and object id is e69f92...
Definition. We define Union_closed to be λU : set ⇒ ∀X : set, X ∈ U → ⋃ X ∈ U of type set → prop.
In Proofgold the corresponding term root is 57561d... and object id is e45ed9...
Definition. We define Power_closed to be λU : set ⇒ ∀X : set, X ∈ U → 𝒫 X ∈ U of type set → prop.
In Proofgold the corresponding term root is 8b9cc0... and object id is d2954e...
Definition. We define Repl_closed to be λU : set ⇒ ∀X : set, X ∈ U → ∀F : set → set, (∀x : set, x ∈ X → F x ∈ U) → {F x|x ∈ X} ∈ U of type set → prop.
In Proofgold the corresponding term root is 5574bb... and object id is 513534...
Definition. We define ZF_closed to be λU : set ⇒ Union_closed U ∧ Power_closed U ∧ Repl_closed U of type set → prop.
In Proofgold the corresponding term root is 1bd4aa... and object id is e1eccf...
Primitive. The name UnivOf is a term of type set → set.
Axiom. (UnivOf_In) We take the following as an axiom:
∀N : set, N ∈ UnivOf N
In Proofgold the corresponding term root is efcc16... and proposition id is e805d4...
Axiom. (UnivOf_TransSet) We take the following as an axiom:
∀N : set, TransSet (UnivOf N)
In Proofgold the corresponding term root is 053575... and proposition id is cf4f5c...
Axiom. (UnivOf_ZF_closed) We take the following as an axiom:
∀N : set, ZF_closed (UnivOf N)
In Proofgold the corresponding term root is 770e35... and proposition id is 0e7363...
Axiom. (UnivOf_Min) We take the following as an axiom:
∀N U : set, N ∈ U → TransSet U → ZF_closed U → UnivOf N ⊆ U
In Proofgold the corresponding term root is 042dda... and proposition id is 1f7933...
Theorem. (FalseE)
False → ∀p : prop, p
In Proofgold the corresponding term root is 0dcfe3... and proposition id is f9a230...
Proof:
An exact proof term for the current goal is (λH ⇒ H).
∎
Theorem. (TrueI)
True
In Proofgold the corresponding term root is f81b39... and proposition id is eb5cbf...
Proof:
An exact proof term for the current goal is (λp H ⇒ H).
∎
Theorem. (andI)
∀A B : prop, A → B → A ∧ B
In Proofgold the corresponding term root is 3c4f59... and proposition id is 3bf198...
Proof:
An exact proof term for the current goal is (λA B a b P H ⇒ H a b).
∎
Theorem. (andEL)
∀A B : prop, A ∧ B → A
In Proofgold the corresponding term root is c2f006... and proposition id is 718be4...
Proof:
An exact proof term for the current goal is (λA B H ⇒ H A (λa b ⇒ a)).
∎
Theorem. (andER)
∀A B : prop, A ∧ B → B
In Proofgold the corresponding term root is f66b82... and proposition id is 5caea9...
Proof:
An exact proof term for the current goal is (λA B H ⇒ H B (λa b ⇒ b)).
∎
Theorem. (orIL)
∀A B : prop, A → A ∨ B
In Proofgold the corresponding term root is 55be8c... and proposition id is 382673...
Proof:
An exact proof term for the current goal is (λA B a P H1 H2 ⇒ H1 a).
∎
Theorem. (orIR)
∀A B : prop, B → A ∨ B
In Proofgold the corresponding term root is 3c63fc... and proposition id is 8d8e40...
Proof:
An exact proof term for the current goal is (λA B b P H1 H2 ⇒ H2 b).
∎
Beginning of Section PropN
Variable P1 P2 P3 : prop
Theorem. (and3I)
P1 → P2 → P3 → P1 ∧ P2 ∧ P3
In Proofgold the corresponding term root is 23ac9c... and proposition id is cb999b...
Proof:
An exact proof term for the current goal is (λH1 H2 H3 ⇒ andI (P1 ∧ P2) P3 (andI P1 P2 H1 H2) H3).
∎
Theorem. (and3E)
P1 ∧ P2 ∧ P3 → (∀p : prop, (P1 → P2 → P3 → p) → p)
In Proofgold the corresponding term root is 445f76... and proposition id is 50fabe...
Proof:
An exact proof term for the current goal is (λu p H ⇒ u p (λu u3 ⇒ u p (λu1 u2 ⇒ H u1 u2 u3))).
∎
Theorem. (or3I1)
P1 → P1 ∨ P2 ∨ P3
In Proofgold the corresponding term root is e44f34... and proposition id is ef9071...
Proof:
An exact proof term for the current goal is (λu ⇒ orIL (P1 ∨ P2) P3 (orIL P1 P2 u)).
∎
Theorem. (or3I2)
P2 → P1 ∨ P2 ∨ P3
In Proofgold the corresponding term root is 8df6e1... and proposition id is 827a91...
Proof:
An exact proof term for the current goal is (λu ⇒ orIL (P1 ∨ P2) P3 (orIR P1 P2 u)).
∎
Theorem. (or3I3)
P3 → P1 ∨ P2 ∨ P3
In Proofgold the corresponding term root is 017a00... and proposition id is 3b7f5a...
Proof:
An exact proof term for the current goal is (orIR (P1 ∨ P2) P3).
∎
Theorem. (or3E)
P1 ∨ P2 ∨ P3 → (∀p : prop, (P1 → p) → (P2 → p) → (P3 → p) → p)
In Proofgold the corresponding term root is 5d069d... and proposition id is 2120c9...
Proof:
An exact proof term for the current goal is (λu p H1 H2 H3 ⇒ u p (λu ⇒ u p H1 H2) H3).
∎
Variable P4 : prop
Theorem. (and4I)
P1 → P2 → P3 → P4 → P1 ∧ P2 ∧ P3 ∧ P4
In Proofgold the corresponding term root is ed32cb... and proposition id is 906aaf...
Proof:
An exact proof term for the current goal is (λH1 H2 H3 H4 ⇒ andI (P1 ∧ P2 ∧ P3) P4 (and3I H1 H2 H3) H4).
∎
Variable P5 : prop
Theorem. (and5I)
P1 → P2 → P3 → P4 → P5 → P1 ∧ P2 ∧ P3 ∧ P4 ∧ P5
In Proofgold the corresponding term root is 041ced... and proposition id is 11fc47...
Proof:
An exact proof term for the current goal is (λH1 H2 H3 H4 H5 ⇒ andI (P1 ∧ P2 ∧ P3 ∧ P4) P5 (and4I H1 H2 H3 H4) H5).
∎
End of Section PropN
Theorem. (not_or_and_demorgan)
∀A B : prop, ¬ (A ∨ B) → ¬ A ∧ ¬ B
In Proofgold the corresponding term root is 8733b0... and proposition id is 27db91...
Proof:
Let A and B be given.
Assume u: ¬ (A ∨ B).
Apply andI to the current goal.
We will prove ¬ A.
Assume a: A.
An exact proof term for the current goal is (u (orIL A B a)).
We will prove ¬ B.
Assume b: B.
An exact proof term for the current goal is (u (orIR A B b)).
∎
Theorem. (not_ex_all_demorgan_i)
∀P : set → prop, (¬ ∃x, P x) → ∀x, ¬ P x
In Proofgold the corresponding term root is 73523e... and proposition id is 075ddf...
Proof:
Let P be given.
Assume H1.
Let x be given.
Assume H2.
Apply H1 to the current goal.
We use x to witness the existential quantifier.
An exact proof term for the current goal is H2.
∎
Theorem. (iffI)
∀A B : prop, (A → B) → (B → A) → (A ↔ B)
In Proofgold the corresponding term root is 7516b9... and proposition id is 4d966a...
Proof:
An exact proof term for the current goal is (λA B ⇒ andI (A → B) (B → A)).
∎
Theorem. (iffEL)
∀A B : prop, (A ↔ B) → A → B
In Proofgold the corresponding term root is 0e94c6... and proposition id is 9a07f6...
Proof:
An exact proof term for the current goal is (λA B ⇒ andEL (A → B) (B → A)).
∎
Theorem. (iffER)
∀A B : prop, (A ↔ B) → B → A
In Proofgold the corresponding term root is 54702a... and proposition id is f5dfc0...
Proof:
An exact proof term for the current goal is (λA B ⇒ andER (A → B) (B → A)).
∎
Theorem. (iff_refl)
∀A : prop, A ↔ A
In Proofgold the corresponding term root is 01338d... and proposition id is 7b71b5...
Proof:
An exact proof term for the current goal is (λA : prop ⇒ andI (A → A) (A → A) (λH : A ⇒ H) (λH : A ⇒ H)).
∎
Theorem. (iff_sym)
∀A B : prop, (A ↔ B) → (B ↔ A)
In Proofgold the corresponding term root is 00bac1... and proposition id is 11f51b...
Proof:
Let A and B be given.
Assume H1: (A → B) ∧ (B → A).
Apply H1 to the current goal.
Assume H2: A → B.
Assume H3: B → A.
An exact proof term for the current goal is iffI B A H3 H2.
∎
Theorem. (iff_trans)
∀A B C : prop, (A ↔ B) → (B ↔ C) → (A ↔ C)
In Proofgold the corresponding term root is 363285... and proposition id is c3b8f1...
Proof:
Let A, B and C be given.
Assume H1: A ↔ B.
Assume H2: B ↔ C.
Apply H1 to the current goal.
Assume H3: A → B.
Assume H4: B → A.
Apply H2 to the current goal.
Assume H5: B → C.
Assume H6: C → B.
An exact proof term for the current goal is (iffI A C (λH ⇒ H5 (H3 H)) (λH ⇒ H4 (H6 H))).
∎
Theorem. (eq_i_tra)
∀x y z, x = y → y = z → x = z
In Proofgold the corresponding term root is 193bbf... and proposition id is 7ecbc4...
Proof:
Let x, y and z be given.
Assume H1 H2.
rewrite the current goal using H2 (from right to left).
An exact proof term for the current goal is H1.
∎
Theorem. (f_eq_i)
∀f : set → set, ∀x y, x = y → f x = f y
In Proofgold the corresponding term root is b0871b... and proposition id is 2a60fd...
Proof:
Let f, x and y be given.
Assume H1.
rewrite the current goal using H1 (from left to right).
Use reflexivity.
∎
Theorem. (neq_i_sym)
∀x y, x ≠ y → y ≠ x
In Proofgold the corresponding term root is 5c0ec9... and proposition id is 1a9898...
Proof:
Let x and y be given.
Assume H1 H2.
Apply H1 to the current goal.
Use symmetry.
An exact proof term for the current goal is H2.
∎
Definition. We define nIn to be λx X ⇒ ¬ In x X of type set → set → prop.
In Proofgold the corresponding term root is 36808c... and object id is 253451...
Notation. We use ∉ as an infix operator with priority 502 and no associativity corresponding to applying term nIn.
Theorem. (Eps_i_ex)
∀P : set → prop, (∃x, P x) → P (Eps_i P)
In Proofgold the corresponding term root is be36f0... and proposition id is 6a2832...
Proof:
Let P be given.
Assume H1.
Apply H1 to the current goal.
Let x be given.
Assume H2.
An exact proof term for the current goal is Eps_i_ax P x H2.
∎
Theorem. (pred_ext)
∀P Q : set → prop, (∀x, P x ↔ Q x) → P = Q
In Proofgold the corresponding term root is e9e64b... and proposition id is d4b2d3...
Proof:
Let P and Q be given.
Assume H1.
Apply func_ext set prop to the current goal.
Let x be given.
Apply prop_ext to the current goal.
We will prove P x ↔ Q x.
An exact proof term for the current goal is H1 x.
∎
Theorem. (prop_ext_2)
∀p q : prop, (p → q) → (q → p) → p = q
In Proofgold the corresponding term root is a9ede2... and proposition id is a56fc7...
Proof:
Let p and q be given.
Assume H1 H2.
Apply prop_ext to the current goal.
Apply iffI to the current goal.
An exact proof term for the current goal is H1.
An exact proof term for the current goal is H2.
∎
Theorem. (Subq_ref)
∀X : set, X ⊆ X
In Proofgold the corresponding term root is 4c5395... and proposition id is 2efbac...
Proof:
An exact proof term for the current goal is (λ(X x : set)(H : x ∈ X) ⇒ H).
∎
Theorem. (Subq_tra)
∀X Y Z : set, X ⊆ Y → Y ⊆ Z → X ⊆ Z
In Proofgold the corresponding term root is 79accb... and proposition id is fed662...
Proof:
An exact proof term for the current goal is (λ(X Y Z : set)(H1 : X ⊆ Y)(H2 : Y ⊆ Z)(x : set)(H : x ∈ X) ⇒ (H2 x (H1 x H))).
∎
Theorem. (Subq_contra)
∀X Y z : set, X ⊆ Y → z ∉ Y → z ∉ X
In Proofgold the corresponding term root is d15e87... and proposition id is bd5a2f...
Proof:
An exact proof term for the current goal is (λX Y z H1 H2 H3 ⇒ H2 (H1 z H3)).
∎
Theorem. (EmptyE)
∀x : set, x ∉ Empty
In Proofgold the corresponding term root is 93a77a... and proposition id is 8349ab...
Proof:
Let x be given.
Assume H.
Apply EmptyAx to the current goal.
We use x to witness the existential quantifier.
An exact proof term for the current goal is H.
∎
Theorem. (Subq_Empty)
∀X : set, Empty ⊆ X
In Proofgold the corresponding term root is a517cd... and proposition id is 2faec3...
Proof:
An exact proof term for the current goal is (λ(X x : set)(H : x ∈ Empty) ⇒ EmptyE x H (x ∈ X)).
∎
Theorem. (Empty_Subq_eq)
∀X : set, X ⊆ Empty → X = Empty
In Proofgold the corresponding term root is 2dd959... and proposition id is 988691...
Proof:
Let X be given.
Assume H1: X ⊆ Empty.
Apply set_ext to the current goal.
An exact proof term for the current goal is H1.
An exact proof term for the current goal is (Subq_Empty X).
∎
Theorem. (Empty_eq)
∀X : set, (∀x, x ∉ X) → X = Empty
In Proofgold the corresponding term root is 3de0fd... and proposition id is 8b915b...
Proof:
Let X be given.
Assume H1: ∀x, x ∉ X.
Apply Empty_Subq_eq to the current goal.
Let x be given.
Assume H2: x ∈ X.
We will prove False.
An exact proof term for the current goal is (H1 x H2).
∎
Theorem. (UnionI)
∀X x Y : set, x ∈ Y → Y ∈ X → x ∈ ⋃ X
In Proofgold the corresponding term root is 51b549... and proposition id is bade05...
Proof:
Let X, x and Y be given.
Assume H1: x ∈ Y.
Assume H2: Y ∈ X.
Apply UnionEq X x to the current goal.
Assume _ H3.
Apply H3 to the current goal.
We will prove ∃Y : set, x ∈ Y ∧ Y ∈ X.
We use Y to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is H1.
An exact proof term for the current goal is H2.
∎
Theorem. (UnionE)
∀X x : set, x ∈ ⋃ X → ∃Y : set, x ∈ Y ∧ Y ∈ X
In Proofgold the corresponding term root is db4bc1... and proposition id is ab781b...
Proof:
An exact proof term for the current goal is (λX x : set ⇒ iffEL (x ∈ ⋃ X) (∃Y : set, x ∈ Y ∧ Y ∈ X) (UnionEq X x)).
∎
Theorem. (UnionE_impred)
∀X x : set, x ∈ ⋃ X → ∀p : prop, (∀Y : set, x ∈ Y → Y ∈ X → p) → p
In Proofgold the corresponding term root is 8efd57... and proposition id is d621b2...
Proof:
Let X and x be given.
Assume H1.
Let p be given.
Assume Hp.
Apply UnionE X x H1 to the current goal.
Let x be given.
Assume H2.
Apply H2 to the current goal.
An exact proof term for the current goal is Hp x.
∎
Theorem. (PowerI)
∀X Y : set, Y ⊆ X → Y ∈ 𝒫 X
In Proofgold the corresponding term root is 7143d7... and proposition id is 3c48d3...
Proof:
Let X and Y be given.
Apply PowerEq X Y to the current goal.
An exact proof term for the current goal is (λ_ H ⇒ H).
∎
Theorem. (PowerE)
∀X Y : set, Y ∈ 𝒫 X → Y ⊆ X
In Proofgold the corresponding term root is 149030... and proposition id is 647070...
Proof:
Let X and Y be given.
Apply PowerEq X Y to the current goal.
An exact proof term for the current goal is (λH _ ⇒ H).
∎
Theorem. (Empty_In_Power)
∀X : set, Empty ∈ 𝒫 X
In Proofgold the corresponding term root is d2656b... and proposition id is 487855...
Proof:
An exact proof term for the current goal is (λX : set ⇒ PowerI X Empty (Subq_Empty X)).
∎
Theorem. (Self_In_Power)
∀X : set, X ∈ 𝒫 X
In Proofgold the corresponding term root is f81ef2... and proposition id is 8336ff...
Proof:
An exact proof term for the current goal is (λX : set ⇒ PowerI X X (Subq_ref X)).
∎
Theorem. (xm)
∀P : prop, P ∨ ¬ P
In Proofgold the corresponding term root is 62e4c3... and proposition id is 7b6737...
Proof:
Let P of type prop be given.
Set p1 to be the term λx : set ⇒ x = Empty ∨ P.
Set p2 to be the term λx : set ⇒ x ≠ Empty ∨ P.
We prove the intermediate claim L1: p1 Empty.
We will prove (Empty = Empty ∨ P).
Apply orIL to the current goal.
An exact proof term for the current goal is (λq H ⇒ H).
We prove the intermediate claim L2: (Eps_i p1) = Empty ∨ P.
An exact proof term for the current goal is (Eps_i_ax p1 Empty L1).
We prove the intermediate claim L3: p2 (𝒫 Empty).
We will prove ¬ (𝒫 Empty = Empty) ∨ P.
Apply orIL to the current goal.
Assume H1: 𝒫 Empty = Empty.
Apply EmptyE Empty to the current goal.
We will prove Empty ∈ Empty.
rewrite the current goal using H1 (from right to left) at position 2.
Apply Empty_In_Power to the current goal.
We prove the intermediate claim L4: Eps_i p2 ≠ Empty ∨ P.
An exact proof term for the current goal is (Eps_i_ax p2 (𝒫 Empty) L3).
Apply L2 to the current goal.
Assume H1: Eps_i p1 = Empty.
Apply L4 to the current goal.
Assume H2: Eps_i p2 ≠ Empty.
We will prove P ∨ ¬ P.
Apply orIR to the current goal.
We will prove ¬ P.
Assume H3: P.
We prove the intermediate claim L5: p1 = p2.
Apply pred_ext to the current goal.
Let x be given.
Apply iffI to the current goal.
Assume H4.
We will prove (¬ (x = Empty) ∨ P).
Apply orIR to the current goal.
We will prove P.
An exact proof term for the current goal is H3.
Assume H4.
We will prove (x = Empty ∨ P).
Apply orIR to the current goal.
We will prove P.
An exact proof term for the current goal is H3.
Apply H2 to the current goal.
rewrite the current goal using L5 (from right to left).
An exact proof term for the current goal is H1.
Assume H2: P.
We will prove P ∨ ¬ P.
Apply orIL to the current goal.
We will prove P.
An exact proof term for the current goal is H2.
Assume H1: P.
We will prove P ∨ ¬ P.
Apply orIL to the current goal.
We will prove P.
An exact proof term for the current goal is H1.
∎
Theorem. (dneg)
∀P : prop, ¬ ¬ P → P
In Proofgold the corresponding term root is 852f34... and proposition id is b7cc63...
Proof:
Let P be given.
Assume H1.
Apply xm P to the current goal.
An exact proof term for the current goal is (λH ⇒ H).
Assume H2: ¬ P.
We will prove False.
An exact proof term for the current goal is H1 H2.
∎
Theorem. (not_all_ex_demorgan_i)
∀P : set → prop, ¬ (∀x, P x) → ∃x, ¬ P x
In Proofgold the corresponding term root is 9bfa06... and proposition id is 4b1474...
Proof:
Let P be given.
Assume u: ¬ ∀x, P x.
Apply dneg to the current goal.
Assume v: ¬ ∃x, ¬ P x.
Apply u to the current goal.
Let x be given.
Apply dneg to the current goal.
Assume w: ¬ P x.
An exact proof term for the current goal is (not_ex_all_demorgan_i (λx ⇒ ¬ P x) v x w).
∎
Theorem. (eq_or_nand)
or = (λx y : prop ⇒ ¬ (¬ x ∧ ¬ y))
In Proofgold the corresponding term root is 24ac29... and proposition id is d212ef...
Proof:
Apply func_ext prop (prop → prop) to the current goal.
Let x be given.
Apply func_ext prop prop to the current goal.
Let y be given.
Apply prop_ext_2 to the current goal.
Assume H1: x ∨ y.
Assume H2: ¬ x ∧ ¬ y.
Apply H2 to the current goal.
Assume H3 H4.
An exact proof term for the current goal is (H1 False H3 H4).
Assume H1: ¬ (¬ x ∧ ¬ y).
Apply (xm x) to the current goal.
Assume H2: x.
Apply orIL to the current goal.
An exact proof term for the current goal is H2.
Assume H2: ¬ x.
Apply (xm y) to the current goal.
Assume H3: y.
Apply orIR to the current goal.
An exact proof term for the current goal is H3.
Assume H3: ¬ y.
Apply H1 to the current goal.
An exact proof term for the current goal is (andI (¬ x) (¬ y) H2 H3).
∎
Definition. We define exactly1of2 to be λA B : prop ⇒ A ∧ ¬ B ∨ ¬ A ∧ B of type prop → prop → prop.
In Proofgold the corresponding term root is 163602... and object id is a0b944...
Theorem. (exactly1of2_I1)
∀A B : prop, A → ¬ B → exactly1of2 A B
In Proofgold the corresponding term root is 8e1e45... and proposition id is 4f94fa...
Proof:
Let A and B be given.
Assume HA: A.
Assume HB: ¬ B.
We will prove A ∧ ¬ B ∨ ¬ A ∧ B.
Apply orIL to the current goal.
We will prove A ∧ ¬ B.
An exact proof term for the current goal is (andI A (¬ B) HA HB).
∎
Theorem. (exactly1of2_I2)
∀A B : prop, ¬ A → B → exactly1of2 A B
In Proofgold the corresponding term root is c9fd90... and proposition id is 6c1ad9...
Proof:
Let A and B be given.
Assume HA: ¬ A.
Assume HB: B.
We will prove A ∧ ¬ B ∨ ¬ A ∧ B.
Apply orIR to the current goal.
We will prove ¬ A ∧ B.
An exact proof term for the current goal is (andI (¬ A) B HA HB).
∎
Theorem. (exactly1of2_E)
∀A B : prop, exactly1of2 A B → ∀p : prop, (A → ¬ B → p) → (¬ A → B → p) → p
In Proofgold the corresponding term root is 29b901... and proposition id is c67434...
Proof:
Let A and B be given.
Assume H1: exactly1of2 A B.
Let p be given.
Assume H2: A → ¬ B → p.
Assume H3: ¬ A → B → p.
Apply (H1 p) to the current goal.
An exact proof term for the current goal is (λH4 : A ∧ ¬ B ⇒ H4 p H2).
An exact proof term for the current goal is (λH4 : ¬ A ∧ B ⇒ H4 p H3).
∎
Theorem. (exactly1of2_or)
∀A B : prop, exactly1of2 A B → A ∨ B
In Proofgold the corresponding term root is 18af73... and proposition id is e092f6...
Proof:
Let A and B be given.
Assume H1: exactly1of2 A B.
Apply (exactly1of2_E A B H1 (A ∨ B)) to the current goal.
An exact proof term for the current goal is (λ(HA : A)(_ : ¬ B) ⇒ orIL A B HA).
An exact proof term for the current goal is (λ(_ : ¬ A)(HB : B) ⇒ orIR A B HB).
∎
Theorem. (ReplI)
∀A : set, ∀F : set → set, ∀x : set, x ∈ A → F x ∈ {F x|x ∈ A}
In Proofgold the corresponding term root is b146c5... and proposition id is f84982...
Proof:
Let A, F and x be given.
Assume H1.
Apply ReplEq A F (F x) to the current goal.
Assume _ H2.
Apply H2 to the current goal.
We will prove ∃x' ∈ A, F x = F x'.
We use x to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is H1.
An exact proof term for the current goal is (λq H ⇒ H).
∎
Theorem. (ReplE)
∀A : set, ∀F : set → set, ∀y : set, y ∈ {F x|x ∈ A} → ∃x ∈ A, y = F x
In Proofgold the corresponding term root is 311cdf... and proposition id is d07e78...
Proof:
Let A, F and y be given.
Apply ReplEq A F y to the current goal.
An exact proof term for the current goal is (λH _ ⇒ H).
∎
Theorem. (ReplE_impred)
∀A : set, ∀F : set → set, ∀y : set, y ∈ {F x|x ∈ A} → ∀p : prop, (∀x : set, x ∈ A → y = F x → p) → p
In Proofgold the corresponding term root is 31e90a... and proposition id is 34fefd...
Proof:
Let A, F and y be given.
Assume H1.
Apply ReplE A F y H1 to the current goal.
Let x be given.
Assume H2.
Apply H2 to the current goal.
Assume H3 H4.
Let p be given.
Assume Hp.
An exact proof term for the current goal is Hp x H3 H4.
∎
Theorem. (ReplE')
∀X, ∀f : set → set, ∀p : set → prop, (∀x ∈ X, p (f x)) → ∀y ∈ {f x|x ∈ X}, p y
In Proofgold the corresponding term root is e374fe... and proposition id is d4b7b0...
Proof:
Let X, f and p be given.
Assume H1.
Let y be given.
Assume Hy.
Apply ReplE_impred X f y Hy to the current goal.
Let x be given.
Assume Hx: x ∈ X.
Assume Hx2: y = f x.
We will prove p y.
rewrite the current goal using Hx2 (from left to right).
An exact proof term for the current goal is H1 x Hx.
∎
Theorem. (Repl_Empty)
∀F : set → set, {F x|x ∈ Empty} = Empty
In Proofgold the corresponding term root is 2af2ac... and proposition id is 0835bc...
Proof:
Let F be given.
Apply (Empty_eq {F x|x ∈ Empty}) to the current goal.
Let y be given.
Assume H1: y ∈ {F x|x ∈ Empty}.
Apply (ReplE_impred Empty F y H1) to the current goal.
Let x be given.
Assume H2: x ∈ Empty.
Assume _.
An exact proof term for the current goal is (EmptyE x H2).
∎
Theorem. (ReplEq_ext_sub)
∀X, ∀F G : set → set, (∀x ∈ X, F x = G x) → {F x|x ∈ X} ⊆ {G x|x ∈ X}
In Proofgold the corresponding term root is eb4c1d... and proposition id is 56aad0...
Proof:
Let X, F and G be given.
Assume H1: ∀x ∈ X, F x = G x.
Let y be given.
Assume Hy: y ∈ {F x|x ∈ X}.
Apply ReplE_impred X F y Hy to the current goal.
Let x be given.
Assume Hx: x ∈ X.
Assume H2: y = F x.
We will prove y ∈ {G x|x ∈ X}.
rewrite the current goal using H2 (from left to right).
We will prove F x ∈ {G x|x ∈ X}.
rewrite the current goal using H1 x Hx (from left to right).
We will prove G x ∈ {G x|x ∈ X}.
Apply ReplI to the current goal.
An exact proof term for the current goal is Hx.
∎
Theorem. (ReplEq_ext)
∀X, ∀F G : set → set, (∀x ∈ X, F x = G x) → {F x|x ∈ X} = {G x|x ∈ X}
In Proofgold the corresponding term root is 81eb90... and proposition id is 781846...
Proof:
Let X, F and G be given.
Assume H1: ∀x ∈ X, F x = G x.
Apply set_ext to the current goal.
An exact proof term for the current goal is ReplEq_ext_sub X F G H1.
Apply ReplEq_ext_sub X G F to the current goal.
Let x be given.
Assume Hx.
Use symmetry.
An exact proof term for the current goal is H1 x Hx.
∎
Theorem. (Repl_inv_eq)
∀P : set → prop, ∀f g : set → set, (∀x, P x → g (f x) = x) → ∀X, (∀x ∈ X, P x) → {g y|y ∈ {f x|x ∈ X}} = X
In Proofgold the corresponding term root is 968c7a... and proposition id is 9ce73d...
Proof:
Let P, f and g be given.
Assume H1.
Let X be given.
Assume HX.
Apply set_ext to the current goal.
Let w be given.
Assume Hw: w ∈ {g y|y ∈ {f x|x ∈ X}}.
Apply ReplE_impred {f x|x ∈ X} g w Hw to the current goal.
Let y be given.
Assume Hy: y ∈ {f x|x ∈ X}.
Assume Hwy: w = g y.
Apply ReplE_impred X f y Hy to the current goal.
Let x be given.
Assume Hx: x ∈ X.
Assume Hyx: y = f x.
We will prove w ∈ X.
rewrite the current goal using Hwy (from left to right).
rewrite the current goal using Hyx (from left to right).
We will prove g (f x) ∈ X.
rewrite the current goal using H1 x (HX x Hx) (from left to right).
An exact proof term for the current goal is Hx.
Let x be given.
Assume Hx: x ∈ X.
rewrite the current goal using H1 x (HX x Hx) (from right to left).
We will prove g (f x) ∈ {g y|y ∈ {f x|x ∈ X}}.
Apply ReplI to the current goal.
We will prove f x ∈ {f x|x ∈ X}.
Apply ReplI to the current goal.
An exact proof term for the current goal is Hx.
∎
Theorem. (Repl_invol_eq)
∀P : set → prop, ∀f : set → set, (∀x, P x → f (f x) = x) → ∀X, (∀x ∈ X, P x) → {f y|y ∈ {f x|x ∈ X}} = X
In Proofgold the corresponding term root is e15a6b... and proposition id is 5faff0...
Proof:
Let P and f be given.
Assume H1.
An exact proof term for the current goal is Repl_inv_eq P f f H1.
∎
Definition. We define If_i to be (λp x y ⇒ Eps_i (λz : set ⇒ p ∧ z = x ∨ ¬ p ∧ z = y)) of type prop → set → set → set.
In Proofgold the corresponding term root is b8ff52... and object id is 76c4c0...
Notation. if cond then T else E is notation corresponding to If_i type cond T E where type is the inferred type of T.
Theorem. (If_i_correct)
∀p : prop, ∀x y : set, p ∧ (if p then x else y) = x ∨ ¬ p ∧ (if p then x else y) = y
In Proofgold the corresponding term root is 2a17ae... and proposition id is b5c26b...
Proof:
Let p, x and y be given.
Apply (xm p) to the current goal.
Assume H1: p.
We prove the intermediate claim L1: p ∧ x = x ∨ ¬ p ∧ x = y.
Apply orIL to the current goal.
Apply andI to the current goal.
An exact proof term for the current goal is H1.
Use reflexivity.
An exact proof term for the current goal is (Eps_i_ax (λz : set ⇒ p ∧ z = x ∨ ¬ p ∧ z = y) x L1).
Assume H1: ¬ p.
We prove the intermediate claim L1: p ∧ y = x ∨ ¬ p ∧ y = y.
Apply orIR to the current goal.
Apply andI to the current goal.
An exact proof term for the current goal is H1.
Use reflexivity.
An exact proof term for the current goal is (Eps_i_ax (λz : set ⇒ p ∧ z = x ∨ ¬ p ∧ z = y) y L1).
∎
Theorem. (If_i_0)
∀p : prop, ∀x y : set, ¬ p → (if p then x else y) = y
In Proofgold the corresponding term root is fde1a8... and proposition id is a9abd3...
Proof:
Let p, x and y be given.
Assume H1: ¬ p.
Apply (If_i_correct p x y) to the current goal.
Assume H2: p ∧ (if p then x else y) = x.
An exact proof term for the current goal is (H1 (andEL p ((if p then x else y) = x) H2) ((if p then x else y) = y)).
Assume H2: ¬ p ∧ (if p then x else y) = y.
An exact proof term for the current goal is (andER (¬ p) ((if p then x else y) = y) H2).
∎
Theorem. (If_i_1)
∀p : prop, ∀x y : set, p → (if p then x else y) = x
In Proofgold the corresponding term root is b1aa3f... and proposition id is a93c2c...
Proof:
Let p, x and y be given.
Assume H1: p.
Apply (If_i_correct p x y) to the current goal.
Assume H2: p ∧ (if p then x else y) = x.
An exact proof term for the current goal is (andER p ((if p then x else y) = x) H2).
Assume H2: ¬ p ∧ (if p then x else y) = y.
An exact proof term for the current goal is (andEL (¬ p) ((if p then x else y) = y) H2 H1 ((if p then x else y) = x)).
∎
Theorem. (If_i_or)
∀p : prop, ∀x y : set, (if p then x else y) = x ∨ (if p then x else y) = y
In Proofgold the corresponding term root is 3e1439... and proposition id is 4de742...
Proof:
Let p, x and y be given.
Apply (xm p) to the current goal.
Assume H1: p.
Apply orIL to the current goal.
An exact proof term for the current goal is (If_i_1 p x y H1).
Assume H1: ¬ p.
Apply orIR to the current goal.
An exact proof term for the current goal is (If_i_0 p x y H1).
∎
Definition. We define UPair to be λy z ⇒ {if Empty ∈ X then y else z|X ∈ 𝒫 (𝒫 Empty)} of type set → set → set.
In Proofgold the corresponding term root is 742438... and object id is dc3134...
Notation. {x,y} is notation for UPair x y.
Theorem. (UPairE)
∀x y z : set, x ∈ {y,z} → x = y ∨ x = z
In Proofgold the corresponding term root is 5009cc... and proposition id is e67fc5...
Proof:
Let x, y and z be given.
Assume H1: x ∈ {y,z}.
Apply (ReplE (𝒫 (𝒫 Empty)) (λX ⇒ if Empty ∈ X then y else z) x H1) to the current goal.
Let X be given.
Assume H2: X ∈ 𝒫 (𝒫 Empty) ∧ x = if Empty ∈ X then y else z.
We prove the intermediate claim L1: x = if Empty ∈ X then y else z.
An exact proof term for the current goal is (andER (X ∈ 𝒫 (𝒫 Empty)) (x = if Empty ∈ X then y else z) H2).
Apply (If_i_or (Empty ∈ X) y z) to the current goal.
Assume H3: (if Empty ∈ X then y else z) = y.
Apply orIL to the current goal.
We will prove x = y.
rewrite the current goal using H3 (from right to left).
An exact proof term for the current goal is L1.
Assume H3: (if Empty ∈ X then y else z) = z.
Apply orIR to the current goal.
We will prove x = z.
rewrite the current goal using H3 (from right to left).
An exact proof term for the current goal is L1.
∎
Theorem. (UPairI1)
∀y z : set, y ∈ {y,z}
In Proofgold the corresponding term root is b1eea2... and proposition id is bf3649...
Proof:
Let y and z be given.
We will prove y ∈ {y,z}.
rewrite the current goal using (If_i_1 (Empty ∈ 𝒫 Empty) y z (Empty_In_Power Empty)) (from right to left) at position 1.
We will prove (if Empty ∈ 𝒫 Empty then y else z) ∈ {y,z}.
We will prove (if Empty ∈ 𝒫 Empty then y else z) ∈ {if Empty ∈ X then y else z|X ∈ 𝒫 (𝒫 Empty)}.
Apply (ReplI (𝒫 (𝒫 Empty)) (λX : set ⇒ if (Empty ∈ X) then y else z) (𝒫 Empty)) to the current goal.
We will prove 𝒫 Empty ∈ 𝒫 (𝒫 Empty).
An exact proof term for the current goal is (Self_In_Power (𝒫 Empty)).
∎
Theorem. (UPairI2)
∀y z : set, z ∈ {y,z}
In Proofgold the corresponding term root is 27455b... and proposition id is a0a5b8...
Proof:
Let y and z be given.
We will prove z ∈ {y,z}.
rewrite the current goal using (If_i_0 (Empty ∈ Empty) y z (EmptyE Empty)) (from right to left) at position 1.
We will prove (if Empty ∈ Empty then y else z) ∈ {y,z}.
We will prove (if Empty ∈ Empty then y else z) ∈ {if Empty ∈ X then y else z|X ∈ 𝒫 (𝒫 Empty)}.
Apply (ReplI (𝒫 (𝒫 Empty)) (λX : set ⇒ if (Empty ∈ X) then y else z) Empty) to the current goal.
We will prove Empty ∈ 𝒫 (𝒫 Empty).
An exact proof term for the current goal is (Empty_In_Power (𝒫 Empty)).
∎
Definition. We define Sing to be λx ⇒ {x,x} of type set → set.
In Proofgold the corresponding term root is bd01a8... and object id is b7dfc1...
Notation. {x} is notation for Sing x.
Theorem. (SingI)
∀x : set, x ∈ {x}
In Proofgold the corresponding term root is 6591bd... and proposition id is 2f93be...
Proof:
An exact proof term for the current goal is (λx : set ⇒ UPairI1 x x).
∎
Theorem. (SingE)
∀x y : set, y ∈ {x} → y = x
In Proofgold the corresponding term root is 9341fe... and proposition id is 3b0dd8...
Proof:
An exact proof term for the current goal is (λx y H ⇒ UPairE y x x H (y = x) (λH ⇒ H) (λH ⇒ H)).
∎
Theorem. (Sing_inj)
∀x y, {x} = {y} → x = y
In Proofgold the corresponding term root is dea1af... and proposition id is 96eb72...
Proof:
Let x and y be given.
Assume H1: {x} = {y}.
Apply SingE to the current goal.
We will prove x ∈ {y}.
rewrite the current goal using H1 (from right to left).
Apply SingI to the current goal.
∎
Definition. We define binunion to be λX Y ⇒ ⋃ {X,Y} of type set → set → set.
In Proofgold the corresponding term root is 5e1ac4... and object id is e14989...
Notation. We use ∪ as an infix operator with priority 345 and which associates to the left corresponding to applying term binunion.
Theorem. (binunionI1)
∀X Y z : set, z ∈ X → z ∈ X ∪ Y
In Proofgold the corresponding term root is 753b6e... and proposition id is e43c8d...
Proof:
Let X, Y and z be given.
Assume H1: z ∈ X.
We will prove z ∈ X ∪ Y.
We will prove z ∈ ⋃ {X,Y}.
Apply (UnionI {X,Y} z X) to the current goal.
We will prove z ∈ X.
An exact proof term for the current goal is H1.
We will prove X ∈ {X,Y}.
An exact proof term for the current goal is (UPairI1 X Y).
∎
Theorem. (binunionI2)
∀X Y z : set, z ∈ Y → z ∈ X ∪ Y
In Proofgold the corresponding term root is b8821c... and proposition id is 959535...
Proof:
Let X, Y and z be given.
Assume H1: z ∈ Y.
We will prove z ∈ X ∪ Y.
We will prove z ∈ ⋃ {X,Y}.
Apply (UnionI {X,Y} z Y) to the current goal.
We will prove z ∈ Y.
An exact proof term for the current goal is H1.
We will prove Y ∈ {X,Y}.
An exact proof term for the current goal is (UPairI2 X Y).
∎
Theorem. (binunionE)
∀X Y z : set, z ∈ X ∪ Y → z ∈ X ∨ z ∈ Y
In Proofgold the corresponding term root is 5a3195... and proposition id is 1ceeb0...
Proof:
Let X, Y and z be given.
Assume H1: z ∈ X ∪ Y.
We will prove z ∈ X ∨ z ∈ Y.
Apply (UnionE_impred {X,Y} z H1) to the current goal.
Let Z be given.
Assume H2: z ∈ Z.
Assume H3: Z ∈ {X,Y}.
Apply (UPairE Z X Y H3) to the current goal.
Assume H4: Z = X.
Apply orIL to the current goal.
We will prove z ∈ X.
rewrite the current goal using H4 (from right to left).
We will prove z ∈ Z.
An exact proof term for the current goal is H2.
Assume H4: Z = Y.
Apply orIR to the current goal.
We will prove z ∈ Y.
rewrite the current goal using H4 (from right to left).
We will prove z ∈ Z.
An exact proof term for the current goal is H2.
∎
Theorem. (binunionE')
∀X Y z, ∀p : prop, (z ∈ X → p) → (z ∈ Y → p) → (z ∈ X ∪ Y → p)
In Proofgold the corresponding term root is e5b951... and proposition id is 570dd1...
Proof:
Let X, Y, z and p be given.
Assume H1 H2 Hz.
Apply binunionE X Y z Hz to the current goal.
Assume H3: z ∈ X.
An exact proof term for the current goal is H1 H3.
Assume H3: z ∈ Y.
An exact proof term for the current goal is H2 H3.
∎
Theorem. (binunion_asso)
∀X Y Z : set, X ∪ (Y ∪ Z) = (X ∪ Y) ∪ Z
In Proofgold the corresponding term root is e0f701... and proposition id is 61dfba...
Proof:
Let X, Y and Z be given.
Apply set_ext to the current goal.
Let w be given.
Assume H1: w ∈ X ∪ (Y ∪ Z).
We will prove w ∈ (X ∪ Y) ∪ Z.
Apply (binunionE X (Y ∪ Z) w H1) to the current goal.
Assume H2: w ∈ X.
Apply binunionI1 to the current goal.
Apply binunionI1 to the current goal.
An exact proof term for the current goal is H2.
Assume H2: w ∈ Y ∪ Z.
Apply (binunionE Y Z w H2) to the current goal.
Assume H3: w ∈ Y.
Apply binunionI1 to the current goal.
Apply binunionI2 to the current goal.
An exact proof term for the current goal is H3.
Assume H3: w ∈ Z.
Apply binunionI2 to the current goal.
An exact proof term for the current goal is H3.
Let w be given.
Assume H1: w ∈ (X ∪ Y) ∪ Z.
We will prove w ∈ X ∪ (Y ∪ Z).
Apply (binunionE (X ∪ Y) Z w H1) to the current goal.
Assume H2: w ∈ X ∪ Y.
Apply (binunionE X Y w H2) to the current goal.
Assume H3: w ∈ X.
Apply binunionI1 to the current goal.
An exact proof term for the current goal is H3.
Assume H3: w ∈ Y.
Apply binunionI2 to the current goal.
Apply binunionI1 to the current goal.
An exact proof term for the current goal is H3.
Assume H2: w ∈ Z.
Apply binunionI2 to the current goal.
Apply binunionI2 to the current goal.
An exact proof term for the current goal is H2.
∎
Theorem. (binunion_com_Subq)
∀X Y : set, X ∪ Y ⊆ Y ∪ X
In Proofgold the corresponding term root is ee8f5a... and proposition id is d84076...
Proof:
Let X, Y and w be given.
Assume H1: w ∈ X ∪ Y.
We will prove w ∈ Y ∪ X.
Apply (binunionE X Y w H1) to the current goal.
Assume H2: w ∈ X.
Apply binunionI2 to the current goal.
An exact proof term for the current goal is H2.
Assume H2: w ∈ Y.
Apply binunionI1 to the current goal.
An exact proof term for the current goal is H2.
∎
Theorem. (binunion_com)
∀X Y : set, X ∪ Y = Y ∪ X
In Proofgold the corresponding term root is e20f4f... and proposition id is bed33e...
Proof:
Let X and Y be given.
Apply set_ext to the current goal.
An exact proof term for the current goal is (binunion_com_Subq X Y).
An exact proof term for the current goal is (binunion_com_Subq Y X).
∎
Theorem. (binunion_idl)
∀X : set, Empty ∪ X = X
In Proofgold the corresponding term root is 8dc616... and proposition id is 52cd24...
Proof:
Let X be given.
Apply set_ext to the current goal.
Let x be given.
Assume H1: x ∈ Empty ∪ X.
Apply (binunionE Empty X x H1) to the current goal.
Assume H2: x ∈ Empty.
We will prove False.
An exact proof term for the current goal is (EmptyE x H2).
Assume H2: x ∈ X.
An exact proof term for the current goal is H2.
Let x be given.
Assume H2: x ∈ X.
We will prove x ∈ Empty ∪ X.
Apply binunionI2 to the current goal.
An exact proof term for the current goal is H2.
∎
Theorem. (binunion_idr)
∀X : set, X ∪ Empty = X
In Proofgold the corresponding term root is 67e82f... and proposition id is d5a005...
Proof:
Let X be given.
rewrite the current goal using (binunion_com X Empty) (from left to right).
An exact proof term for the current goal is (binunion_idl X).
∎
Theorem. (binunion_Subq_1)
∀X Y : set, X ⊆ X ∪ Y
In Proofgold the corresponding term root is b2c658... and proposition id is a9b3a8...
Proof:
An exact proof term for the current goal is binunionI1.
∎
Theorem. (binunion_Subq_2)
∀X Y : set, Y ⊆ X ∪ Y
In Proofgold the corresponding term root is 3cf229... and proposition id is 1507ce...
Proof:
An exact proof term for the current goal is binunionI2.
∎
Theorem. (binunion_Subq_min)
∀X Y Z : set, X ⊆ Z → Y ⊆ Z → X ∪ Y ⊆ Z
In Proofgold the corresponding term root is cc32a5... and proposition id is d64da4...
Proof:
Let X, Y and Z be given.
Assume H1: X ⊆ Z.
Assume H2: Y ⊆ Z.
Let w be given.
Assume H3: w ∈ X ∪ Y.
Apply (binunionE X Y w H3) to the current goal.
Assume H4: w ∈ X.
An exact proof term for the current goal is (H1 w H4).
Assume H4: w ∈ Y.
An exact proof term for the current goal is (H2 w H4).
∎
Theorem. (Subq_binunion_eq)
∀X Y, (X ⊆ Y) = (X ∪ Y = Y)
In Proofgold the corresponding term root is f8ee53... and proposition id is c2d393...
Proof:
Let X and Y be given.
Apply prop_ext_2 to the current goal.
Assume H1: X ⊆ Y.
We will prove X ∪ Y = Y.
Apply set_ext to the current goal.
We will prove X ∪ Y ⊆ Y.
Apply (binunion_Subq_min X Y Y) to the current goal.
We will prove X ⊆ Y.
An exact proof term for the current goal is H1.
We will prove Y ⊆ Y.
An exact proof term for the current goal is (Subq_ref Y).
We will prove Y ⊆ X ∪ Y.
An exact proof term for the current goal is (binunion_Subq_2 X Y).
Assume H1: X ∪ Y = Y.
We will prove X ⊆ Y.
rewrite the current goal using H1 (from right to left).
We will prove X ⊆ X ∪ Y.
An exact proof term for the current goal is (binunion_Subq_1 X Y).
∎
Definition. We define SetAdjoin to be λX y ⇒ X ∪ {y} of type set → set → set.
In Proofgold the corresponding term root is 1f3a09... and object id is 5bdda8...
Notation. We now use the set enumeration notation {...,...,...} in general. If 0 elements are given, then Empty is used to form the corresponding term. If 1 element is given, then Sing is used to form the corresponding term. If 2 elements are given, then UPair is used to form the corresponding term. If more than elements are given, then SetAdjoin is used to reduce to the case with one fewer elements.
Definition. We define famunion to be λX F ⇒ ⋃ {F x|x ∈ X} of type set → (set → set) → set.
In Proofgold the corresponding term root is b3e3bf... and object id is 241349...
Notation. We use ⋃ x [possibly with ascriptions] , B as a binder notation corresponding to a term constructed using famunion.
Theorem. (famunionI)
∀X : set, ∀F : (set → set), ∀x y : set, x ∈ X → y ∈ F x → y ∈ ⋃x ∈ XF x
In Proofgold the corresponding term root is 0c23d0... and proposition id is e4f490...
Proof:
An exact proof term for the current goal is (λX F x y H1 H2 ⇒ UnionI (Repl X F) y (F x) H2 (ReplI X F x H1)).
∎
Theorem. (famunionE)
∀X : set, ∀F : (set → set), ∀y : set, y ∈ (⋃x ∈ XF x) → ∃x ∈ X, y ∈ F x
In Proofgold the corresponding term root is 48683a... and proposition id is ede6e5...
Proof:
Let X, F and y be given.
Assume H1: y ∈ (⋃x ∈ XF x).
We will prove ∃x ∈ X, y ∈ F x.
Apply (UnionE_impred {F x|x ∈ X} y H1) to the current goal.
Let Y be given.
Assume H2: y ∈ Y.
Assume H3: Y ∈ {F x|x ∈ X}.
Apply (ReplE_impred X F Y H3) to the current goal.
Let x be given.
Assume H4: x ∈ X.
Assume H5: Y = F x.
We use x to witness the existential quantifier.
We will prove x ∈ X ∧ y ∈ F x.
Apply andI to the current goal.
An exact proof term for the current goal is H4.
We will prove y ∈ F x.
rewrite the current goal using H5 (from right to left).
An exact proof term for the current goal is H2.
∎
Theorem. (famunionE_impred)
∀X : set, ∀F : (set → set), ∀y : set, y ∈ (⋃x ∈ XF x) → ∀p : prop, (∀x, x ∈ X → y ∈ F x → p) → p
In Proofgold the corresponding term root is ab2266... and proposition id is 562023...
Proof:
Let X, F and y be given.
Assume Hy.
Let p be given.
Assume Hp.
Apply famunionE X F y Hy to the current goal.
Let x be given.
Assume H1.
Apply H1 to the current goal.
An exact proof term for the current goal is Hp x.
∎
Theorem. (famunion_Empty)
∀F : set → set, (⋃x ∈ EmptyF x) = Empty
In Proofgold the corresponding term root is fc1bdd... and proposition id is 22bd10...
Proof:
Let F be given.
Apply Empty_Subq_eq to the current goal.
Let y be given.
Assume Hy: y ∈ ⋃x ∈ EmptyF x.
Apply famunionE_impred Empty F y Hy to the current goal.
Let x be given.
Assume Hx: x ∈ Empty.
We will prove False.
An exact proof term for the current goal is EmptyE x Hx.
∎
Theorem. (famunion_Subq)
∀X, ∀f g : set → set, (∀x ∈ X, f x ⊆ g x) → famunion X f ⊆ famunion X g
In Proofgold the corresponding term root is 5f0b79... and proposition id is b1b158...
Proof:
Let X, f and g be given.
Assume Hfg.
Let y be given.
Assume Hy.
Apply famunionE_impred X f y Hy to the current goal.
Let x be given.
Assume Hx.
Assume H1: y ∈ f x.
Apply famunionI X g x y Hx to the current goal.
We will prove y ∈ g x.
An exact proof term for the current goal is Hfg x Hx y H1.
∎
Theorem. (famunion_ext)
∀X, ∀f g : set → set, (∀x ∈ X, f x = g x) → famunion X f = famunion X g
In Proofgold the corresponding term root is 8dc4c3... and proposition id is 5bc238...
Proof:
Let X, f and g be given.
Assume Hfg.
Apply set_ext to the current goal.
Apply famunion_Subq to the current goal.
Let x be given.
Assume Hx.
rewrite the current goal using Hfg x Hx (from left to right).
Apply Subq_ref to the current goal.
Apply famunion_Subq to the current goal.
Let x be given.
Assume Hx.
rewrite the current goal using Hfg x Hx (from left to right).
Apply Subq_ref to the current goal.
∎
Beginning of Section SepSec
Variable X : set
Variable P : set → prop
Let z : set ≝ Eps_i (λz ⇒ z ∈ X ∧ P z)
Let F : set → set ≝ λx ⇒ if P x then x else z
Definition. We define Sep to be if (∃z ∈ X, P z) then {F x|x ∈ X} else Empty of type set.
In Proofgold the corresponding term root is f336a4... and object id is 7008ec...
End of Section SepSec
Notation. {x ∈ A | B} is notation for Sep A (λ x . B).
Theorem. (SepI)
∀X : set, ∀P : (set → prop), ∀x : set, x ∈ X → P x → x ∈ {x ∈ X|P x}
In Proofgold the corresponding term root is ea56bb... and proposition id is 147a6a...
Proof:
Let X, P and x be given.
Set z to be the term Eps_i (λz ⇒ z ∈ X ∧ P z) of type set.
Set F to be the term λx ⇒ if P x then x else z of type set → set.
Assume H1: x ∈ X.
Assume H2: P x.
We prove the intermediate claim L1: ∃z ∈ X, P z.
We use x to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is H1.
An exact proof term for the current goal is H2.
We will prove x ∈ {x ∈ X|P x}.
We will prove x ∈ if (∃z ∈ X, P z) then {F x|x ∈ X} else Empty.
We prove the intermediate claim L2: (if (∃z ∈ X, P z) then {F x|x ∈ X} else Empty) = {F x|x ∈ X}.
An exact proof term for the current goal is (If_i_1 (∃z ∈ X, P z) {F x|x ∈ X} Empty L1).
rewrite the current goal using L2 (from left to right).
We will prove x ∈ {F x|x ∈ X}.
We prove the intermediate claim L3: F x = x.
We will prove (if P x then x else z) = x.
An exact proof term for the current goal is (If_i_1 (P x) x z H2).
rewrite the current goal using L3 (from right to left).
We will prove F x ∈ {F x|x ∈ X}.
An exact proof term for the current goal is (ReplI X F x H1).
∎
Theorem. (SepE)
∀X : set, ∀P : (set → prop), ∀x : set, x ∈ {x ∈ X|P x} → x ∈ X ∧ P x
In Proofgold the corresponding term root is 102830... and proposition id is 00899a...
Proof:
Let X, P and x be given.
Set z to be the term Eps_i (λz ⇒ z ∈ X ∧ P z) of type set.
Set F to be the term λx ⇒ if P x then x else z of type set → set.
Apply (xm (∃z ∈ X, P z)) to the current goal.
Assume H1: ∃z ∈ X, P z.
We will prove (x ∈ (if (∃z ∈ X, P z) then {F x|x ∈ X} else Empty) → x ∈ X ∧ P x).
We prove the intermediate claim L1: (if (∃z ∈ X, P z) then {F x|x ∈ X} else Empty) = {F x|x ∈ X}.
An exact proof term for the current goal is (If_i_1 (∃z ∈ X, P z) {F x|x ∈ X} Empty H1).
rewrite the current goal using L1 (from left to right).
We will prove x ∈ {F x|x ∈ X} → x ∈ X ∧ P x.
Assume H2: x ∈ {F x|x ∈ X}.
Apply (ReplE_impred X F x H2) to the current goal.
Let y be given.
Assume H3: y ∈ X.
Assume H4: x = F y.
We will prove x ∈ X ∧ P x.
Apply (xm (P y)) to the current goal.
Assume H5: P y.
We prove the intermediate claim L2: x = y.
rewrite the current goal using (If_i_1 (P y) y z H5) (from right to left).
An exact proof term for the current goal is H4.
rewrite the current goal using L2 (from left to right).
We will prove y ∈ X ∧ P y.
Apply andI to the current goal.
An exact proof term for the current goal is H3.
An exact proof term for the current goal is H5.
Assume H5: ¬ P y.
We prove the intermediate claim L2: x = z.
rewrite the current goal using (If_i_0 (P y) y z H5) (from right to left).
An exact proof term for the current goal is H4.
rewrite the current goal using L2 (from left to right).
We will prove z ∈ X ∧ P z.
An exact proof term for the current goal is (Eps_i_ex (λz ⇒ z ∈ X ∧ P z) H1).
Assume H1: ¬ ∃z ∈ X, P z.
We will prove (x ∈ (if (∃z ∈ X, P z) then {F x|x ∈ X} else Empty) → x ∈ X ∧ P x).
We prove the intermediate claim L1: (if (∃z ∈ X, P z) then {F x|x ∈ X} else Empty) = Empty.
An exact proof term for the current goal is (If_i_0 (∃z ∈ X, P z) {F x|x ∈ X} Empty H1).
rewrite the current goal using L1 (from left to right).
We will prove x ∈ Empty → x ∈ X ∧ P x.
Assume H2: x ∈ Empty.
We will prove False.
An exact proof term for the current goal is (EmptyE x H2).
∎
Theorem. (SepE1)
∀X : set, ∀P : (set → prop), ∀x : set, x ∈ {x ∈ X|P x} → x ∈ X
In Proofgold the corresponding term root is d83445... and proposition id is 6399b0...
Proof:
An exact proof term for the current goal is (λX P x H ⇒ SepE X P x H (x ∈ X) (λH _ ⇒ H)).
∎
Theorem. (SepE2)
∀X : set, ∀P : (set → prop), ∀x : set, x ∈ {x ∈ X|P x} → P x
In Proofgold the corresponding term root is aa61ca... and proposition id is 104227...
Proof:
An exact proof term for the current goal is (λX P x H ⇒ SepE X P x H (P x) (λ_ H ⇒ H)).
∎
Theorem. (Sep_Empty)
∀P : set → prop, {x ∈ Empty|P x} = Empty
In Proofgold the corresponding term root is f36949... and proposition id is e3473a...
Proof:
Let P be given.
Apply Empty_eq to the current goal.
Let x be given.
Assume Hx.
An exact proof term for the current goal is EmptyE x (SepE1 Empty P x Hx).
∎
Theorem. (Sep_Subq)
∀X : set, ∀P : set → prop, {x ∈ X|P x} ⊆ X
In Proofgold the corresponding term root is 97d080... and proposition id is c17d45...
Proof:
An exact proof term for the current goal is SepE1.
∎
Theorem. (Sep_In_Power)
∀X : set, ∀P : set → prop, {x ∈ X|P x} ∈ 𝒫 X
In Proofgold the corresponding term root is 612c94... and proposition id is fba53e...
Proof:
An exact proof term for the current goal is (λX P ⇒ PowerI X (Sep X P) (Sep_Subq X P)).
∎
Definition. We define ReplSep to be λX P F ⇒ {F x|x ∈ {z ∈ X|P z}} of type set → (set → prop) → (set → set) → set.
In Proofgold the corresponding term root is ec807b... and object id is 9bdb56...
Notation. {B| x ∈ A, C} is notation for ReplSep A (λ x . C) (λ x . B).
Theorem. (ReplSepI)
∀X : set, ∀P : set → prop, ∀F : set → set, ∀x : set, x ∈ X → P x → F x ∈ {F x|x ∈ X, P x}
In Proofgold the corresponding term root is 35b8ec... and proposition id is aa7b2b...
Proof:
An exact proof term for the current goal is (λX P F x u v ⇒ ReplI (Sep X P) F x (SepI X P x u v)).
∎
Theorem. (ReplSepE)
∀X : set, ∀P : set → prop, ∀F : set → set, ∀y : set, y ∈ {F x|x ∈ X, P x} → ∃x : set, x ∈ X ∧ P x ∧ y = F x
In Proofgold the corresponding term root is 415ff1... and proposition id is b1bc76...
Proof:
Let X, P, F and y be given.
Assume H1: y ∈ {F x|x ∈ {z ∈ X|P z}}.
Apply (ReplE {z ∈ X|P z} F y H1) to the current goal.
Let x be given.
Assume H2: x ∈ {z ∈ X|P z} ∧ y = F x.
Apply H2 to the current goal.
Assume H3: x ∈ {z ∈ X|P z}.
Assume H4: y = F x.
Apply (SepE X P x H3) to the current goal.
Assume H5: x ∈ X.
Assume H6: P x.
We use x to witness the existential quantifier.
Apply and3I to the current goal.
An exact proof term for the current goal is H5.
An exact proof term for the current goal is H6.
An exact proof term for the current goal is H4.
∎
Theorem. (ReplSepE_impred)
∀X : set, ∀P : set → prop, ∀F : set → set, ∀y : set, y ∈ {F x|x ∈ X, P x} → ∀p : prop, (∀x ∈ X, P x → y = F x → p) → p
In Proofgold the corresponding term root is 5d26e1... and proposition id is 063cea...
Proof:
Let X, P, F and y be given.
Assume H1: y ∈ {F x|x ∈ X, P x}.
Let p be given.
Assume H2: ∀x ∈ X, P x → y = F x → p.
We will prove p.
Apply ReplSepE X P F y H1 to the current goal.
Let x be given.
Assume H3.
Apply H3 to the current goal.
Assume H3.
Apply H3 to the current goal.
An exact proof term for the current goal is H2 x.
∎
Definition. We define binintersect to be λX Y ⇒ {x ∈ X|x ∈ Y} of type set → set → set.
In Proofgold the corresponding term root is b2abd2... and object id is 2c68d8...
Notation. We use ∩ as an infix operator with priority 340 and which associates to the left corresponding to applying term binintersect.
Theorem. (binintersectI)
∀X Y z, z ∈ X → z ∈ Y → z ∈ X ∩ Y
In Proofgold the corresponding term root is fd656d... and proposition id is 9081a0...
Proof:
An exact proof term for the current goal is (λX Y z H1 H2 ⇒ SepI X (λx : set ⇒ x ∈ Y) z H1 H2).
∎
Theorem. (binintersectE)
∀X Y z, z ∈ X ∩ Y → z ∈ X ∧ z ∈ Y
In Proofgold the corresponding term root is 463b5a... and proposition id is ca0636...
Proof:
An exact proof term for the current goal is (λX Y z H1 ⇒ SepE X (λx : set ⇒ x ∈ Y) z H1).
∎
Theorem. (binintersectE1)
∀X Y z, z ∈ X ∩ Y → z ∈ X
In Proofgold the corresponding term root is b12123... and proposition id is 51005e...
Proof:
An exact proof term for the current goal is (λX Y z H1 ⇒ SepE1 X (λx : set ⇒ x ∈ Y) z H1).
∎
Theorem. (binintersectE2)
∀X Y z, z ∈ X ∩ Y → z ∈ Y
In Proofgold the corresponding term root is 39656f... and proposition id is 599de4...
Proof:
An exact proof term for the current goal is (λX Y z H1 ⇒ SepE2 X (λx : set ⇒ x ∈ Y) z H1).
∎
Theorem. (binintersect_Subq_1)
∀X Y : set, X ∩ Y ⊆ X
In Proofgold the corresponding term root is f296ad... and proposition id is 303833...
Proof:
An exact proof term for the current goal is binintersectE1.
∎
Theorem. (binintersect_Subq_2)
∀X Y : set, X ∩ Y ⊆ Y
In Proofgold the corresponding term root is fcd5f5... and proposition id is 8b9fd2...
Proof:
An exact proof term for the current goal is binintersectE2.
∎
Theorem. (binintersect_Subq_eq_1)
∀X Y, X ⊆ Y → X ∩ Y = X
In Proofgold the corresponding term root is 277eec... and proposition id is ec1cc3...
Proof:
Let X and Y be given.
Assume H1: X ⊆ Y.
Apply set_ext to the current goal.
Apply binintersect_Subq_1 to the current goal.
Let x be given.
Assume H2: x ∈ X.
Apply binintersectI to the current goal.
An exact proof term for the current goal is H2.
Apply H1 to the current goal.
An exact proof term for the current goal is H2.
∎
Theorem. (binintersect_Subq_max)
∀X Y Z : set, Z ⊆ X → Z ⊆ Y → Z ⊆ X ∩ Y
In Proofgold the corresponding term root is 95624d... and proposition id is 64636b...
Proof:
Let X, Y and Z be given.
Assume H1: Z ⊆ X.
Assume H2: Z ⊆ Y.
Let w be given.
Assume H3: w ∈ Z.
Apply (binintersectI X Y w) to the current goal.
We will prove w ∈ X.
An exact proof term for the current goal is (H1 w H3).
We will prove w ∈ Y.
An exact proof term for the current goal is (H2 w H3).
∎
Theorem. (binintersect_com_Subq)
∀X Y : set, X ∩ Y ⊆ Y ∩ X
In Proofgold the corresponding term root is 410255... and proposition id is 5977b8...
Proof:
Let X and Y be given.
Apply (binintersect_Subq_max Y X (X ∩ Y)) to the current goal.
We will prove X ∩ Y ⊆ Y.
Apply binintersect_Subq_2 to the current goal.
We will prove X ∩ Y ⊆ X.
Apply binintersect_Subq_1 to the current goal.
∎
Theorem. (binintersect_com)
∀X Y : set, X ∩ Y = Y ∩ X
In Proofgold the corresponding term root is 1a2c6f... and proposition id is ea8a1c...
Proof:
Let X and Y be given.
Apply set_ext to the current goal.
An exact proof term for the current goal is (binintersect_com_Subq X Y).
An exact proof term for the current goal is (binintersect_com_Subq Y X).
∎
Definition. We define setminus to be λX Y ⇒ Sep X (λx ⇒ x ∉ Y) of type set → set → set.
In Proofgold the corresponding term root is c68e5a... and object id is 913da1...
Notation. We use ∖ as an infix operator with priority 350 and no associativity corresponding to applying term setminus.
Theorem. (setminusI)
∀X Y z, (z ∈ X) → (z ∉ Y) → z ∈ X ∖ Y
In Proofgold the corresponding term root is 0feeb8... and proposition id is 80db6a...
Proof:
An exact proof term for the current goal is (λX Y z H1 H2 ⇒ SepI X (λx : set ⇒ x ∉ Y) z H1 H2).
∎
Theorem. (setminusE)
∀X Y z, (z ∈ X ∖ Y) → z ∈ X ∧ z ∉ Y
In Proofgold the corresponding term root is 6774f4... and proposition id is 71763a...
Proof:
An exact proof term for the current goal is (λX Y z H ⇒ SepE X (λx : set ⇒ x ∉ Y) z H).
∎
Theorem. (setminusE1)
∀X Y z, (z ∈ X ∖ Y) → z ∈ X
In Proofgold the corresponding term root is 064bd5... and proposition id is d74baf...
Proof:
An exact proof term for the current goal is (λX Y z H ⇒ SepE1 X (λx : set ⇒ x ∉ Y) z H).
∎
Theorem. (setminusE2)
∀X Y z, (z ∈ X ∖ Y) → z ∉ Y
In Proofgold the corresponding term root is 3e6969... and proposition id is 36b273...
Proof:
An exact proof term for the current goal is (λX Y z H ⇒ SepE2 X (λx : set ⇒ x ∉ Y) z H).
∎
Theorem. (setminus_Subq)
∀X Y : set, X ∖ Y ⊆ X
In Proofgold the corresponding term root is 5a497a... and proposition id is 66f982...
Proof:
An exact proof term for the current goal is setminusE1.
∎
Theorem. (setminus_Subq_contra)
∀X Y Z : set, Z ⊆ Y → X ∖ Y ⊆ X ∖ Z
In Proofgold the corresponding term root is b1f440... and proposition id is 9a06b2...
Proof:
Let X, Y and Z be given.
Assume H1: Z ⊆ Y.
Let x be given.
Assume H2: x ∈ X ∖ Y.
Apply (setminusE X Y x H2) to the current goal.
Assume H3: x ∈ X.
Assume H4: x ∉ Y.
We will prove x ∈ X ∖ Z.
Apply setminusI to the current goal.
An exact proof term for the current goal is H3.
We will prove x ∉ Z.
An exact proof term for the current goal is (Subq_contra Z Y x H1 H4).
∎
Theorem. (setminus_In_Power)
∀A U, A ∖ U ∈ 𝒫 A
In Proofgold the corresponding term root is 062275... and proposition id is f503e1...
Proof:
Let A and U be given.
Apply PowerI to the current goal.
Apply setminus_Subq to the current goal.
∎
Theorem. (setminus_idr)
∀X, X ∖ Empty = X
In Proofgold the corresponding term root is 20e108... and proposition id is ebb8ec...
Proof:
Let X be given.
Apply set_ext to the current goal.
An exact proof term for the current goal is setminus_Subq X Empty.
Let u be given.
Assume Hu: u ∈ X.
Apply setminusI to the current goal.
An exact proof term for the current goal is Hu.
An exact proof term for the current goal is EmptyE u.
∎
Theorem. (In_irref)
∀x, x ∉ x
In Proofgold the corresponding term root is 12629a... and proposition id is 7bfd46...
Proof:
Apply In_ind to the current goal.
We will prove (∀X : set, (∀x : set, x ∈ X → x ∉ x) → X ∉ X).
Let X be given.
Assume IH: ∀x : set, x ∈ X → x ∉ x.
Assume H: X ∈ X.
An exact proof term for the current goal is IH X H H.
∎
Theorem. (In_no2cycle)
∀x y, x ∈ y → y ∈ x → False
In Proofgold the corresponding term root is 6d0cb3... and proposition id is 344f30...
Proof:
Apply In_ind to the current goal.
Let x be given.
Assume IH: ∀z, z ∈ x → ∀y, z ∈ y → y ∈ z → False.
Let y be given.
Assume H1: x ∈ y.
Assume H2: y ∈ x.
An exact proof term for the current goal is IH y H2 x H2 H1.
∎
Definition. We define ordsucc to be λx : set ⇒ x ∪ {x} of type set → set.
In Proofgold the corresponding term root is 65d883... and object id is 1452f7...
Theorem. (ordsuccI1)
∀x : set, x ⊆ ordsucc x
In Proofgold the corresponding term root is 5a2f15... and proposition id is 367d80...
Proof:
Let x be given.
An exact proof term for the current goal is (λ(y : set)(H1 : y ∈ x) ⇒ binunionI1 x {x} y H1).
∎
Theorem. (ordsuccI2)
∀x : set, x ∈ ordsucc x
In Proofgold the corresponding term root is 2f6efd... and proposition id is 2be6be...
Proof:
An exact proof term for the current goal is (λx : set ⇒ binunionI2 x {x} x (SingI x)).
∎
Theorem. (ordsuccE)
∀x y : set, y ∈ ordsucc x → y ∈ x ∨ y = x
In Proofgold the corresponding term root is 01c4ee... and proposition id is 3849f5...
Proof:
Let x and y be given.
Assume H1: y ∈ x ∪ {x}.
Apply (binunionE x {x} y H1) to the current goal.
Assume H2: y ∈ x.
Apply orIL to the current goal.
An exact proof term for the current goal is H2.
Assume H2: y ∈ {x}.
Apply orIR to the current goal.
An exact proof term for the current goal is (SingE x y H2).
∎
Notation. Natural numbers 0,1,2,... are notation for the terms formed using Empty as 0 and forming successors with ordsucc.
Theorem. (neq_0_ordsucc)
∀a : set, 0 ≠ ordsucc a
In Proofgold the corresponding term root is 1b20cb... and proposition id is 04c410...
Proof:
Let a be given.
We will prove ¬ (0 = ordsucc a).
Assume H1: 0 = ordsucc a.
We prove the intermediate claim L1: a ∈ ordsucc a → False.
rewrite the current goal using H1 (from right to left).
An exact proof term for the current goal is (EmptyE a).
An exact proof term for the current goal is (L1 (ordsuccI2 a)).
∎
Theorem. (neq_ordsucc_0)
∀a : set, ordsucc a ≠ 0
In Proofgold the corresponding term root is 9b391e... and proposition id is 346ee0...
Proof:
Let a be given.
An exact proof term for the current goal is neq_i_sym 0 (ordsucc a) (neq_0_ordsucc a).
∎
Theorem. (ordsucc_inj)
∀a b : set, ordsucc a = ordsucc b → a = b
In Proofgold the corresponding term root is 40cad6... and proposition id is 4869a1...
Proof:
Let a and b be given.
Assume H1: ordsucc a = ordsucc b.
We prove the intermediate claim L1: a ∈ ordsucc b.
rewrite the current goal using H1 (from right to left).
An exact proof term for the current goal is (ordsuccI2 a).
Apply (ordsuccE b a L1) to the current goal.
Assume H2: a ∈ b.
We prove the intermediate claim L2: b ∈ ordsucc a.
rewrite the current goal using H1 (from left to right).
An exact proof term for the current goal is (ordsuccI2 b).
Apply (ordsuccE a b L2) to the current goal.
Assume H3: b ∈ a.
We will prove False.
An exact proof term for the current goal is (In_no2cycle a b H2 H3).
Assume H3: b = a.
Use symmetry.
An exact proof term for the current goal is H3.
Assume H2: a = b.
An exact proof term for the current goal is H2.
∎
Theorem. (ordsucc_inj_contra)
∀a b : set, a ≠ b → ordsucc a ≠ ordsucc b
In Proofgold the corresponding term root is 791304... and proposition id is 66f4df...
Proof:
Let a and b be given.
Assume H1.
We will prove ¬ (ordsucc a = ordsucc b).
Assume H2: ordsucc a = ordsucc b.
An exact proof term for the current goal is H1 (ordsucc_inj a b H2).
∎
Theorem. (In_0_1)
0 ∈ 1
In Proofgold the corresponding term root is e984b0... and proposition id is 3be80a...
Proof:
An exact proof term for the current goal is (ordsuccI2 0).
∎
Theorem. (In_0_2)
0 ∈ 2
In Proofgold the corresponding term root is 56dc2c... and proposition id is 5a202e...
Proof:
An exact proof term for the current goal is (ordsuccI1 1 0 In_0_1).
∎
Theorem. (In_1_2)
1 ∈ 2
In Proofgold the corresponding term root is 8704b3... and proposition id is 431e9e...
Proof:
An exact proof term for the current goal is (ordsuccI2 1).
∎
Theorem. (In_1_3)
1 ∈ 3
In Proofgold the corresponding term root is 85201c... and proposition id is 8f3cec...
Proof:
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is In_1_2.
∎
Theorem. (In_2_3)
2 ∈ 3
In Proofgold the corresponding term root is 4d67ce... and proposition id is ebdb5d...
Proof:
Apply ordsuccI2 to the current goal.
∎
Theorem. (In_1_4)
1 ∈ 4
In Proofgold the corresponding term root is 5cabbb... and proposition id is 1e2d5b...
Proof:
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is In_1_3.
∎
Theorem. (In_2_4)
2 ∈ 4
In Proofgold the corresponding term root is 33c2ff... and proposition id is 2cbb94...
Proof:
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is In_2_3.
∎
Theorem. (In_3_4)
3 ∈ 4
In Proofgold the corresponding term root is f25e44... and proposition id is 22a0af...
Proof:
Apply ordsuccI2 to the current goal.
∎
Theorem. (In_1_5)
1 ∈ 5
In Proofgold the corresponding term root is be96a3... and proposition id is 34c02d...
Proof:
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is In_1_4.
∎
Theorem. (In_2_5)
2 ∈ 5
In Proofgold the corresponding term root is ec68b2... and proposition id is 7ed8e4...
Proof:
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is In_2_4.
∎
Theorem. (In_3_5)
3 ∈ 5
In Proofgold the corresponding term root is eddd36... and proposition id is df92fe...
Proof:
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is In_3_4.
∎
Theorem. (In_4_5)
4 ∈ 5
In Proofgold the corresponding term root is e72ad7... and proposition id is 03b9ff...
Proof:
Apply ordsuccI2 to the current goal.
∎
Theorem. (In_1_6)
1 ∈ 6
In Proofgold the corresponding term root is 4c8428... and proposition id is 327244...
Proof:
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is In_1_5.
∎
Theorem. (In_1_7)
1 ∈ 7
In Proofgold the corresponding term root is aa5cde... and proposition id is 392ce8...
Proof:
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is In_1_6.
∎
Theorem. (In_1_8)
1 ∈ 8
In Proofgold the corresponding term root is fbef0f... and proposition id is 0d7bd5...
Proof:
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is In_1_7.
∎
Definition. We define nat_p to be λn : set ⇒ ∀p : set → prop, p 0 → (∀x : set, p x → p (ordsucc x)) → p n of type set → prop.
In Proofgold the corresponding term root is 458be3... and object id is cfd8eb...
Theorem. (nat_0)
nat_p 0
In Proofgold the corresponding term root is 1617bc... and proposition id is 457a72...
Proof:
An exact proof term for the current goal is (λp H _ ⇒ H).
∎
Theorem. (nat_ordsucc)
∀n : set, nat_p n → nat_p (ordsucc n)
In Proofgold the corresponding term root is 87f001... and proposition id is 57e940...
Proof:
An exact proof term for the current goal is (λn H1 p H2 H3 ⇒ H3 n (H1 p H2 H3)).
∎
Theorem. (nat_1)
nat_p 1
In Proofgold the corresponding term root is 2f83fa... and proposition id is 672d0d...
Proof:
An exact proof term for the current goal is (nat_ordsucc 0 nat_0).
∎
Theorem. (nat_2)
nat_p 2
In Proofgold the corresponding term root is a8be15... and proposition id is 9112e9...
Proof:
An exact proof term for the current goal is (nat_ordsucc 1 nat_1).
∎
Theorem. (nat_3)
nat_p 3
In Proofgold the corresponding term root is c0849f... and proposition id is 133bb7...
Proof:
An exact proof term for the current goal is nat_ordsucc 2 nat_2.
∎
Theorem. (nat_4)
nat_p 4
In Proofgold the corresponding term root is 2c8e1a... and proposition id is 69ffc3...
Proof:
An exact proof term for the current goal is nat_ordsucc 3 nat_3.
∎
Theorem. (nat_5)
nat_p 5
In Proofgold the corresponding term root is 1a2477... and proposition id is b84b25...
Proof:
An exact proof term for the current goal is nat_ordsucc 4 nat_4.
∎
Theorem. (nat_6)
nat_p 6
In Proofgold the corresponding term root is 25de88... and proposition id is b4ec67...
Proof:
An exact proof term for the current goal is nat_ordsucc 5 nat_5.
∎
Theorem. (nat_7)
nat_p 7
In Proofgold the corresponding term root is c2e74e... and proposition id is 86ad85...
Proof:
An exact proof term for the current goal is nat_ordsucc 6 nat_6.
∎
Theorem. (nat_8)
nat_p 8
In Proofgold the corresponding term root is 18883c... and proposition id is f93125...
Proof:
An exact proof term for the current goal is nat_ordsucc 7 nat_7.
∎
Theorem. (nat_0_in_ordsucc)
∀n, nat_p n → 0 ∈ ordsucc n
In Proofgold the corresponding term root is 9b50b9... and proposition id is 251f3d...
Proof:
Let n be given.
Assume H1.
Apply H1 (λn ⇒ 0 ∈ ordsucc n) to the current goal.
We will prove 0 ∈ ordsucc 0.
An exact proof term for the current goal is In_0_1.
Let n be given.
Assume IH: 0 ∈ ordsucc n.
We will prove 0 ∈ ordsucc (ordsucc n).
An exact proof term for the current goal is (ordsuccI1 (ordsucc n) 0 IH).
∎
Theorem. (nat_ordsucc_in_ordsucc)
∀n, nat_p n → ∀m ∈ n, ordsucc m ∈ ordsucc n
In Proofgold the corresponding term root is 819399... and proposition id is 4c2c95...
Proof:
Let n be given.
Assume H1.
Apply (H1 (λn ⇒ ∀m ∈ n, ordsucc m ∈ ordsucc n)) to the current goal.
We will prove ∀m ∈ 0, ordsucc m ∈ ordsucc 0.
Let m be given.
Assume Hm: m ∈ 0.
We will prove False.
An exact proof term for the current goal is (EmptyE m Hm).
Let n be given.
Assume IH: ∀m ∈ n, ordsucc m ∈ ordsucc n.
We will prove ∀m ∈ ordsucc n, ordsucc m ∈ ordsucc (ordsucc n).
Let m be given.
Assume H2: m ∈ ordsucc n.
We will prove ordsucc m ∈ ordsucc (ordsucc n).
Apply (ordsuccE n m H2) to the current goal.
Assume H3: m ∈ n.
We prove the intermediate claim L1: ordsucc m ∈ ordsucc n.
An exact proof term for the current goal is (IH m H3).
An exact proof term for the current goal is (ordsuccI1 (ordsucc n) (ordsucc m) L1).
Assume H3: m = n.
rewrite the current goal using H3 (from left to right).
We will prove ordsucc n ∈ ordsucc (ordsucc n).
An exact proof term for the current goal is (ordsuccI2 (ordsucc n)).
∎
Theorem. (nat_ind)
∀p : set → prop, p 0 → (∀n, nat_p n → p n → p (ordsucc n)) → ∀n, nat_p n → p n
In Proofgold the corresponding term root is 5d06f5... and proposition id is ecdce8...
Proof:
Let p be given.
Assume H1: p 0.
Assume H2: ∀n, nat_p n → p n → p (ordsucc n).
We prove the intermediate claim L1: nat_p 0 ∧ p 0.
An exact proof term for the current goal is (andI (nat_p 0) (p 0) nat_0 H1).
We prove the intermediate claim L2: ∀n, nat_p n ∧ p n → nat_p (ordsucc n) ∧ p (ordsucc n).
Let n be given.
Assume H3: nat_p n ∧ p n.
Apply H3 to the current goal.
Assume H4: nat_p n.
Assume H5: p n.
Apply andI to the current goal.
We will prove nat_p (ordsucc n).
An exact proof term for the current goal is (nat_ordsucc n H4).
We will prove p (ordsucc n).
An exact proof term for the current goal is (H2 n H4 H5).
Let n be given.
Assume H3.
We prove the intermediate claim L3: nat_p n ∧ p n.
An exact proof term for the current goal is (H3 (λn ⇒ nat_p n ∧ p n) L1 L2).
An exact proof term for the current goal is (andER (nat_p n) (p n) L3).
∎
Theorem. (nat_inv_impred)
∀p : set → prop, p 0 → (∀n, nat_p n → p (ordsucc n)) → ∀n, nat_p n → p n
In Proofgold the corresponding term root is eddc5f... and proposition id is 175e22...
Proof:
Let p be given.
Assume H1 H2.
An exact proof term for the current goal is nat_ind p H1 (λn H _ ⇒ H2 n H).
∎
Theorem. (nat_inv)
∀n, nat_p n → n = 0 ∨ ∃x, nat_p x ∧ n = ordsucc x
In Proofgold the corresponding term root is 5d4296... and proposition id is 4f4167...
Proof:
Apply nat_inv_impred to the current goal.
Apply orIL to the current goal.
Use reflexivity.
Let n be given.
Assume Hn.
Apply orIR to the current goal.
We use n to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is Hn.
Use reflexivity.
∎
Theorem. (nat_complete_ind)
∀p : set → prop, (∀n, nat_p n → (∀m ∈ n, p m) → p n) → ∀n, nat_p n → p n
In Proofgold the corresponding term root is d32e5b... and proposition id is 1e1608...
Proof:
Let p be given.
Assume H1: ∀n, nat_p n → (∀m ∈ n, p m) → p n.
We prove the intermediate claim L1: ∀n : set, nat_p n → ∀m ∈ n, p m.
Apply nat_ind to the current goal.
We will prove ∀m ∈ 0, p m.
Let m be given.
Assume Hm: m ∈ 0.
We will prove False.
An exact proof term for the current goal is (EmptyE m Hm).
Let n be given.
Assume Hn: nat_p n.
Assume IHn: ∀m ∈ n, p m.
We will prove ∀m ∈ ordsucc n, p m.
Let m be given.
Assume Hm: m ∈ ordsucc n.
We will prove p m.
Apply (ordsuccE n m Hm) to the current goal.
Assume H2: m ∈ n.
An exact proof term for the current goal is (IHn m H2).
Assume H2: m = n.
We will prove p m.
rewrite the current goal using H2 (from left to right).
We will prove p n.
An exact proof term for the current goal is (H1 n Hn IHn).
We will prove ∀n, nat_p n → p n.
An exact proof term for the current goal is (λn Hn ⇒ H1 n Hn (L1 n Hn)).
∎
Theorem. (nat_p_trans)
∀n, nat_p n → ∀m ∈ n, nat_p m
In Proofgold the corresponding term root is e88449... and proposition id is 03b874...
Proof:
Apply nat_ind to the current goal.
We will prove ∀m ∈ 0, nat_p m.
Let m be given.
Assume Hm: m ∈ 0.
We will prove False.
An exact proof term for the current goal is (EmptyE m Hm).
Let n be given.
Assume Hn: nat_p n.
Assume IHn: ∀m ∈ n, nat_p m.
We will prove ∀m ∈ ordsucc n, nat_p m.
Let m be given.
Assume Hm: m ∈ ordsucc n.
Apply (ordsuccE n m Hm) to the current goal.
Assume H1: m ∈ n.
An exact proof term for the current goal is (IHn m H1).
Assume H1: m = n.
rewrite the current goal using H1 (from left to right).
An exact proof term for the current goal is Hn.
∎
Theorem. (nat_trans)
∀n, nat_p n → ∀m ∈ n, m ⊆ n
In Proofgold the corresponding term root is 6bc6e1... and proposition id is cad6fe...
Proof:
Apply nat_ind to the current goal.
We will prove ∀m ∈ 0, m ⊆ 0.
Let m be given.
Assume Hm: m ∈ 0.
We will prove False.
An exact proof term for the current goal is (EmptyE m Hm).
Let n be given.
Assume Hn: nat_p n.
Assume IHn: ∀m ∈ n, m ⊆ n.
We will prove ∀m ∈ ordsucc n, m ⊆ ordsucc n.
Let m be given.
Assume Hm: m ∈ ordsucc n.
Apply (ordsuccE n m Hm) to the current goal.
Assume H1: m ∈ n.
We will prove m ⊆ ordsucc n.
Apply (Subq_tra m n (ordsucc n)) to the current goal.
An exact proof term for the current goal is (IHn m H1).
An exact proof term for the current goal is (ordsuccI1 n).
Assume H1: m = n.
We will prove m ⊆ ordsucc n.
rewrite the current goal using H1 (from left to right).
We will prove n ⊆ ordsucc n.
An exact proof term for the current goal is (ordsuccI1 n).
∎
Theorem. (nat_ordsucc_trans)
∀n, nat_p n → ∀m ∈ ordsucc n, m ⊆ n
In Proofgold the corresponding term root is e42e58... and proposition id is bc868d...
Proof:
Let n be given.
Assume Hn: nat_p n.
Let m be given.
Assume Hm: m ∈ ordsucc n.
Let k be given.
Assume Hk: k ∈ m.
We will prove k ∈ n.
Apply (ordsuccE n m Hm) to the current goal.
Assume H1: m ∈ n.
An exact proof term for the current goal is nat_trans n Hn m H1 k Hk.
Assume H1: m = n.
rewrite the current goal using H1 (from right to left).
An exact proof term for the current goal is Hk.
∎
Theorem. (Union_ordsucc_eq)
∀n, nat_p n → ⋃ (ordsucc n) = n
In Proofgold the corresponding term root is 87308c... and proposition id is e8cdeb...
Proof:
Apply nat_complete_ind to the current goal.
Let n be given.
Assume Hn: nat_p n.
Assume IHn: ∀m ∈ n, ⋃ (ordsucc m) = m.
We will prove ⋃ (ordsucc n) = n.
Apply set_ext to the current goal.
Let m be given.
Assume Hm: m ∈ ⋃ (ordsucc n).
Apply (UnionE_impred (ordsucc n) m Hm) to the current goal.
Let k be given.
Assume H1: m ∈ k.
Assume H2: k ∈ ordsucc n.
We will prove m ∈ n.
An exact proof term for the current goal is nat_ordsucc_trans n Hn k H2 m H1.
Let m be given.
Assume Hm: m ∈ n.
We will prove m ∈ ⋃ (ordsucc n).
Apply (UnionI (ordsucc n) m n) to the current goal.
An exact proof term for the current goal is Hm.
An exact proof term for the current goal is (ordsuccI2 n).
∎
Theorem. (cases_1)
∀i ∈ 1, ∀p : set → prop, p 0 → p i
In Proofgold the corresponding term root is 706605... and proposition id is c6f0c0...
Proof:
Let i be given.
Assume Hi.
Let p be given.
Assume Hp0.
Apply ordsuccE 0 i Hi to the current goal.
Assume Hil: i ∈ 0.
Apply EmptyE i Hil to the current goal.
Assume Hie: i = 0.
rewrite the current goal using Hie (from left to right).
An exact proof term for the current goal is Hp0.
∎
Theorem. (cases_2)
∀i ∈ 2, ∀p : set → prop, p 0 → p 1 → p i
In Proofgold the corresponding term root is 4693b2... and proposition id is 74563a...
Proof:
Let i be given.
Assume Hi.
Let p be given.
Assume Hp0 Hp1.
Apply ordsuccE 1 i Hi to the current goal.
Assume Hil: i ∈ 1.
An exact proof term for the current goal is cases_1 i Hil p Hp0.
Assume Hie: i = 1.
rewrite the current goal using Hie (from left to right).
An exact proof term for the current goal is Hp1.
∎
Theorem. (cases_3)
∀i ∈ 3, ∀p : set → prop, p 0 → p 1 → p 2 → p i
In Proofgold the corresponding term root is 2a0ca2... and proposition id is 1a8a20...
Proof:
Let i be given.
Assume Hi.
Let p be given.
Assume Hp0 Hp1 Hp2.
Apply ordsuccE 2 i Hi to the current goal.
Assume Hil: i ∈ 2.
An exact proof term for the current goal is cases_2 i Hil p Hp0 Hp1.
Assume Hie: i = 2.
rewrite the current goal using Hie (from left to right).
An exact proof term for the current goal is Hp2.
∎
Theorem. (neq_0_1)
0 ≠ 1
In Proofgold the corresponding term root is 9630fd... and proposition id is 74867a...
Proof:
An exact proof term for the current goal is (neq_0_ordsucc 0).
∎
Theorem. (neq_1_0)
1 ≠ 0
In Proofgold the corresponding term root is 3a0478... and proposition id is 952fed...
Proof:
An exact proof term for the current goal is (neq_ordsucc_0 0).
∎
Theorem. (neq_0_2)
0 ≠ 2
In Proofgold the corresponding term root is 055580... and proposition id is 9c9a3f...
Proof:
An exact proof term for the current goal is (neq_0_ordsucc 1).
∎
Theorem. (neq_2_0)
2 ≠ 0
In Proofgold the corresponding term root is 4d54d8... and proposition id is 00b357...
Proof:
An exact proof term for the current goal is (neq_ordsucc_0 1).
∎
Theorem. (neq_1_2)
1 ≠ 2
In Proofgold the corresponding term root is 1d088a... and proposition id is e5542e...
Proof:
An exact proof term for the current goal is (ordsucc_inj_contra 0 1 neq_0_1).
∎
Theorem. (neq_1_3)
1 ≠ 3
In Proofgold the corresponding term root is d3b293... and proposition id is c0a1f7...
Proof:
Apply ordsucc_inj_contra to the current goal.
An exact proof term for the current goal is neq_0_2.
∎
Theorem. (neq_2_3)
2 ≠ 3
In Proofgold the corresponding term root is 9a904f... and proposition id is 2cd643...
Proof:
Apply ordsucc_inj_contra to the current goal.
An exact proof term for the current goal is neq_1_2.
∎
Theorem. (neq_2_4)
2 ≠ 4
In Proofgold the corresponding term root is 0bb647... and proposition id is c4e5df...
Proof:
Apply ordsucc_inj_contra to the current goal.
An exact proof term for the current goal is neq_1_3.
∎
Theorem. (neq_3_4)
3 ≠ 4
In Proofgold the corresponding term root is 031ec9... and proposition id is 66a378...
Proof:
Apply ordsucc_inj_contra to the current goal.
An exact proof term for the current goal is neq_2_3.
∎
Theorem. (ZF_closed_E)
∀U, ZF_closed U → ∀p : prop, (Union_closed U → Power_closed U → Repl_closed U → p) → p
In Proofgold the corresponding term root is d795d4... and proposition id is c78c79...
Proof:
An exact proof term for the current goal is (λU C p v ⇒ C p (λC H3 ⇒ C p (λH1 H2 ⇒ v H1 H2 H3))).
∎
Theorem. (ZF_Union_closed)
∀U, ZF_closed U → ∀X ∈ U, ⋃ X ∈ U
In Proofgold the corresponding term root is a7cdf1... and proposition id is 250496...
Proof:
An exact proof term for the current goal is (λU C ⇒ ZF_closed_E U C (∀X ∈ U, ⋃ X ∈ U) (λH _ _ ⇒ H)).
∎
Theorem. (ZF_Power_closed)
∀U, ZF_closed U → ∀X ∈ U, 𝒫 X ∈ U
In Proofgold the corresponding term root is 5d781f... and proposition id is 7411dd...
Proof:
An exact proof term for the current goal is (λU C ⇒ ZF_closed_E U C (∀X ∈ U, 𝒫 X ∈ U) (λ_ H _ ⇒ H)).
∎
Theorem. (ZF_Repl_closed)
∀U, ZF_closed U → ∀X ∈ U, ∀F : set → set, (∀x ∈ X, F x ∈ U) → {F x|x ∈ X} ∈ U
In Proofgold the corresponding term root is 312268... and proposition id is 18f8ac...
Proof:
An exact proof term for the current goal is (λU C ⇒ ZF_closed_E U C (∀X ∈ U, ∀F : set → set, (∀x ∈ X, F x ∈ U) → {F x|x ∈ X} ∈ U) (λ_ _ H ⇒ H)).
∎
Theorem. (ZF_UPair_closed)
∀U, ZF_closed U → ∀x y ∈ U, {x,y} ∈ U
In Proofgold the corresponding term root is 54256a... and proposition id is ef42d3...
Proof:
Let U be given.
Assume C: ZF_closed U.
Let x be given.
Assume Hx: x ∈ U.
Let y be given.
Assume Hy: y ∈ U.
We will prove {x,y} ∈ U.
We prove the intermediate claim L1: {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)} = {x,y}.
Apply set_ext to the current goal.
We will prove {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)} ⊆ {x,y}.
Let z be given.
Assume Hz: z ∈ {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)}.
We will prove z ∈ {x,y}.
Apply (ReplE_impred (𝒫 (𝒫 x)) (λX ⇒ if x ∈ X then x else y) z Hz) to the current goal.
Let X be given.
Assume _.
We will prove z = (if x ∈ X then x else y) → z ∈ {x,y}.
Apply (xm (x ∈ X)) to the current goal.
Assume H2: x ∈ X.
rewrite the current goal using (If_i_1 (x ∈ X) x y H2) (from left to right).
We will prove (z = x → z ∈ {x,y}).
Assume H3: z = x.
rewrite the current goal using H3 (from left to right).
An exact proof term for the current goal is (UPairI1 x y).
Assume H2: x ∉ X.
rewrite the current goal using (If_i_0 (x ∈ X) x y H2) (from left to right).
We will prove (z = y → z ∈ {x,y}).
Assume H3: z = y.
rewrite the current goal using H3 (from left to right).
An exact proof term for the current goal is (UPairI2 x y).
We will prove {x,y} ⊆ {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)}.
Let z be given.
Assume Hz: z ∈ {x,y}.
We will prove z ∈ {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)}.
We prove the intermediate claim L1a: (if x ∈ (𝒫 x) then x else y) ∈ {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)}.
Apply (ReplI (𝒫 (𝒫 x)) (λX ⇒ if x ∈ X then x else y) (𝒫 x)) to the current goal.
We will prove 𝒫 x ∈ 𝒫 (𝒫 x).
An exact proof term for the current goal is (Self_In_Power (𝒫 x)).
We prove the intermediate claim L1b: (if x ∈ Empty then x else y) ∈ {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)}.
Apply (ReplI (𝒫 (𝒫 x)) (λX ⇒ if x ∈ X then x else y) Empty) to the current goal.
We will prove Empty ∈ 𝒫 (𝒫 x).
An exact proof term for the current goal is (Empty_In_Power (𝒫 x)).
Apply (UPairE z x y Hz) to the current goal.
Assume H1: z = x.
rewrite the current goal using H1 (from left to right).
We will prove x ∈ {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)}.
rewrite the current goal using (If_i_1 (x ∈ (𝒫 x)) x y (Self_In_Power x)) (from right to left) at position 1.
An exact proof term for the current goal is L1a.
Assume H1: z = y.
rewrite the current goal using H1 (from left to right).
We will prove y ∈ {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)}.
rewrite the current goal using (If_i_0 (x ∈ Empty) x y (EmptyE x)) (from right to left) at position 1.
An exact proof term for the current goal is L1b.
We will prove {x,y} ∈ U.
rewrite the current goal using L1 (from right to left).
We will prove {if x ∈ X then x else y|X ∈ 𝒫 (𝒫 x)} ∈ U.
We prove the intermediate claim L2: 𝒫 (𝒫 x) ∈ U.
An exact proof term for the current goal is (ZF_Power_closed U C (𝒫 x) (ZF_Power_closed U C x Hx)).
We prove the intermediate claim L3: ∀X ∈ 𝒫 (𝒫 x), (if x ∈ X then x else y) ∈ U.
Let X be given.
Assume _.
We will prove (if x ∈ X then x else y) ∈ U.
Apply (xm (x ∈ X)) to the current goal.
Assume H1: x ∈ X.
rewrite the current goal using (If_i_1 (x ∈ X) x y H1) (from left to right).
We will prove x ∈ U.
An exact proof term for the current goal is Hx.
Assume H1: x ∉ X.
rewrite the current goal using (If_i_0 (x ∈ X) x y H1) (from left to right).
We will prove y ∈ U.
An exact proof term for the current goal is Hy.
An exact proof term for the current goal is (ZF_Repl_closed U C (𝒫 (𝒫 x)) L2 (λX ⇒ if x ∈ X then x else y) L3).
∎
Theorem. (ZF_Sing_closed)
∀U, ZF_closed U → ∀x ∈ U, {x} ∈ U
In Proofgold the corresponding term root is 5d8a83... and proposition id is d02d66...
Proof:
An exact proof term for the current goal is (λU C x H ⇒ ZF_UPair_closed U C x H x H).
∎
Theorem. (ZF_binunion_closed)
∀U, ZF_closed U → ∀X Y ∈ U, (X ∪ Y) ∈ U
In Proofgold the corresponding term root is 8a6240... and proposition id is 1bacfd...
Proof:
An exact proof term for the current goal is (λU C X HX Y HY ⇒ ZF_Union_closed U C {X,Y} (ZF_UPair_closed U C X HX Y HY)).
∎
Theorem. (ZF_ordsucc_closed)
∀U, ZF_closed U → ∀x ∈ U, ordsucc x ∈ U
In Proofgold the corresponding term root is 68a6ba... and proposition id is 946e61...
Proof:
An exact proof term for the current goal is (λU C x H ⇒ ZF_binunion_closed U C x H {x} (ZF_Sing_closed U C x H)).
∎
Theorem. (nat_p_UnivOf_Empty)
∀n : set, nat_p n → n ∈ UnivOf Empty
In Proofgold the corresponding term root is c885b4... and proposition id is 645aee...
Proof:
Apply nat_ind to the current goal.
We will prove 0 ∈ UnivOf Empty.
An exact proof term for the current goal is (UnivOf_In Empty).
Let n be given.
Assume Hn: nat_p n.
Assume IHn: n ∈ UnivOf Empty.
We will prove ordsucc n ∈ UnivOf Empty.
An exact proof term for the current goal is (ZF_ordsucc_closed (UnivOf Empty) (UnivOf_ZF_closed Empty) n IHn).
∎
Definition. We define ω to be {n ∈ UnivOf Empty|nat_p n} of type set.
In Proofgold the corresponding term root is 6fc30a... and object id is ad9ad8...
Theorem. (omega_nat_p)
∀n ∈ ω, nat_p n
In Proofgold the corresponding term root is f44a30... and proposition id is c2c61b...
Proof:
An exact proof term for the current goal is (λn H ⇒ SepE2 (UnivOf Empty) nat_p n H).
∎
Theorem. (nat_p_omega)
∀n : set, nat_p n → n ∈ ω
In Proofgold the corresponding term root is 6a1b44... and proposition id is d5080f...
Proof:
Let n be given.
Assume H: nat_p n.
We will prove n ∈ {n ∈ UnivOf Empty|nat_p n}.
Apply SepI to the current goal.
We will prove n ∈ UnivOf Empty.
An exact proof term for the current goal is (nat_p_UnivOf_Empty n H).
We will prove nat_p n.
An exact proof term for the current goal is H.
∎
Theorem. (omega_ordsucc)
∀n ∈ ω, ordsucc n ∈ ω
In Proofgold the corresponding term root is 246d63... and proposition id is 17fb64...
Proof:
Let n be given.
Assume Hn.
Apply nat_p_omega to the current goal.
Apply nat_ordsucc to the current goal.
Apply omega_nat_p to the current goal.
An exact proof term for the current goal is Hn.
∎
Definition. We define ordinal to be λalpha : set ⇒ TransSet alpha ∧ ∀beta ∈ alpha, TransSet beta of type set → prop.
In Proofgold the corresponding term root is ee28d9... and object id is 53ee8f...
Theorem. (ordinal_TransSet)
∀alpha : set, ordinal alpha → TransSet alpha
In Proofgold the corresponding term root is 0dbaa3... and proposition id is c9e708...
Proof:
An exact proof term for the current goal is (λalpha H ⇒ andEL (TransSet alpha) (∀beta ∈ alpha, TransSet beta) H).
∎
Theorem. (ordinal_Empty)
ordinal Empty
In Proofgold the corresponding term root is ce6bea... and proposition id is 7d95d1...
Proof:
We will prove TransSet Empty ∧ ∀x ∈ Empty, TransSet x.
Apply andI to the current goal.
Let x be given.
Assume Hx: x ∈ Empty.
We will prove False.
An exact proof term for the current goal is (EmptyE x Hx).
Let x be given.
Assume Hx: x ∈ Empty.
We will prove False.
An exact proof term for the current goal is (EmptyE x Hx).
∎
Theorem. (ordinal_Hered)
∀alpha : set, ordinal alpha → ∀beta ∈ alpha, ordinal beta
In Proofgold the corresponding term root is a7e78b... and proposition id is 43b0f3...
Proof:
Let alpha be given.
Assume H1: TransSet alpha ∧ ∀x ∈ alpha, TransSet x.
Let beta be given.
Assume H2: beta ∈ alpha.
We will prove TransSet beta ∧ ∀x ∈ beta, TransSet x.
Apply H1 to the current goal.
Assume H3: TransSet alpha.
Assume H4: ∀x ∈ alpha, TransSet x.
Apply andI to the current goal.
An exact proof term for the current goal is (H4 beta H2).
We will prove ∀x ∈ beta, TransSet x.
Let x be given.
Assume Hx: x ∈ beta.
We prove the intermediate claim L1: x ∈ alpha.
An exact proof term for the current goal is (H3 beta H2 x Hx).
We will prove TransSet x.
An exact proof term for the current goal is (H4 x L1).
∎
Theorem. (TransSet_ordsucc)
∀X : set, TransSet X → TransSet (ordsucc X)
In Proofgold the corresponding term root is fc2157... and proposition id is ebf9e3...
Proof:
Let X be given.
Assume H1: TransSet X.
Let x be given.
Assume H2: x ∈ ordsucc X.
Let y be given.
Assume H3: y ∈ x.
We will prove y ∈ ordsucc X.
Apply (ordsuccE X x H2) to the current goal.
Assume H4: x ∈ X.
Apply ordsuccI1 to the current goal.
We will prove y ∈ X.
An exact proof term for the current goal is (H1 x H4 y H3).
Assume H4: x = X.
Apply ordsuccI1 to the current goal.
We will prove y ∈ X.
rewrite the current goal using H4 (from right to left).
We will prove y ∈ x.
An exact proof term for the current goal is H3.
∎
Theorem. (ordinal_ordsucc)
∀alpha : set, ordinal alpha → ordinal (ordsucc alpha)
In Proofgold the corresponding term root is 0ba97f... and proposition id is ab0f7f...
Proof:
Let alpha be given.
Assume H1: TransSet alpha ∧ ∀beta ∈ alpha, TransSet beta.
Apply H1 to the current goal.
Assume H2: TransSet alpha.
Assume H3: ∀beta ∈ alpha, TransSet beta.
We will prove TransSet (ordsucc alpha) ∧ ∀beta ∈ ordsucc alpha, TransSet beta.
Apply andI to the current goal.
An exact proof term for the current goal is (TransSet_ordsucc alpha H2).
Let beta be given.
Assume H4: beta ∈ ordsucc alpha.
We will prove TransSet beta.
Apply (ordsuccE alpha beta H4) to the current goal.
Assume H5: beta ∈ alpha.
An exact proof term for the current goal is (H3 beta H5).
Assume H5: beta = alpha.
rewrite the current goal using H5 (from left to right).
An exact proof term for the current goal is H2.
∎
Theorem. (nat_p_ordinal)
∀n : set, nat_p n → ordinal n
In Proofgold the corresponding term root is 274fc4... and proposition id is 47cd49...
Proof:
Apply nat_ind to the current goal.
We will prove ordinal 0.
An exact proof term for the current goal is ordinal_Empty.
Let n be given.
Assume Hn: nat_p n.
Assume IHn: ordinal n.
We will prove ordinal (ordsucc n).
An exact proof term for the current goal is (ordinal_ordsucc n IHn).
∎
Theorem. (ordinal_1)
ordinal 1
In Proofgold the corresponding term root is c94d41... and proposition id is e325a5...
Proof:
An exact proof term for the current goal is nat_p_ordinal 1 nat_1.
∎
Theorem. (ordinal_2)
ordinal 2
In Proofgold the corresponding term root is a57075... and proposition id is e64a0d...
Proof:
An exact proof term for the current goal is nat_p_ordinal 2 nat_2.
∎
Theorem. (omega_TransSet)
TransSet ω
In Proofgold the corresponding term root is 8e89e6... and proposition id is c79e75...
Proof:
Let n be given.
Assume Hn: n ∈ ω.
Let m be given.
Assume Hm: m ∈ n.
We will prove m ∈ ω.
Apply nat_p_omega to the current goal.
We will prove nat_p m.
Apply (nat_p_trans n) to the current goal.
We will prove nat_p n.
An exact proof term for the current goal is (omega_nat_p n Hn).
We will prove m ∈ n.
An exact proof term for the current goal is Hm.
∎
Theorem. (omega_ordinal)
ordinal ω
In Proofgold the corresponding term root is 3d621a... and proposition id is 177484...
Proof:
We will prove TransSet ω ∧ ∀n ∈ ω, TransSet n.
Apply andI to the current goal.
An exact proof term for the current goal is omega_TransSet.
Let n be given.
Assume Hn: n ∈ ω.
We will prove TransSet n.
Apply ordinal_TransSet to the current goal.
We will prove ordinal n.
Apply nat_p_ordinal to the current goal.
We will prove nat_p n.
An exact proof term for the current goal is (omega_nat_p n Hn).
∎
Theorem. (ordsucc_omega_ordinal)
ordinal (ordsucc ω)
In Proofgold the corresponding term root is 66659c... and proposition id is ecfad7...
Proof:
An exact proof term for the current goal is ordinal_ordsucc ω omega_ordinal.
∎
Theorem. (TransSet_ordsucc_In_Subq)
∀X : set, TransSet X → ∀x ∈ X, ordsucc x ⊆ X
In Proofgold the corresponding term root is d5d882... and proposition id is 3c2e45...
Proof:
Let X be given.
Assume H1: TransSet X.
Let x be given.
Assume H2: x ∈ X.
Let y be given.
Assume H3: y ∈ ordsucc x.
We will prove y ∈ X.
Apply (ordsuccE x y H3) to the current goal.
Assume H4: y ∈ x.
An exact proof term for the current goal is (H1 x H2 y H4).
Assume H4: y = x.
rewrite the current goal using H4 (from left to right).
We will prove x ∈ X.
An exact proof term for the current goal is H2.
∎
Theorem. (ordinal_ordsucc_In_Subq)
∀alpha, ordinal alpha → ∀beta ∈ alpha, ordsucc beta ⊆ alpha
In Proofgold the corresponding term root is 57345e... and proposition id is b00304...
Proof:
Let alpha be given.
Assume H: ordinal alpha.
An exact proof term for the current goal is (TransSet_ordsucc_In_Subq alpha (ordinal_TransSet alpha H)).
∎
Theorem. (ordinal_trichotomy_or)
∀alpha beta : set, ordinal alpha → ordinal beta → alpha ∈ beta ∨ alpha = beta ∨ beta ∈ alpha
In Proofgold the corresponding term root is 078a34... and proposition id is 21f2d6...
Proof:
Apply In_ind to the current goal.
Let alpha be given.
Assume IHalpha: ∀gamma ∈ alpha, ∀beta : set, ordinal gamma → ordinal beta → gamma ∈ beta ∨ gamma = beta ∨ beta ∈ gamma.
We will prove ∀beta : set, ordinal alpha → ordinal beta → alpha ∈ beta ∨ alpha = beta ∨ beta ∈ alpha.
Apply In_ind to the current goal.
Let beta be given.
Assume IHbeta: ∀delta ∈ beta, ordinal alpha → ordinal delta → alpha ∈ delta ∨ alpha = delta ∨ delta ∈ alpha.
Assume Halpha: ordinal alpha.
Assume Hbeta: ordinal beta.
We will prove alpha ∈ beta ∨ alpha = beta ∨ beta ∈ alpha.
Apply (xm (alpha ∈ beta)) to the current goal.
Assume H1: alpha ∈ beta.
Apply or3I1 to the current goal.
An exact proof term for the current goal is H1.
Assume H1: alpha ∉ beta.
Apply (xm (beta ∈ alpha)) to the current goal.
Assume H2: beta ∈ alpha.
Apply or3I3 to the current goal.
An exact proof term for the current goal is H2.
Assume H2: beta ∉ alpha.
Apply or3I2 to the current goal.
We will prove alpha = beta.
Apply set_ext to the current goal.
We will prove alpha ⊆ beta.
Let gamma be given.
Assume H3: gamma ∈ alpha.
We will prove gamma ∈ beta.
We prove the intermediate claim Lgamma: ordinal gamma.
An exact proof term for the current goal is (ordinal_Hered alpha Halpha gamma H3).
Apply (or3E (gamma ∈ beta) (gamma = beta) (beta ∈ gamma) (IHalpha gamma H3 beta Lgamma Hbeta)) to the current goal.
Assume H4: gamma ∈ beta.
An exact proof term for the current goal is H4.
Assume H4: gamma = beta.
We will prove False.
Apply H2 to the current goal.
We will prove beta ∈ alpha.
rewrite the current goal using H4 (from right to left).
An exact proof term for the current goal is H3.
Assume H4: beta ∈ gamma.
We will prove False.
Apply H2 to the current goal.
We will prove beta ∈ alpha.
An exact proof term for the current goal is (ordinal_TransSet alpha Halpha gamma H3 beta H4).
We will prove beta ⊆ alpha.
Let delta be given.
Assume H3: delta ∈ beta.
We will prove delta ∈ alpha.
We prove the intermediate claim Ldelta: ordinal delta.
An exact proof term for the current goal is (ordinal_Hered beta Hbeta delta H3).
Apply (or3E (alpha ∈ delta) (alpha = delta) (delta ∈ alpha) (IHbeta delta H3 Halpha Ldelta)) to the current goal.
Assume H4: alpha ∈ delta.
We will prove False.
Apply H1 to the current goal.
We will prove alpha ∈ beta.
An exact proof term for the current goal is (ordinal_TransSet beta Hbeta delta H3 alpha H4).
Assume H4: alpha = delta.
We will prove False.
Apply H1 to the current goal.
We will prove alpha ∈ beta.
rewrite the current goal using H4 (from left to right).
An exact proof term for the current goal is H3.
Assume H4: delta ∈ alpha.
An exact proof term for the current goal is H4.
∎
Theorem. (ordinal_trichotomy_or_impred)
∀alpha beta : set, ordinal alpha → ordinal beta → ∀p : prop, (alpha ∈ beta → p) → (alpha = beta → p) → (beta ∈ alpha → p) → p
In Proofgold the corresponding term root is 7c0f0b... and proposition id is bb6d2d...
Proof:
Let alpha and beta be given.
Assume H1 H2.
An exact proof term for the current goal is (or3E (alpha ∈ beta) (alpha = beta) (beta ∈ alpha) (ordinal_trichotomy_or alpha beta H1 H2)).
∎
Theorem. (ordinal_In_Or_Subq)
∀alpha beta, ordinal alpha → ordinal beta → alpha ∈ beta ∨ beta ⊆ alpha
In Proofgold the corresponding term root is 70fe0d... and proposition id is 7b99b0...
Proof:
Let alpha and beta be given.
Assume H1: ordinal alpha.
Assume H2: ordinal beta.
Apply (or3E (alpha ∈ beta) (alpha = beta) (beta ∈ alpha) (ordinal_trichotomy_or alpha beta H1 H2)) to the current goal.
Assume H3: alpha ∈ beta.
Apply orIL to the current goal.
An exact proof term for the current goal is H3.
Assume H3: alpha = beta.
Apply orIR to the current goal.
rewrite the current goal using H3 (from left to right).
Apply Subq_ref to the current goal.
Assume H3: beta ∈ alpha.
Apply orIR to the current goal.
An exact proof term for the current goal is (ordinal_TransSet alpha H1 beta H3).
∎
Theorem. (ordinal_linear)
∀alpha beta, ordinal alpha → ordinal beta → alpha ⊆ beta ∨ beta ⊆ alpha
In Proofgold the corresponding term root is 3f30a5... and proposition id is 3b1fce...
Proof:
Let alpha and beta be given.
Assume H1: ordinal alpha.
Assume H2: ordinal beta.
Apply (ordinal_In_Or_Subq alpha beta H1 H2) to the current goal.
Assume H3: alpha ∈ beta.
Apply orIL to the current goal.
An exact proof term for the current goal is (ordinal_TransSet beta H2 alpha H3).
Assume H3: beta ⊆ alpha.
Apply orIR to the current goal.
An exact proof term for the current goal is H3.
∎
Theorem. (ordinal_ordsucc_In_eq)
∀alpha beta, ordinal alpha → beta ∈ alpha → ordsucc beta ∈ alpha ∨ alpha = ordsucc beta
In Proofgold the corresponding term root is 500378... and proposition id is 0c19e7...
Proof:
Let alpha and beta be given.
Assume Ha Hb.
We prove the intermediate claim L1: ordinal (ordsucc beta).
An exact proof term for the current goal is ordinal_ordsucc beta (ordinal_Hered alpha Ha beta Hb).
Apply ordinal_trichotomy_or alpha (ordsucc beta) Ha L1 to the current goal.
Assume H1.
Apply H1 to the current goal.
Assume H2: alpha ∈ ordsucc beta.
We will prove False.
Apply ordsuccE beta alpha H2 to the current goal.
Assume H3: alpha ∈ beta.
An exact proof term for the current goal is In_no2cycle alpha beta H3 Hb.
Assume H3: alpha = beta.
Apply In_irref alpha to the current goal.
rewrite the current goal using H3 (from left to right) at position 1.
An exact proof term for the current goal is Hb.
Assume H2: alpha = ordsucc beta.
Apply orIR to the current goal.
An exact proof term for the current goal is H2.
Assume H2: ordsucc beta ∈ alpha.
Apply orIL to the current goal.
An exact proof term for the current goal is H2.
∎
Theorem. (ordinal_lim_or_succ)
∀alpha, ordinal alpha → (∀beta ∈ alpha, ordsucc beta ∈ alpha) ∨ (∃beta ∈ alpha, alpha = ordsucc beta)
In Proofgold the corresponding term root is 4450c0... and proposition id is a12697...
Proof:
Let alpha be given.
Assume Ha.
Apply xm (∃beta ∈ alpha, alpha = ordsucc beta) to the current goal.
Assume H1.
Apply orIR to the current goal.
An exact proof term for the current goal is H1.
Assume H1: ¬ ∃beta ∈ alpha, alpha = ordsucc beta.
Apply orIL to the current goal.
Let beta be given.
Assume H2: beta ∈ alpha.
Apply ordinal_ordsucc_In_eq alpha beta Ha H2 to the current goal.
Assume H3: ordsucc beta ∈ alpha.
An exact proof term for the current goal is H3.
Assume H3: alpha = ordsucc beta.
We will prove False.
Apply H1 to the current goal.
We use beta to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is H2.
An exact proof term for the current goal is H3.
∎
Theorem. (ordinal_ordsucc_In)
∀alpha, ordinal alpha → ∀beta ∈ alpha, ordsucc beta ∈ ordsucc alpha
In Proofgold the corresponding term root is f4b1ed... and proposition id is 8ab1dc...
Proof:
Let alpha be given.
Assume Ha.
Let beta be given.
Assume Hb.
We prove the intermediate claim L1: ordsucc beta ⊆ alpha.
An exact proof term for the current goal is ordinal_ordsucc_In_Subq alpha Ha beta Hb.
Apply ordinal_In_Or_Subq (ordsucc beta) alpha (ordinal_ordsucc beta (ordinal_Hered alpha Ha beta Hb)) Ha to the current goal.
Assume H1: ordsucc beta ∈ alpha.
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is H1.
Assume H1: alpha ⊆ ordsucc beta.
We prove the intermediate claim L2: ordsucc beta = alpha.
Apply set_ext to the current goal.
An exact proof term for the current goal is L1.
An exact proof term for the current goal is H1.
rewrite the current goal using L2 (from left to right).
Apply ordsuccI2 to the current goal.
∎
Theorem. (ordinal_famunion)
∀X, ∀F : set → set, (∀x ∈ X, ordinal (F x)) → ordinal (⋃x ∈ XF x)
In Proofgold the corresponding term root is fc0359... and proposition id is b3f1e1...
Proof:
Let X and F be given.
Assume HXF.
We will prove TransSet (⋃x ∈ XF x) ∧ ∀y ∈ (⋃x ∈ XF x), TransSet y.
Apply andI to the current goal.
Let y be given.
Assume Hy: y ∈ ⋃x ∈ XF x.
Apply famunionE X F y Hy to the current goal.
Let x be given.
Assume H1.
Apply H1 to the current goal.
Assume Hx: x ∈ X.
Assume Hy: y ∈ F x.
We will prove y ⊆ ⋃x ∈ XF x.
We prove the intermediate claim LFx: ordinal (F x).
An exact proof term for the current goal is HXF x Hx.
Apply LFx to the current goal.
Assume HFx1 _.
Let z be given.
Assume Hz: z ∈ y.
We will prove z ∈ ⋃x ∈ XF x.
We prove the intermediate claim LzFx: z ∈ F x.
An exact proof term for the current goal is HFx1 y Hy z Hz.
An exact proof term for the current goal is famunionI X F x z Hx LzFx.
Let y be given.
Assume Hy: y ∈ ⋃x ∈ XF x.
We will prove TransSet y.
Apply famunionE X F y Hy to the current goal.
Let x be given.
Assume H1.
Apply H1 to the current goal.
Assume Hx: x ∈ X.
Assume Hy: y ∈ F x.
We prove the intermediate claim LFx: ordinal (F x).
An exact proof term for the current goal is HXF x Hx.
We prove the intermediate claim Ly: ordinal y.
An exact proof term for the current goal is ordinal_Hered (F x) LFx y Hy.
Apply Ly to the current goal.
Assume Hy1 _.
An exact proof term for the current goal is Hy1.
∎
Theorem. (ordinal_binintersect)
∀alpha beta, ordinal alpha → ordinal beta → ordinal (alpha ∩ beta)
In Proofgold the corresponding term root is 43d507... and proposition id is 025328...
Proof:
Let alpha and beta be given.
Assume Ha Hb.
Apply Ha to the current goal.
Assume Ha1 _.
Apply Hb to the current goal.
Assume Hb1 _.
Apply ordinal_In_Or_Subq alpha beta Ha Hb to the current goal.
Assume H1: alpha ∈ beta.
rewrite the current goal using binintersect_Subq_eq_1 alpha beta (Hb1 alpha H1) (from left to right).
An exact proof term for the current goal is Ha.
Assume H1: beta ⊆ alpha.
rewrite the current goal using binintersect_com (from left to right).
rewrite the current goal using binintersect_Subq_eq_1 beta alpha H1 (from left to right).
An exact proof term for the current goal is Hb.
∎
Theorem. (ordinal_binunion)
∀alpha beta, ordinal alpha → ordinal beta → ordinal (alpha ∪ beta)
In Proofgold the corresponding term root is 861ae3... and proposition id is b24989...
Proof:
Let alpha and beta be given.
Assume Ha Hb.
Apply Ha to the current goal.
Assume Ha1 _.
Apply Hb to the current goal.
Assume Hb1 _.
Apply ordinal_linear alpha beta Ha Hb to the current goal.
rewrite the current goal using Subq_binunion_eq (from left to right).
Assume H1: alpha ∪ beta = beta.
rewrite the current goal using H1 (from left to right).
An exact proof term for the current goal is Hb.
rewrite the current goal using Subq_binunion_eq (from left to right).
rewrite the current goal using binunion_com (from left to right).
Assume H1: alpha ∪ beta = alpha.
rewrite the current goal using H1 (from left to right).
An exact proof term for the current goal is Ha.
∎
Theorem. (ordinal_ind)
∀p : set → prop, (∀alpha, ordinal alpha → (∀beta ∈ alpha, p beta) → p alpha) → ∀alpha, ordinal alpha → p alpha
In Proofgold the corresponding term root is bb5575... and proposition id is 973cd8...
Proof:
Let p be given.
Assume H1: ∀alpha, ordinal alpha → (∀beta ∈ alpha, p beta) → p alpha.
Apply In_ind to the current goal.
Let alpha be given.
Assume IH: ∀beta ∈ alpha, ordinal beta → p beta.
Assume H2: ordinal alpha.
We will prove p alpha.
Apply (H1 alpha H2) to the current goal.
Let beta be given.
Assume H3: beta ∈ alpha.
We will prove p beta.
Apply (IH beta H3) to the current goal.
We will prove ordinal beta.
An exact proof term for the current goal is (ordinal_Hered alpha H2 beta H3).
∎
Theorem. (least_ordinal_ex)
∀p : set → prop, (∃alpha, ordinal alpha ∧ p alpha) → ∃alpha, ordinal alpha ∧ p alpha ∧ ∀beta ∈ alpha, ¬ p beta
In Proofgold the corresponding term root is df6c77... and proposition id is ab6fe1...
Proof:
Let p be given.
Assume H1.
Apply dneg to the current goal.
Assume H2: ¬ ∃alpha, ordinal alpha ∧ p alpha ∧ ∀beta ∈ alpha, ¬ p beta.
We prove the intermediate claim L1: ∀alpha, ordinal alpha → ¬ p alpha.
Apply ordinal_ind to the current goal.
Let alpha be given.
Assume Ha.
Assume IH: ∀beta ∈ alpha, ¬ p beta.
Assume H3: p alpha.
Apply H2 to the current goal.
We use alpha to witness the existential quantifier.
Apply and3I to the current goal.
An exact proof term for the current goal is Ha.
An exact proof term for the current goal is H3.
An exact proof term for the current goal is IH.
Apply H1 to the current goal.
Let alpha be given.
Assume H1a.
Apply H1a to the current goal.
Assume Ha: ordinal alpha.
Assume Hp: p alpha.
An exact proof term for the current goal is L1 alpha Ha Hp.
∎
Definition. We define inj to be λX Y f ⇒ (∀u ∈ X, f u ∈ Y) ∧ (∀u v ∈ X, f u = f v → u = v) of type set → set → (set → set) → prop.
In Proofgold the corresponding term root is 264b04... and object id is 442958...
Definition. We define bij to be λX Y f ⇒ (∀u ∈ X, f u ∈ Y) ∧ (∀u v ∈ X, f u = f v → u = v) ∧ (∀w ∈ Y, ∃u ∈ X, f u = w) of type set → set → (set → set) → prop.
In Proofgold the corresponding term root is 76bef6... and object id is f7d93b...
Theorem. (bijI)
∀X Y, ∀f : set → set, (∀u ∈ X, f u ∈ Y) → (∀u v ∈ X, f u = f v → u = v) → (∀w ∈ Y, ∃u ∈ X, f u = w) → bij X Y f
In Proofgold the corresponding term root is a9064f... and proposition id is 3229f9...
Proof:
Let X, Y and f be given.
Assume Hf1 Hf2 Hf3.
We will prove (∀u ∈ X, f u ∈ Y) ∧ (∀u v ∈ X, f u = f v → u = v) ∧ (∀w ∈ Y, ∃u ∈ X, f u = w).
Apply and3I to the current goal.
An exact proof term for the current goal is Hf1.
An exact proof term for the current goal is Hf2.
An exact proof term for the current goal is Hf3.
∎
Theorem. (bijE)
∀X Y, ∀f : set → set, bij X Y f → ∀p : prop, ((∀u ∈ X, f u ∈ Y) → (∀u v ∈ X, f u = f v → u = v) → (∀w ∈ Y, ∃u ∈ X, f u = w) → p) → p
In Proofgold the corresponding term root is 2cf278... and proposition id is cf0759...
Proof:
Let X, Y and f be given.
Assume Hf.
Let p be given.
Assume Hp.
Apply Hf to the current goal.
Assume Hf.
Apply Hf to the current goal.
Assume Hf1 Hf2 Hf3.
An exact proof term for the current goal is Hp Hf1 Hf2 Hf3.
∎
Definition. We define inv to be λX f ⇒ λy : set ⇒ Eps_i (λx ⇒ x ∈ X ∧ f x = y) of type set → (set → set) → set → set.
In Proofgold the corresponding term root is 896fa9... and object id is f3a015...
Theorem. (surj_rinv)
∀X Y, ∀f : set → set, (∀w ∈ Y, ∃u ∈ X, f u = w) → ∀y ∈ Y, inv X f y ∈ X ∧ f (inv X f y) = y
In Proofgold the corresponding term root is ebf371... and proposition id is 6489d2...
Proof:
Let X, Y and f be given.
Assume H1.
Let y be given.
Assume Hy: y ∈ Y.
Apply H1 y Hy to the current goal.
Let x be given.
Assume H2.
An exact proof term for the current goal is Eps_i_ax (λx ⇒ x ∈ X ∧ f x = y) x H2.
∎
Theorem. (inj_linv)
∀X, ∀f : set → set, (∀u v ∈ X, f u = f v → u = v) → ∀x ∈ X, inv X f (f x) = x
In Proofgold the corresponding term root is 18ecce... and proposition id is d16557...
Proof:
Let X and f be given.
Assume H1.
Let x be given.
Assume Hx.
We prove the intermediate claim L1: inv X f (f x) ∈ X ∧ f (inv X f (f x)) = f x.
Apply Eps_i_ax (λx' ⇒ x' ∈ X ∧ f x' = f x) x to the current goal.
Apply andI to the current goal.
An exact proof term for the current goal is Hx.
Use reflexivity.
Apply L1 to the current goal.
Assume H2 H3.
An exact proof term for the current goal is H1 (inv X f (f x)) H2 x Hx H3.
∎
Theorem. (bij_inv)
∀X Y, ∀f : set → set, bij X Y f → bij Y X (inv X f)
In Proofgold the corresponding term root is ff1332... and proposition id is 69c537...
Proof:
Let X, Y and f be given.
Assume H1.
Apply H1 to the current goal.
Assume H2.
Apply H2 to the current goal.
Assume H3: ∀u ∈ X, f u ∈ Y.
Assume H4: ∀u v ∈ X, f u = f v → u = v.
Assume H5: ∀w ∈ Y, ∃u ∈ X, f u = w.
Set g to be the term λy ⇒ Eps_i (λx ⇒ x ∈ X ∧ f x = y) of type set → set.
We prove the intermediate claim L1: ∀y ∈ Y, g y ∈ X ∧ f (g y) = y.
An exact proof term for the current goal is surj_rinv X Y f H5.
We will prove (∀u ∈ Y, g u ∈ X) ∧ (∀u v ∈ Y, g u = g v → u = v) ∧ (∀w ∈ X, ∃u ∈ Y, g u = w).
Apply and3I to the current goal.
We will prove ∀u ∈ Y, g u ∈ X.
Let u be given.
Assume Hu.
Apply L1 u Hu to the current goal.
Assume H _.
An exact proof term for the current goal is H.
We will prove ∀u v ∈ Y, g u = g v → u = v.
Let u be given.
Assume Hu.
Let v be given.
Assume Hv H6.
We will prove u = v.
Apply L1 u Hu to the current goal.
Assume H7: g u ∈ X.
Assume H8: f (g u) = u.
Apply L1 v Hv to the current goal.
Assume H9: g v ∈ X.
Assume H10: f (g v) = v.
rewrite the current goal using H8 (from right to left).
rewrite the current goal using H10 (from right to left).
rewrite the current goal using H6 (from right to left).
Use reflexivity.
We will prove ∀w ∈ X, ∃u ∈ Y, g u = w.
Let w be given.
Assume Hw.
We prove the intermediate claim Lfw: f w ∈ Y.
An exact proof term for the current goal is H3 w Hw.
We use f w to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is Lfw.
An exact proof term for the current goal is inj_linv X f H4 w Hw.
∎
Theorem. (bij_id)
∀X, bij X X (λx ⇒ x)
In Proofgold the corresponding term root is a8cefb... and proposition id is 707f37...
Proof:
Let X be given.
We will prove (∀u ∈ X, u ∈ X) ∧ (∀u v ∈ X, u = v → u = v) ∧ (∀w ∈ X, ∃u ∈ X, u = w).
Apply and3I to the current goal.
An exact proof term for the current goal is (λu Hu ⇒ Hu).
An exact proof term for the current goal is (λu Hu v Hv H1 ⇒ H1).
Let w be given.
Assume Hw.
We use w to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is Hw.
Use reflexivity.
∎
Theorem. (bij_comp)
∀X Y Z : set, ∀f g : set → set, bij X Y f → bij Y Z g → bij X Z (λx ⇒ g (f x))
In Proofgold the corresponding term root is abf851... and proposition id is 4c9045...
Proof:
Let X, Y, Z, f and g be given.
Assume Hf.
Apply Hf to the current goal.
Assume Hf12 Hf3.
Apply Hf12 to the current goal.
Assume Hf1 Hf2.
Assume Hg.
Apply Hg to the current goal.
Assume Hg12 Hg3.
Apply Hg12 to the current goal.
Assume Hg1 Hg2.
We will prove (∀u ∈ X, g (f u) ∈ Z) ∧ (∀u v ∈ X, g (f u) = g (f v) → u = v) ∧ (∀w ∈ Z, ∃u ∈ X, g (f u) = w).
Apply and3I to the current goal.
Let u be given.
Assume Hu: u ∈ X.
An exact proof term for the current goal is (Hg1 (f u) (Hf1 u Hu)).
Let u be given.
Assume Hu: u ∈ X.
Let v be given.
Assume Hv: v ∈ X.
Assume H1: g (f u) = g (f v).
We will prove u = v.
Apply Hf2 u Hu v Hv to the current goal.
We will prove f u = f v.
Apply Hg2 (f u) (Hf1 u Hu) (f v) (Hf1 v Hv) to the current goal.
We will prove g (f u) = g (f v).
An exact proof term for the current goal is H1.
Let w be given.
Assume Hw: w ∈ Z.
Apply Hg3 w Hw to the current goal.
Let y be given.
Assume Hy12.
Apply Hy12 to the current goal.
Assume Hy1: y ∈ Y.
Assume Hy2: g y = w.
Apply Hf3 y Hy1 to the current goal.
Let u be given.
Assume Hu12.
Apply Hu12 to the current goal.
Assume Hu1: u ∈ X.
Assume Hu2: f u = y.
We will prove ∃u ∈ X, g (f u) = w.
We use u to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is Hu1.
rewrite the current goal using Hu2 (from left to right).
An exact proof term for the current goal is Hy2.
∎
Definition. We define equip to be λX Y : set ⇒ ∃f : set → set, bij X Y f of type set → set → prop.
In Proofgold the corresponding term root is eb4419... and object id is 7f2bf6...
Theorem. (equip_ref)
∀X, equip X X
In Proofgold the corresponding term root is cc7e80... and proposition id is c41fb1...
Proof:
Let X be given.
We will prove ∃f : set → set, bij X X f.
We use (λx : set ⇒ x) to witness the existential quantifier.
An exact proof term for the current goal is bij_id X.
∎
Theorem. (equip_sym)
∀X Y, equip X Y → equip Y X
In Proofgold the corresponding term root is f28954... and proposition id is 272f04...
Proof:
Let X and Y be given.
Assume H1.
Apply H1 to the current goal.
Let f be given.
Assume H2: bij X Y f.
We will prove ∃g : set → set, bij Y X g.
We use (inv X f) to witness the existential quantifier.
We will prove bij Y X (inv X f).
An exact proof term for the current goal is bij_inv X Y f H2.
∎
Theorem. (equip_tra)
∀X Y Z, equip X Y → equip Y Z → equip X Z
In Proofgold the corresponding term root is 322bfb... and proposition id is d616c1...
Proof:
Let X, Y and Z be given.
Assume H1 H2.
Apply H1 to the current goal.
Let f be given.
Assume H3: bij X Y f.
Apply H2 to the current goal.
Let g be given.
Assume H4: bij Y Z g.
We will prove ∃h : set → set, bij X Z h.
We use (λx : set ⇒ g (f x)) to witness the existential quantifier.
An exact proof term for the current goal is bij_comp X Y Z f g H3 H4.
∎
Theorem. (equip_0_Empty)
∀X, equip X 0 → X = 0
In Proofgold the corresponding term root is 83ab2f... and proposition id is 4c101c...
Proof:
Let X be given.
Assume H1.
Apply Empty_eq to the current goal.
Let x be given.
Assume Hx.
We will prove False.
Apply H1 to the current goal.
Let f be given.
Assume Hf: bij X 0 f.
Apply bijE X 0 f Hf to the current goal.
Assume Hf1: ∀x ∈ X, f x ∈ 0.
We will prove False.
An exact proof term for the current goal is EmptyE (f x) (Hf1 x Hx).
∎
Beginning of Section SchroederBernstein
Theorem. (KnasterTarski_set)
∀A, ∀F : set → set, (∀U ∈ 𝒫 A, F U ∈ 𝒫 A) → (∀U V ∈ 𝒫 A, U ⊆ V → F U ⊆ F V) → ∃Y ∈ 𝒫 A, F Y = Y
In Proofgold the corresponding term root is 2ffb9f... and proposition id is 31ef9d...
Proof:
Let A and F be given.
Assume H1 H2.
Set Y to be the term {u ∈ A|∀X ∈ 𝒫 A, F X ⊆ X → u ∈ X} of type set.
We prove the intermediate claim L1: Y ∈ 𝒫 A.
Apply Sep_In_Power to the current goal.
We prove the intermediate claim L2: F Y ∈ 𝒫 A.
An exact proof term for the current goal is H1 Y L1.
We prove the intermediate claim L3: ∀X ∈ 𝒫 A, F X ⊆ X → Y ⊆ X.
Let X be given.
Assume HX: X ∈ 𝒫 A.
Assume H3: F X ⊆ X.
Let y be given.
Assume Hy: y ∈ Y.
An exact proof term for the current goal is SepE2 A (λu ⇒ ∀X ∈ 𝒫 A, F X ⊆ X → u ∈ X) y Hy X HX H3.
We prove the intermediate claim L4: F Y ⊆ Y.
Let u be given.
Assume H3: u ∈ F Y.
We will prove u ∈ Y.
Apply SepI to the current goal.
We will prove u ∈ A.
An exact proof term for the current goal is PowerE A (F Y) L2 u H3.
Let X be given.
Assume HX: X ∈ 𝒫 A.
Assume H4: F X ⊆ X.
We will prove u ∈ X.
We prove the intermediate claim L4a: Y ⊆ X.
An exact proof term for the current goal is L3 X HX H4.
We prove the intermediate claim L4b: F Y ⊆ F X.
An exact proof term for the current goal is H2 Y L1 X HX L4a.
We will prove u ∈ X.
Apply H4 to the current goal.
We will prove u ∈ F X.
Apply L4b to the current goal.
An exact proof term for the current goal is H3.
We prove the intermediate claim L5: F (F Y) ⊆ F Y.
An exact proof term for the current goal is H2 (F Y) L2 Y L1 L4.
We use Y to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is L1.
Apply set_ext to the current goal.
An exact proof term for the current goal is L4.
We will prove Y ⊆ F Y.
Apply L3 to the current goal.
An exact proof term for the current goal is L2.
An exact proof term for the current goal is L5.
∎
Theorem. (image_In_Power)
∀A B, ∀f : set → set, (∀x ∈ A, f x ∈ B) → ∀U ∈ 𝒫 A, {f x|x ∈ U} ∈ 𝒫 B
In Proofgold the corresponding term root is 6799e8... and proposition id is 43ed06...
Proof:
Let A, B and f be given.
Assume Hf.
Let U be given.
Assume HU.
Apply PowerI to the current goal.
Let y be given.
Assume Hy: y ∈ {f x|x ∈ U}.
Apply ReplE_impred U f y Hy to the current goal.
Let x be given.
Assume Hx: x ∈ U.
Assume H1: y = f x.
rewrite the current goal using H1 (from left to right).
Apply Hf to the current goal.
We will prove x ∈ A.
Apply PowerE A U HU to the current goal.
An exact proof term for the current goal is Hx.
∎
Theorem. (image_monotone)
∀f : set → set, ∀U V, U ⊆ V → {f x|x ∈ U} ⊆ {f x|x ∈ V}
In Proofgold the corresponding term root is ac6fe8... and proposition id is f923c6...
Proof:
Let f, U and V be given.
Assume HUV.
Let y be given.
Assume Hy: y ∈ {f x|x ∈ U}.
Apply ReplE_impred U f y Hy to the current goal.
Let x be given.
Assume Hx: x ∈ U.
Assume H1: y = f x.
rewrite the current goal using H1 (from left to right).
We will prove f x ∈ {f x|x ∈ V}.
Apply ReplI to the current goal.
Apply HUV to the current goal.
An exact proof term for the current goal is Hx.
∎
Theorem. (setminus_antimonotone)
∀A U V, U ⊆ V → A ∖ V ⊆ A ∖ U
In Proofgold the corresponding term root is 67bd0e... and proposition id is f9b735...
Proof:
Let A, U and V be given.
Assume HUV.
Let x be given.
Assume Hx.
Apply setminusE A V x Hx to the current goal.
Assume H1 H2.
Apply setminusI to the current goal.
An exact proof term for the current goal is H1.
Assume H3: x ∈ U.
Apply H2 to the current goal.
We will prove x ∈ V.
An exact proof term for the current goal is HUV x H3.
∎
Theorem. (SchroederBernstein)
∀A B, ∀f g : set → set, inj A B f → inj B A g → equip A B
In Proofgold the corresponding term root is e247a8... and proposition id is 03a2cd...
Proof:
Let A, B, f and g be given.
Assume Hf Hg.
Apply Hf to the current goal.
Assume Hf1 Hf2.
Apply Hg to the current goal.
Assume Hg1 Hg2.
Set F to be the term λX ⇒ {g y|y ∈ B ∖ {f x|x ∈ A ∖ X}} of type set → set.
We prove the intermediate claim L1: ∀U ∈ 𝒫 A, F U ∈ 𝒫 A.
Let U be given.
Assume HU.
We will prove {g y|y ∈ B ∖ {f x|x ∈ A ∖ U}} ∈ 𝒫 A.
Apply image_In_Power B A g Hg1 to the current goal.
We will prove B ∖ {f x|x ∈ A ∖ U} ∈ 𝒫 B.
Apply setminus_In_Power to the current goal.
We prove the intermediate claim L2: ∀U V ∈ 𝒫 A, U ⊆ V → F U ⊆ F V.
Let U be given.
Assume HU.
Let V be given.
Assume HV HUV.
We will prove {g y|y ∈ B ∖ {f x|x ∈ A ∖ U}} ⊆ {g y|y ∈ B ∖ {f x|x ∈ A ∖ V}}.
Apply image_monotone g to the current goal.
We will prove B ∖ {f x|x ∈ A ∖ U} ⊆ B ∖ {f x|x ∈ A ∖ V}.
Apply setminus_antimonotone to the current goal.
We will prove {f x|x ∈ A ∖ V} ⊆ {f x|x ∈ A ∖ U}.
Apply image_monotone f to the current goal.
We will prove A ∖ V ⊆ A ∖ U.
Apply setminus_antimonotone to the current goal.
An exact proof term for the current goal is HUV.
Apply KnasterTarski_set A F L1 L2 to the current goal.
Let Y be given.
Assume H1.
Apply H1 to the current goal.
Assume H1: Y ∈ 𝒫 A.
Assume H2: F Y = Y.
Set h to be the term λx ⇒ if x ∈ Y then inv B g x else f x of type set → set.
We will prove ∃f : set → set, bij A B f.
We use h to witness the existential quantifier.
Apply bijI to the current goal.
Let x be given.
Assume Hx.
We will prove (if x ∈ Y then inv B g x else f x) ∈ B.
Apply xm (x ∈ Y) to the current goal.
Assume H3: x ∈ Y.
rewrite the current goal using If_i_1 (x ∈ Y) (inv B g x) (f x) H3 (from left to right).
We will prove inv B g x ∈ B.
We prove the intermediate claim L1: x ∈ F Y.
rewrite the current goal using H2 (from left to right).
An exact proof term for the current goal is H3.
Apply ReplE_impred (B ∖ {f x|x ∈ A ∖ Y}) g x L1 to the current goal.
Let y be given.
Assume H4: y ∈ B ∖ {f x|x ∈ A ∖ Y}.
Assume H5: x = g y.
We prove the intermediate claim L2: y ∈ B.
An exact proof term for the current goal is setminusE1 B {f x|x ∈ A ∖ Y} y H4.
rewrite the current goal using H5 (from left to right).
We will prove inv B g (g y) ∈ B.
rewrite the current goal using inj_linv B g Hg2 y L2 (from left to right).
We will prove y ∈ B.
An exact proof term for the current goal is L2.
Assume H3: x ∉ Y.
rewrite the current goal using If_i_0 (x ∈ Y) (inv B g x) (f x) H3 (from left to right).
We will prove f x ∈ B.
Apply Hf1 to the current goal.
An exact proof term for the current goal is Hx.
Let x be given.
Assume Hx.
Let y be given.
Assume Hy.
We will prove (if x ∈ Y then inv B g x else f x) = (if y ∈ Y then inv B g y else f y) → x = y.
Apply xm (x ∈ Y) to the current goal.
Assume H3: x ∈ Y.
rewrite the current goal using If_i_1 (x ∈ Y) (inv B g x) (f x) H3 (from left to right).
We will prove inv B g x = (if y ∈ Y then inv B g y else f y) → x = y.
We prove the intermediate claim Lx: x ∈ F Y.
rewrite the current goal using H2 (from left to right).
An exact proof term for the current goal is H3.
Apply ReplE_impred (B ∖ {f x|x ∈ A ∖ Y}) g x Lx to the current goal.
Let z be given.
Assume Hz1: z ∈ B ∖ {f x|x ∈ A ∖ Y}.
Assume Hz2: x = g z.
Apply setminusE B {f x|x ∈ A ∖ Y} z Hz1 to the current goal.
Assume Hz1a Hz1b.
Apply xm (y ∈ Y) to the current goal.
Assume H4: y ∈ Y.
rewrite the current goal using If_i_1 (y ∈ Y) (inv B g y) (f y) H4 (from left to right).
We will prove inv B g x = inv B g y → x = y.
We prove the intermediate claim Ly: y ∈ F Y.
rewrite the current goal using H2 (from left to right).
An exact proof term for the current goal is H4.
Apply ReplE_impred (B ∖ {f x|x ∈ A ∖ Y}) g y Ly to the current goal.
Let w be given.
Assume Hw1: w ∈ B ∖ {f x|x ∈ A ∖ Y}.
Assume Hw2: y = g w.
rewrite the current goal using Hz2 (from left to right).
rewrite the current goal using Hw2 (from left to right).
rewrite the current goal using inj_linv B g Hg2 z Hz1a (from left to right).
rewrite the current goal using inj_linv B g Hg2 w (setminusE1 B {f x|x ∈ A ∖ Y} w Hw1) (from left to right).
Assume H5: z = w.
We will prove g z = g w.
Use f_equal.
An exact proof term for the current goal is H5.
Assume H4: y ∉ Y.
rewrite the current goal using If_i_0 (y ∈ Y) (inv B g y) (f y) H4 (from left to right).
We will prove inv B g x = f y → x = y.
rewrite the current goal using Hz2 (from left to right).
rewrite the current goal using inj_linv B g Hg2 z Hz1a (from left to right).
We will prove z = f y → g z = y.
Assume H5: z = f y.
We will prove False.
Apply Hz1b to the current goal.
We will prove z ∈ {f x|x ∈ A ∖ Y}.
rewrite the current goal using H5 (from left to right).
Apply ReplI to the current goal.
We will prove y ∈ A ∖ Y.
Apply setminusI to the current goal.
An exact proof term for the current goal is Hy.
An exact proof term for the current goal is H4.
Assume H3: x ∉ Y.
rewrite the current goal using If_i_0 (x ∈ Y) (inv B g x) (f x) H3 (from left to right).
We will prove f x = (if y ∈ Y then inv B g y else f y) → x = y.
Apply xm (y ∈ Y) to the current goal.
Assume H4: y ∈ Y.
rewrite the current goal using If_i_1 (y ∈ Y) (inv B g y) (f y) H4 (from left to right).
We will prove f x = inv B g y → x = y.
We prove the intermediate claim Ly: y ∈ F Y.
rewrite the current goal using H2 (from left to right).
An exact proof term for the current goal is H4.
Apply ReplE_impred (B ∖ {f x|x ∈ A ∖ Y}) g y Ly to the current goal.
Let w be given.
Assume Hw1: w ∈ B ∖ {f x|x ∈ A ∖ Y}.
Assume Hw2: y = g w.
Apply setminusE B {f x|x ∈ A ∖ Y} w Hw1 to the current goal.
Assume Hw1a Hw1b.
rewrite the current goal using Hw2 (from left to right).
rewrite the current goal using inj_linv B g Hg2 w Hw1a (from left to right).
Assume H5: f x = w.
We will prove False.
Apply Hw1b to the current goal.
We will prove w ∈ {f x|x ∈ A ∖ Y}.
rewrite the current goal using H5 (from right to left).
Apply ReplI to the current goal.
Apply setminusI to the current goal.
An exact proof term for the current goal is Hx.
An exact proof term for the current goal is H3.
Assume H4: y ∉ Y.
rewrite the current goal using If_i_0 (y ∈ Y) (inv B g y) (f y) H4 (from left to right).
We will prove f x = f y → x = y.
An exact proof term for the current goal is Hf2 x Hx y Hy.
Let w be given.
Assume Hw: w ∈ B.
Apply xm (w ∈ {f x|x ∈ A ∖ Y}) to the current goal.
Assume H3: w ∈ {f x|x ∈ A ∖ Y}.
Apply ReplE_impred (A ∖ Y) f w H3 to the current goal.
Let x be given.
Assume H4: x ∈ A ∖ Y.
Assume H5: w = f x.
Apply setminusE A Y x H4 to the current goal.
Assume H6: x ∈ A.
Assume H7: x ∉ Y.
We use x to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is H6.
We will prove (if x ∈ Y then inv B g x else f x) = w.
rewrite the current goal using If_i_0 (x ∈ Y) (inv B g x) (f x) H7 (from left to right).
Use symmetry.
An exact proof term for the current goal is H5.
Assume H3: w ∉ {f x|x ∈ A ∖ Y}.
We prove the intermediate claim Lgw: g w ∈ Y.
rewrite the current goal using H2 (from right to left).
We will prove g w ∈ F Y.
We will prove g w ∈ {g y|y ∈ B ∖ {f x|x ∈ A ∖ Y}}.
Apply ReplI to the current goal.
Apply setminusI to the current goal.
We will prove w ∈ B.
An exact proof term for the current goal is Hw.
We will prove w ∉ {f x|x ∈ A ∖ Y}.
An exact proof term for the current goal is H3.
We use (g w) to witness the existential quantifier.
Apply andI to the current goal.
We will prove g w ∈ A.
Apply Hg1 to the current goal.
An exact proof term for the current goal is Hw.
We will prove (if g w ∈ Y then inv B g (g w) else f (g w)) = w.
rewrite the current goal using If_i_1 (g w ∈ Y) (inv B g (g w)) (f (g w)) Lgw (from left to right).
An exact proof term for the current goal is inj_linv B g Hg2 w Hw.
∎
End of Section SchroederBernstein
Beginning of Section PigeonHole
Theorem. (PigeonHole_nat)
∀n, nat_p n → ∀f : set → set, (∀i ∈ ordsucc n, f i ∈ n) → ¬ (∀i j ∈ ordsucc n, f i = f j → i = j)
In Proofgold the corresponding term root is d756b7... and proposition id is 604a84...
Proof:
Apply nat_ind (λn ⇒ ∀f : set → set, (∀i ∈ ordsucc n, f i ∈ n) → ¬ (∀i j ∈ ordsucc n, f i = f j → i = j)) to the current goal.
We will prove ∀f : set → set, (∀i ∈ 1, f i ∈ 0) → ¬ (∀i j ∈ 1, f i = f j → i = j).
Let f be given.
Assume H1.
Apply EmptyE (f 0) (H1 0 In_0_1) to the current goal.
Let n be given.
Assume Hn: nat_p n.
Assume IH: ∀f : set → set, (∀i ∈ ordsucc n, f i ∈ n) → ¬ (∀i j ∈ ordsucc n, f i = f j → i = j).
Let f be given.
Assume H1: ∀i ∈ ordsucc (ordsucc n), f i ∈ ordsucc n.
Assume H2: ∀i j ∈ ordsucc (ordsucc n), f i = f j → i = j.
Apply xm (∃i ∈ ordsucc (ordsucc n), f i = n) to the current goal.
Assume H3.
Apply H3 to the current goal.
Let k be given.
Assume Hk.
Apply Hk to the current goal.
Assume Hk1: k ∈ ordsucc (ordsucc n).
Assume Hk2: f k = n.
Set f' to be the term λi : set ⇒ if k ⊆ i then f (ordsucc i) else f i.
Apply IH f' to the current goal.
We will prove ∀i ∈ ordsucc n, f' i ∈ n.
Let i be given.
Assume Hi: i ∈ ordsucc n.
Apply xm (k ⊆ i) to the current goal.
Assume H4: k ⊆ i.
We will prove (if k ⊆ i then f (ordsucc i) else f i) ∈ n.
rewrite the current goal using If_i_1 (k ⊆ i) (f (ordsucc i)) (f i) H4 (from left to right).
We will prove f (ordsucc i) ∈ n.
We prove the intermediate claim L1: ordsucc i ∈ ordsucc (ordsucc n).
Apply nat_ordsucc_in_ordsucc to the current goal.
Apply nat_ordsucc to the current goal.
An exact proof term for the current goal is Hn.
An exact proof term for the current goal is Hi.
Apply ordsuccE n (f (ordsucc i)) (H1 (ordsucc i) L1) to the current goal.
Assume H5: f (ordsucc i) ∈ n.
An exact proof term for the current goal is H5.
Assume H5: f (ordsucc i) = n.
We will prove False.
Apply In_irref i to the current goal.
We will prove i ∈ i.
We prove the intermediate claim L2: k = ordsucc i.
Apply H2 to the current goal.
An exact proof term for the current goal is Hk1.
An exact proof term for the current goal is L1.
rewrite the current goal using H5 (from left to right).
An exact proof term for the current goal is Hk2.
We prove the intermediate claim L3: i ∈ k.
rewrite the current goal using L2 (from left to right).
Apply ordsuccI2 to the current goal.
An exact proof term for the current goal is H4 i L3.
Assume H4: ¬ (k ⊆ i).
We will prove (if k ⊆ i then f (ordsucc i) else f i) ∈ n.
rewrite the current goal using If_i_0 (k ⊆ i) (f (ordsucc i)) (f i) H4 (from left to right).
We will prove f i ∈ n.
Apply ordsuccE n (f i) (H1 i (ordsuccI1 (ordsucc n) i Hi)) to the current goal.
Assume H5: f i ∈ n.
An exact proof term for the current goal is H5.
Assume H5: f i = n.
We will prove False.
Apply H4 to the current goal.
We will prove k ⊆ i.
We prove the intermediate claim L2: k = i.
Apply H2 to the current goal.
An exact proof term for the current goal is Hk1.
An exact proof term for the current goal is ordsuccI1 (ordsucc n) i Hi.
rewrite the current goal using H5 (from left to right).
An exact proof term for the current goal is Hk2.
rewrite the current goal using L2 (from right to left).
An exact proof term for the current goal is (λx Hx ⇒ Hx).
We will prove ∀i j ∈ ordsucc n, f' i = f' j → i = j.
Let i be given.
Assume Hi.
Let j be given.
Assume Hj.
We will prove (if k ⊆ i then f (ordsucc i) else f i) = (if k ⊆ j then f (ordsucc j) else f j) → i = j.
We prove the intermediate claim Li1: i ∈ ordsucc (ordsucc n).
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hi.
We prove the intermediate claim Li2: ordsucc i ∈ ordsucc (ordsucc n).
Apply nat_ordsucc_in_ordsucc to the current goal.
Apply nat_ordsucc to the current goal.
An exact proof term for the current goal is Hn.
An exact proof term for the current goal is Hi.
We prove the intermediate claim Lj1: j ∈ ordsucc (ordsucc n).
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hj.
We prove the intermediate claim Lj2: ordsucc j ∈ ordsucc (ordsucc n).
Apply nat_ordsucc_in_ordsucc to the current goal.
Apply nat_ordsucc to the current goal.
An exact proof term for the current goal is Hn.
An exact proof term for the current goal is Hj.
Apply xm (k ⊆ i) to the current goal.
Assume H4: k ⊆ i.
rewrite the current goal using If_i_1 (k ⊆ i) (f (ordsucc i)) (f i) H4 (from left to right).
Apply xm (k ⊆ j) to the current goal.
Assume H5: k ⊆ j.
rewrite the current goal using If_i_1 (k ⊆ j) (f (ordsucc j)) (f j) H5 (from left to right).
We will prove f (ordsucc i) = f (ordsucc j) → i = j.
Assume H6.
Apply ordsucc_inj to the current goal.
We will prove ordsucc i = ordsucc j.
An exact proof term for the current goal is H2 (ordsucc i) Li2 (ordsucc j) Lj2 H6.
Assume H5: ¬ (k ⊆ j).
rewrite the current goal using If_i_0 (k ⊆ j) (f (ordsucc j)) (f j) H5 (from left to right).
We will prove f (ordsucc i) = f j → i = j.
Assume H6.
We will prove False.
We prove the intermediate claim L3: ordsucc i = j.
Apply H2 to the current goal.
An exact proof term for the current goal is Li2.
An exact proof term for the current goal is Lj1.
An exact proof term for the current goal is H6.
Apply H5 to the current goal.
rewrite the current goal using L3 (from right to left).
Let x be given.
Assume Hx: x ∈ k.
Apply ordsuccI1 to the current goal.
Apply H4 to the current goal.
An exact proof term for the current goal is Hx.
Assume H4: ¬ (k ⊆ i).
rewrite the current goal using If_i_0 (k ⊆ i) (f (ordsucc i)) (f i) H4 (from left to right).
Apply xm (k ⊆ j) to the current goal.
Assume H5: k ⊆ j.
rewrite the current goal using If_i_1 (k ⊆ j) (f (ordsucc j)) (f j) H5 (from left to right).
We will prove f i = f (ordsucc j) → i = j.
Assume H6.
We will prove False.
We prove the intermediate claim L3: i = ordsucc j.
Apply H2 to the current goal.
An exact proof term for the current goal is Li1.
An exact proof term for the current goal is Lj2.
An exact proof term for the current goal is H6.
Apply H4 to the current goal.
rewrite the current goal using L3 (from left to right).
Let x be given.
Assume Hx: x ∈ k.
Apply ordsuccI1 to the current goal.
Apply H5 to the current goal.
An exact proof term for the current goal is Hx.
Assume H5: ¬ (k ⊆ j).
rewrite the current goal using If_i_0 (k ⊆ j) (f (ordsucc j)) (f j) H5 (from left to right).
We will prove f i = f j → i = j.
Apply H2 to the current goal.
An exact proof term for the current goal is Li1.
An exact proof term for the current goal is Lj1.
Assume H3: ¬ (∃i ∈ ordsucc (ordsucc n), f i = n).
Apply IH f to the current goal.
We will prove ∀i ∈ ordsucc n, f i ∈ n.
Let i be given.
Assume Hi: i ∈ ordsucc n.
Apply ordsuccE n (f i) (H1 i (ordsuccI1 (ordsucc n) i Hi)) to the current goal.
Assume Hfi: f i ∈ n.
An exact proof term for the current goal is Hfi.
Assume Hfi: f i = n.
Apply H3 to the current goal.
We use i to witness the existential quantifier.
Apply andI to the current goal.
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hi.
An exact proof term for the current goal is Hfi.
We will prove ∀i j ∈ ordsucc n, f i = f j → i = j.
Let i be given.
Assume Hi.
Let j be given.
Assume Hj.
Apply H2 to the current goal.
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hi.
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hj.
∎
Theorem. (PigeonHole_nat_bij)
∀n, nat_p n → ∀f : set → set, (∀i ∈ n, f i ∈ n) → (∀i j ∈ n, f i = f j → i = j) → bij n n f
In Proofgold the corresponding term root is cd3b8d... and proposition id is 44a08f...
Proof:
Let n be given.
Assume Hn.
Let f be given.
Assume Hf1 Hf2.
We will prove (∀i ∈ n, f i ∈ n) ∧ (∀i j ∈ n, f i = f j → i = j) ∧ (∀j ∈ n, ∃i ∈ n, f i = j).
Apply and3I to the current goal.
An exact proof term for the current goal is Hf1.
An exact proof term for the current goal is Hf2.
Let j be given.
Assume Hj: j ∈ n.
Apply dneg to the current goal.
Assume H1: ¬ ∃i ∈ n, f i = j.
Set f' to be the term λi : set ⇒ if i = n then j else f i.
Apply PigeonHole_nat n Hn f' to the current goal.
We will prove ∀i ∈ ordsucc n, f' i ∈ n.
Let i be given.
Assume Hi.
We will prove (if i = n then j else f i) ∈ n.
Apply xm (i = n) to the current goal.
Assume H1: i = n.
rewrite the current goal using If_i_1 (i = n) j (f i) H1 (from left to right).
We will prove j ∈ n.
An exact proof term for the current goal is Hj.
Assume H1: i ≠ n.
rewrite the current goal using If_i_0 (i = n) j (f i) H1 (from left to right).
We will prove f i ∈ n.
Apply Hf1 to the current goal.
Apply ordsuccE n i Hi to the current goal.
Assume H2: i ∈ n.
An exact proof term for the current goal is H2.
Assume H2: i = n.
Apply H1 H2 to the current goal.
We will prove ∀i k ∈ ordsucc n, f' i = f' k → i = k.
Let i be given.
Assume Hi.
Let k be given.
Assume Hk.
We prove the intermediate claim Li: i ≠ n → i ∈ n.
Assume Hin: i ≠ n.
Apply ordsuccE n i Hi to the current goal.
Assume H2.
An exact proof term for the current goal is H2.
Assume H2.
Apply Hin H2 to the current goal.
We prove the intermediate claim Lk: k ≠ n → k ∈ n.
Assume Hkn: k ≠ n.
Apply ordsuccE n k Hk to the current goal.
Assume H2.
An exact proof term for the current goal is H2.
Assume H2.
Apply Hkn H2 to the current goal.
We will prove (if i = n then j else f i) = (if k = n then j else f k) → i = k.
Apply xm (i = n) to the current goal.
Assume H2: i = n.
Apply xm (k = n) to the current goal.
Assume H3: k = n.
Assume _.
rewrite the current goal using H3 (from left to right).
An exact proof term for the current goal is H2.
Assume H3: k ≠ n.
rewrite the current goal using If_i_1 (i = n) j (f i) H2 (from left to right).
rewrite the current goal using If_i_0 (k = n) j (f k) H3 (from left to right).
Assume H4: j = f k.
Apply H1 to the current goal.
We use k to witness the existential quantifier.
Apply andI to the current goal.
We will prove k ∈ n.
An exact proof term for the current goal is Lk H3.
Use symmetry.
An exact proof term for the current goal is H4.
Assume H2: i ≠ n.
Apply xm (k = n) to the current goal.
Assume H3: k = n.
rewrite the current goal using If_i_0 (i = n) j (f i) H2 (from left to right).
rewrite the current goal using If_i_1 (k = n) j (f k) H3 (from left to right).
Assume H4: f i = j.
Apply H1 to the current goal.
We use i to witness the existential quantifier.
Apply andI to the current goal.
We will prove i ∈ n.
An exact proof term for the current goal is Li H2.
An exact proof term for the current goal is H4.
Assume H3: k ≠ n.
rewrite the current goal using If_i_0 (i = n) j (f i) H2 (from left to right).
rewrite the current goal using If_i_0 (k = n) j (f k) H3 (from left to right).
We will prove f i = f k → i = k.
Apply Hf2 to the current goal.
We will prove i ∈ n.
An exact proof term for the current goal is Li H2.
We will prove k ∈ n.
An exact proof term for the current goal is Lk H3.
∎
End of Section PigeonHole
Definition. We define finite to be λX ⇒ ∃n ∈ ω, equip X n of type set → prop.
In Proofgold the corresponding term root is 04632d... and object id is 4c54a5...
Theorem. (finite_ind)
∀p : set → prop, p Empty → (∀X y, finite X → y ∉ X → p X → p (X ∪ {y})) → ∀X, finite X → p X
In Proofgold the corresponding term root is c53cb1... and proposition id is 19baca...
Proof:
Let p be given.
Assume H1 H2.
We prove the intermediate claim L1: ∀n, nat_p n → ∀X, equip X n → p X.
Apply nat_ind to the current goal.
Let X be given.
Assume H3: equip X 0.
rewrite the current goal using equip_0_Empty X H3 (from left to right).
An exact proof term for the current goal is H1.
Let n be given.
Assume Hn.
Assume IHn: ∀X, equip X n → p X.
Let X be given.
Assume H3: equip X (ordsucc n).
Apply equip_sym X (ordsucc n) H3 to the current goal.
Let f be given.
Assume Hf: bij (ordsucc n) X f.
Apply bijE (ordsucc n) X f Hf to the current goal.
Assume Hf1: ∀i ∈ ordsucc n, f i ∈ X.
Assume Hf2: ∀i j ∈ ordsucc n, f i = f j → i = j.
Assume Hf3: ∀x ∈ X, ∃i ∈ ordsucc n, f i = x.
Set Z to be the term {f i|i ∈ n}.
Set y to be the term f n.
We prove the intermediate claim L1a: X = Z ∪ {y}.
Apply set_ext to the current goal.
Let x be given.
Assume Hx: x ∈ X.
Apply Hf3 x Hx to the current goal.
Let i be given.
Assume H.
Apply H to the current goal.
Assume Hi: i ∈ ordsucc n.
Assume H4: f i = x.
Apply ordsuccE n i Hi to the current goal.
Assume H5: i ∈ n.
Apply binunionI1 to the current goal.
We will prove x ∈ Z.
rewrite the current goal using H4 (from right to left).
We will prove f i ∈ Z.
Apply ReplI to the current goal.
An exact proof term for the current goal is H5.
Assume H5: i = n.
Apply binunionI2 to the current goal.
We will prove x ∈ {y}.
rewrite the current goal using H4 (from right to left).
rewrite the current goal using H5 (from left to right).
An exact proof term for the current goal is SingI (f n).
Let x be given.
Assume Hx: x ∈ Z ∪ {y}.
Apply binunionE Z {y} x Hx to the current goal.
Assume H4: x ∈ Z.
Apply ReplE_impred n f x H4 to the current goal.
Let i be given.
Assume Hi: i ∈ n.
Assume H5: x = f i.
We will prove x ∈ X.
rewrite the current goal using H5 (from left to right).
We will prove f i ∈ X.
An exact proof term for the current goal is Hf1 i (ordsuccI1 n i Hi).
Assume H4: x ∈ {y}.
rewrite the current goal using SingE y x H4 (from left to right).
We will prove f n ∈ X.
An exact proof term for the current goal is Hf1 n (ordsuccI2 n).
We prove the intermediate claim L1b: equip Z n.
Apply equip_sym to the current goal.
We will prove ∃f : set → set, bij n Z f.
We use f to witness the existential quantifier.
Apply bijI to the current goal.
Let i be given.
Assume Hi.
We will prove f i ∈ {f i|i ∈ n}.
Apply ReplI to the current goal.
An exact proof term for the current goal is Hi.
Let i be given.
Assume Hi.
Let j be given.
Assume Hj.
Apply Hf2 to the current goal.
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hi.
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hj.
Apply ReplE' to the current goal.
Let i be given.
Assume Hi: i ∈ n.
We will prove ∃i' ∈ n, f i' = f i.
We use i to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is Hi.
Use reflexivity.
rewrite the current goal using L1a (from left to right).
We will prove p (Z ∪ {y}).
Apply H2 Z y to the current goal.
We will prove finite Z.
We will prove ∃n ∈ ω, equip Z n.
We use n to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is nat_p_omega n Hn.
An exact proof term for the current goal is L1b.
We will prove y ∉ Z.
Assume H4: y ∈ Z.
Apply ReplE_impred n f y H4 to the current goal.
Let i be given.
Assume Hi: i ∈ n.
Assume H5: y = f i.
Apply In_irref n to the current goal.
We will prove n ∈ n.
rewrite the current goal using Hf2 n (ordsuccI2 n) i (ordsuccI1 n i Hi) H5 (from left to right) at position 1.
We will prove i ∈ n.
An exact proof term for the current goal is Hi.
We will prove p Z.
An exact proof term for the current goal is IHn Z L1b.
Let X be given.
Assume H3.
Apply H3 to the current goal.
Let n be given.
Assume H.
Apply H to the current goal.
Assume Hn: n ∈ ω.
Assume H4: equip X n.
An exact proof term for the current goal is L1 n (omega_nat_p n Hn) X H4.
∎
Theorem. (finite_Empty)
finite 0
In Proofgold the corresponding term root is c073c6... and proposition id is ec2c27...
Proof:
We will prove ∃n ∈ ω, equip 0 n.
We use 0 to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is nat_p_omega 0 nat_0.
Apply equip_ref to the current goal.
∎
Theorem. (adjoin_finite)
∀X y, finite X → finite (X ∪ {y})
In Proofgold the corresponding term root is def317... and proposition id is e6aa6d...
Proof:
Let X and y be given.
Assume H1.
Apply H1 to the current goal.
Let n be given.
Assume H.
Apply H to the current goal.
Assume Hn: n ∈ ω.
Assume H2: equip X n.
Apply equip_sym X n H2 to the current goal.
Let f be given.
Assume Hf: bij n X f.
Apply bijE n X f Hf to the current goal.
Assume Hf1: ∀i ∈ n, f i ∈ X.
Assume Hf2: ∀i j ∈ n, f i = f j → i = j.
Assume Hf3: ∀x ∈ X, ∃i ∈ n, f i = x.
Apply xm (y ∈ X) to the current goal.
Assume H3: y ∈ X.
We prove the intermediate claim L1: X ∪ {y} = X.
Apply set_ext to the current goal.
Let x be given.
Assume Hx.
Apply binunionE X {y} x Hx to the current goal.
Assume H4.
An exact proof term for the current goal is H4.
Assume H4: x ∈ {y}.
rewrite the current goal using SingE y x H4 (from left to right).
An exact proof term for the current goal is H3.
Apply binunion_Subq_1 to the current goal.
rewrite the current goal using L1 (from left to right).
An exact proof term for the current goal is H1.
Assume H3: y ∉ X.
We will prove ∃m ∈ ω, equip (X ∪ {y}) m.
We use ordsucc n to witness the existential quantifier.
Apply andI to the current goal.
Apply omega_ordsucc to the current goal.
An exact proof term for the current goal is Hn.
We will prove equip (X ∪ {y}) (ordsucc n).
Apply equip_sym to the current goal.
We will prove ∃g : set → set, bij (ordsucc n) (X ∪ {y}) g.
We prove the intermediate claim Lg: ∃g : set → set, (∀i ∈ n, g i = f i) ∧ g n = y.
We use (λi : set ⇒ if i ∈ n then f i else y) to witness the existential quantifier.
Apply andI to the current goal.
Let i be given.
Assume Hi.
An exact proof term for the current goal is If_i_1 (i ∈ n) (f i) y Hi.
An exact proof term for the current goal is If_i_0 (n ∈ n) (f n) y (In_irref n).
Apply Lg to the current goal.
Let g be given.
Assume H.
Apply H to the current goal.
Assume Hg1 Hg2.
We use g to witness the existential quantifier.
Apply bijI to the current goal.
We will prove ∀i ∈ ordsucc n, g i ∈ X ∪ {y}.
Let i be given.
Assume Hi.
Apply ordsuccE n i Hi to the current goal.
Assume H4: i ∈ n.
Apply binunionI1 to the current goal.
rewrite the current goal using Hg1 i H4 (from left to right).
We will prove f i ∈ X.
An exact proof term for the current goal is Hf1 i H4.
Assume H4: i = n.
Apply binunionI2 to the current goal.
rewrite the current goal using H4 (from left to right).
rewrite the current goal using Hg2 (from left to right).
Apply SingI to the current goal.
We will prove ∀i j ∈ ordsucc n, g i = g j → i = j.
Let i be given.
Assume Hi.
Let j be given.
Assume Hj.
Apply ordsuccE n i Hi to the current goal.
Assume H4: i ∈ n.
rewrite the current goal using Hg1 i H4 (from left to right).
Apply ordsuccE n j Hj to the current goal.
Assume H5: j ∈ n.
rewrite the current goal using Hg1 j H5 (from left to right).
An exact proof term for the current goal is Hf2 i H4 j H5.
Assume H5: j = n.
rewrite the current goal using H5 (from left to right).
rewrite the current goal using Hg2 (from left to right).
Assume H6: f i = y.
We will prove False.
Apply H3 to the current goal.
We will prove y ∈ X.
rewrite the current goal using H6 (from right to left).
An exact proof term for the current goal is Hf1 i H4.
Assume H4: i = n.
rewrite the current goal using H4 (from left to right).
rewrite the current goal using Hg2 (from left to right).
Apply ordsuccE n j Hj to the current goal.
Assume H5: j ∈ n.
rewrite the current goal using Hg1 j H5 (from left to right).
Assume H6: y = f j.
We will prove False.
Apply H3 to the current goal.
We will prove y ∈ X.
rewrite the current goal using H6 (from left to right).
An exact proof term for the current goal is Hf1 j H5.
Assume H5: j = n.
rewrite the current goal using H5 (from left to right).
Assume _.
We will prove n = n.
Use reflexivity.
We will prove ∀x ∈ X ∪ {y}, ∃i ∈ ordsucc n, g i = x.
Let x be given.
Assume Hx.
Apply binunionE X {y} x Hx to the current goal.
Assume H4: x ∈ X.
Apply Hf3 x H4 to the current goal.
Let i be given.
Assume H.
Apply H to the current goal.
Assume Hi: i ∈ n.
Assume H5: f i = x.
We use i to witness the existential quantifier.
Apply andI to the current goal.
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hi.
We will prove g i = x.
rewrite the current goal using Hg1 i Hi (from left to right).
An exact proof term for the current goal is H5.
Assume H4: x ∈ {y}.
We use n to witness the existential quantifier.
Apply andI to the current goal.
Apply ordsuccI2 to the current goal.
We will prove g n = x.
rewrite the current goal using SingE y x H4 (from left to right).
An exact proof term for the current goal is Hg2.
∎
Theorem. (binunion_finite)
∀X, finite X → ∀Y, finite Y → finite (X ∪ Y)
In Proofgold the corresponding term root is fc26bb... and proposition id is 3365a6...
Proof:
Let X be given.
Assume HX.
Apply finite_ind to the current goal.
We will prove finite (X ∪ 0).
rewrite the current goal using binunion_idr X (from left to right).
An exact proof term for the current goal is HX.
Let Y and z be given.
Assume HY: finite Y.
Assume Hz: z ∉ Y.
Assume IH: finite (X ∪ Y).
We will prove finite (X ∪ (Y ∪ {z})).
rewrite the current goal using binunion_asso (from left to right).
We will prove finite ((X ∪ Y) ∪ {z}).
Apply adjoin_finite to the current goal.
An exact proof term for the current goal is IH.
∎
Theorem. (famunion_nat_finite)
∀X : set → set, ∀n, nat_p n → (∀i ∈ n, finite (X i)) → finite (⋃i ∈ nX i)
In Proofgold the corresponding term root is 7f709a... and proposition id is f84ed0...
Proof:
Let X be given.
Apply nat_ind to the current goal.
Assume _.
We will prove finite (⋃i ∈ 0X i).
rewrite the current goal using famunion_Empty (from left to right).
An exact proof term for the current goal is finite_Empty.
Let n be given.
Assume Hn.
Assume IHn: (∀i ∈ n, finite (X i)) → finite (⋃i ∈ nX i).
Assume H1: ∀i ∈ ordsucc n, finite (X i).
We will prove finite (⋃i ∈ ordsucc nX i).
We prove the intermediate claim L1: (⋃i ∈ ordsucc nX i) = (⋃i ∈ nX i) ∪ X n.
Apply set_ext to the current goal.
Let z be given.
Assume Hz: z ∈ ⋃i ∈ ordsucc nX i.
Apply famunionE_impred (ordsucc n) X z Hz to the current goal.
Let i be given.
Assume Hi: i ∈ ordsucc n.
Assume H2: z ∈ X i.
Apply ordsuccE n i Hi to the current goal.
Assume H3: i ∈ n.
Apply binunionI1 to the current goal.
An exact proof term for the current goal is famunionI n X i z H3 H2.
Assume H3: i = n.
Apply binunionI2 to the current goal.
rewrite the current goal using H3 (from right to left).
An exact proof term for the current goal is H2.
Let z be given.
Apply binunionE' to the current goal.
Assume Hz.
Apply famunionE_impred n X z Hz to the current goal.
Let i be given.
Assume Hi: i ∈ n.
Assume H2: z ∈ X i.
Apply famunionI (ordsucc n) X i z to the current goal.
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hi.
An exact proof term for the current goal is H2.
Assume Hz.
Apply famunionI (ordsucc n) X n z to the current goal.
Apply ordsuccI2 to the current goal.
An exact proof term for the current goal is Hz.
rewrite the current goal using L1 (from left to right).
We will prove finite ((⋃i ∈ nX i) ∪ X n).
Apply binunion_finite to the current goal.
Apply IHn to the current goal.
Let i be given.
Assume Hi: i ∈ n.
Apply H1 to the current goal.
Apply ordsuccI1 to the current goal.
An exact proof term for the current goal is Hi.
Apply H1 n to the current goal.
Apply ordsuccI2 to the current goal.
∎
Theorem. (Subq_finite)
∀X, finite X → ∀Y, Y ⊆ X → finite Y
In Proofgold the corresponding term root is 6b062c... and proposition id is e4e3ee...
Proof:
Apply finite_ind to the current goal.
Let Y be given.
Assume H1: Y ⊆ 0.
We will prove finite Y.
rewrite the current goal using Empty_Subq_eq Y H1 (from left to right).
An exact proof term for the current goal is finite_Empty.
Let X and z be given.
Assume HX: finite X.
Assume Hz: z ∉ X.
Assume IH: ∀Y, Y ⊆ X → finite Y.
Let Y be given.
Assume H1: Y ⊆ X ∪ {z}.
We will prove finite Y.
Apply xm (z ∈ Y) to the current goal.
Assume H2: z ∈ Y.
We prove the intermediate claim L1: Y = (Y ∖ {z}) ∪ {z}.
Apply set_ext to the current goal.
Let w be given.
Assume Hw: w ∈ Y.
Apply xm (w ∈ {z}) to the current goal.
Assume H3: w ∈ {z}.
Apply binunionI2 to the current goal.
An exact proof term for the current goal is H3.
Assume H3: w ∉ {z}.
Apply binunionI1 to the current goal.
Apply setminusI to the current goal.
An exact proof term for the current goal is Hw.
An exact proof term for the current goal is H3.
Let w be given.
Assume Hw: w ∈ (Y ∖ {z}) ∪ {z}.
Apply binunionE (Y ∖ {z}) {z} w Hw to the current goal.
Assume H3: w ∈ Y ∖ {z}.
An exact proof term for the current goal is setminusE1 Y {z} w H3.
Assume H3: w ∈ {z}.
We will prove w ∈ Y.
rewrite the current goal using SingE z w H3 (from left to right).
An exact proof term for the current goal is H2.
rewrite the current goal using L1 (from left to right).
Apply adjoin_finite to the current goal.
We will prove finite (Y ∖ {z}).
Apply IH to the current goal.
Let y be given.
Assume Hy: y ∈ Y ∖ {z}.
Apply setminusE Y {z} y Hy to the current goal.
Assume Hy1: y ∈ Y.
Assume Hy2: y ∉ {z}.
We will prove y ∈ X.
Apply binunionE X {z} y (H1 y Hy1) to the current goal.
Assume H3: y ∈ X.
An exact proof term for the current goal is H3.
Assume H3: y ∈ {z}.
We will prove False.
Apply Hy2 to the current goal.
An exact proof term for the current goal is H3.
Assume H2: z ∉ Y.
Apply IH to the current goal.
Let y be given.
Assume Hy: y ∈ Y.
We will prove y ∈ X.
Apply binunionE X {z} y (H1 y Hy) to the current goal.
Assume H3: y ∈ X.
An exact proof term for the current goal is H3.
Assume H3: y ∈ {z}.
We will prove False.
Apply H2 to the current goal.
We will prove z ∈ Y.
rewrite the current goal using SingE z y H3 (from right to left).
An exact proof term for the current goal is Hy.
∎
Theorem. (TransSet_In_ordsucc_Subq)
∀x y, TransSet y → x ∈ ordsucc y → x ⊆ y
In Proofgold the corresponding term root is f80c6e... and proposition id is 341993...
Proof:
Let x and y be given.
Assume H1 H2.
Apply ordsuccE y x H2 to the current goal.
Assume H3: x ∈ y.
An exact proof term for the current goal is H1 x H3.
Assume H3: x = y.
rewrite the current goal using H3 (from left to right).
Apply Subq_ref to the current goal.
∎
Theorem. (exandE_i)
∀P Q : set → prop, (∃x, P x ∧ Q x) → ∀r : prop, (∀x, P x → Q x → r) → r
In Proofgold the corresponding term root is 083f82... and proposition id is b1b26e...
Proof:
Let P and Q be given.
Assume H1.
Let r be given.
Assume Hr.
Apply H1 to the current goal.
Let x be given.
Assume H2.
Apply H2 to the current goal.
An exact proof term for the current goal is Hr x.
∎
Theorem. (exandE_ii)
∀P Q : (set → set) → prop, (∃x : set → set, P x ∧ Q x) → ∀p : prop, (∀x : set → set, P x → Q x → p) → p
In Proofgold the corresponding term root is c3e995... and proposition id is ed9d65...
Proof:
Let P and Q be given.
Assume H.
Let p be given.
Assume Hp.
Apply H to the current goal.
Let x be given.
Assume H0.
Apply H0 to the current goal.
Assume H1 H2.
An exact proof term for the current goal is Hp x H1 H2.
∎
Theorem. (exandE_iii)
∀P Q : (set → set → set) → prop, (∃x : set → set → set, P x ∧ Q x) → ∀p : prop, (∀x : set → set → set, P x → Q x → p) → p
In Proofgold the corresponding term root is 028a8f... and proposition id is 80b3b0...
Proof:
Let P and Q be given.
Assume H.
Let p be given.
Assume Hp.
Apply H to the current goal.
Let x be given.
Assume H0.
Apply H0 to the current goal.
Assume H1 H2.
An exact proof term for the current goal is Hp x H1 H2.
∎
Theorem. (exandE_iiii)
∀P Q : (set → set → set → set) → prop, (∃x : set → set → set → set, P x ∧ Q x) → ∀p : prop, (∀x : set → set → set → set, P x → Q x → p) → p
In Proofgold the corresponding term root is 7807d4... and proposition id is d1536a...
Proof:
Let P and Q be given.
Assume H.
Let p be given.
Assume Hp.
Apply H to the current goal.
Let x be given.
Assume H0.
Apply H0 to the current goal.
Assume H1 H2.
An exact proof term for the current goal is Hp x H1 H2.
∎
Beginning of Section Descr_ii
Variable P : (set → set) → prop
Definition. We define Descr_ii to be λx : set ⇒ Eps_i (λy ⇒ ∀h : set → set, P h → h x = y) of type set → set.
In Proofgold the corresponding term root is 3bae39... and object id is 3a4ad2...
Hypothesis Pex : ∃f : set → set, P f
Hypothesis Puniq : ∀f g : set → set, P f → P g → f = g
Theorem. (Descr_ii_prop)
P Descr_ii
In Proofgold the corresponding term root is 45bfc0... and proposition id is e705d1...
Proof:
Apply Pex to the current goal.
Let f be given.
Assume Hf: P f.
We prove the intermediate claim L1: f = Descr_ii.
Apply func_ext set set to the current goal.
Let x be given.
We will prove f x = Descr_ii x.
We will prove f x = Eps_i (λy ⇒ ∀h : set → set, P h → h x = y).
We prove the intermediate claim L2: ∀h : set → set, P h → h x = f x.
Let h be given.
Assume Hh.
rewrite the current goal using Puniq f h Hf Hh (from left to right).
An exact proof term for the current goal is (λq H ⇒ H).
We prove the intermediate claim L3: ∀h : set → set, P h → h x = Descr_ii x.
An exact proof term for the current goal is Eps_i_ax (λy ⇒ ∀h : set → set, P h → h x = y) (f x) L2.
An exact proof term for the current goal is L3 f Hf.
rewrite the current goal using L1 (from right to left).
An exact proof term for the current goal is Hf.
∎
End of Section Descr_ii
Beginning of Section Descr_iii
Variable P : (set → set → set) → prop
Definition. We define Descr_iii to be λx y : set ⇒ Eps_i (λz ⇒ ∀h : set → set → set, P h → h x y = z) of type set → set → set.
In Proofgold the corresponding term root is ca5fc1... and object id is 877e40...
Hypothesis Pex : ∃f : set → set → set, P f
Hypothesis Puniq : ∀f g : set → set → set, P f → P g → f = g
Theorem. (Descr_iii_prop)
P Descr_iii
In Proofgold the corresponding term root is 261c3d... and proposition id is 160421...
Proof:
Apply Pex to the current goal.
Let f be given.
Assume Hf: P f.
We prove the intermediate claim L1: f = Descr_iii.
Apply func_ext set (set → set) to the current goal.
Let x be given.
Apply func_ext set set to the current goal.
Let y be given.
We will prove f x y = Descr_iii x y.
We will prove f x y = Eps_i (λz ⇒ ∀h : set → set → set, P h → h x y = z).
We prove the intermediate claim L2: ∀h : set → set → set, P h → h x y = f x y.
Let h be given.
Assume Hh.
rewrite the current goal using Puniq f h Hf Hh (from left to right).
An exact proof term for the current goal is (λq H ⇒ H).
We prove the intermediate claim L3: ∀h : set → set → set, P h → h x y = Descr_iii x y.
An exact proof term for the current goal is Eps_i_ax (λz ⇒ ∀h : set → set → set, P h → h x y = z) (f x y) L2.
An exact proof term for the current goal is L3 f Hf.
rewrite the current goal using L1 (from right to left).
An exact proof term for the current goal is Hf.
∎
End of Section Descr_iii
Beginning of Section Descr_Vo1
Variable P : Vo 1 → prop
Definition. We define Descr_Vo1 to be λx : set ⇒ ∀h : set → prop, P h → h x of type Vo 1.
In Proofgold the corresponding term root is 615c0a... and object id is a4ecdd...
Hypothesis Pex : ∃f : Vo 1, P f
Hypothesis Puniq : ∀f g : Vo 1, P f → P g → f = g
Theorem. (Descr_Vo1_prop)
P Descr_Vo1
In Proofgold the corresponding term root is f550ef... and proposition id is 2213e2...
Proof:
Apply Pex to the current goal.
Let f be given.
Assume Hf: P f.
We prove the intermediate claim L1: f = Descr_Vo1.
Apply func_ext set prop to the current goal.
Let x be given.
We will prove f x = Descr_Vo1 x.
Apply prop_ext_2 to the current goal.
Assume H1: f x.
Let h be given.
Assume Hh: P h.
We will prove h x.
rewrite the current goal using Puniq f h Hf Hh (from right to left).
An exact proof term for the current goal is H1.
Assume H1: Descr_Vo1 x.
An exact proof term for the current goal is H1 f Hf.
rewrite the current goal using L1 (from right to left).
An exact proof term for the current goal is Hf.
∎
End of Section Descr_Vo1
Beginning of Section If_ii
Variable p : prop
Variable f g : set → set
Definition. We define If_ii to be λx ⇒ if p then f x else g x of type set → set.
In Proofgold the corresponding term root is 17057f... and object id is e29015...
Theorem. (If_ii_1)
p → If_ii = f
In Proofgold the corresponding term root is 31521e... and proposition id is 342a1f...
Proof:
Assume H1.
Apply func_ext set set to the current goal.
Let x be given.
We will prove If_ii x = f x.
An exact proof term for the current goal is If_i_1 p (f x) (g x) H1.
∎
Theorem. (If_ii_0)
¬ p → If_ii = g
In Proofgold the corresponding term root is 6acab0... and proposition id is 36418b...
Proof:
Assume H1.
Apply func_ext set set to the current goal.
Let x be given.
We will prove If_ii x = g x.
An exact proof term for the current goal is If_i_0 p (f x) (g x) H1.
∎
End of Section If_ii
Beginning of Section If_iii
Variable p : prop
Variable f g : set → set → set
Definition. We define If_iii to be λx y ⇒ if p then f x y else g x y of type set → set → set.
In Proofgold the corresponding term root is 331476... and object id is dd7e2b...
Theorem. (If_iii_1)
p → If_iii = f
In Proofgold the corresponding term root is 2d62ee... and proposition id is 48bc10...
Proof:
Assume H1.
Apply func_ext set (set → set) to the current goal.
Let x be given.
Apply func_ext set set to the current goal.
Let y be given.
We will prove If_iii x y = f x y.
An exact proof term for the current goal is If_i_1 p (f x y) (g x y) H1.
∎
Theorem. (If_iii_0)
¬ p → If_iii = g
In Proofgold the corresponding term root is 9b77d9... and proposition id is d11201...
Proof:
Assume H1.
Apply func_ext set (set → set) to the current goal.
Let x be given.
Apply func_ext set set to the current goal.
Let y be given.
We will prove If_iii x y = g x y.
An exact proof term for the current goal is If_i_0 p (f x y) (g x y) H1.
∎
End of Section If_iii
Beginning of Section EpsilonRec_i
Variable F : set → (set → set) → set
Definition. We define In_rec_i_G to be λX Y ⇒ ∀R : set → set → prop, (∀X : set, ∀f : set → set, (∀x ∈ X, R x (f x)) → R X (F X f)) → R X Y of type set → set → prop.
In Proofgold the corresponding term root is fde509... and object id is a81e18...
Definition. We define In_rec_i to be λX ⇒ Eps_i (In_rec_i_G X) of type set → set.
In Proofgold the corresponding term root is fac413... and object id is 93aac4...
Theorem. (In_rec_i_G_c)
∀X : set, ∀f : set → set, (∀x ∈ X, In_rec_i_G x (f x)) → In_rec_i_G X (F X f)
In Proofgold the corresponding term root is 4da597... and proposition id is bc452e...
Proof:
Let X of type set be given.
Let f of type set → set be given.
Assume H1: ∀x ∈ X, In_rec_i_G x (f x).
We will prove In_rec_i_G X (F X f).
Let R of type set → set → prop be given.
Assume H2: ∀X : set, ∀f : set → set, (∀x ∈ X, R x (f x)) → R X (F X f).
We will prove R X (F X f).
Apply (H2 X f) to the current goal.
We will prove ∀x ∈ X, R x (f x).
Let x of type set be given.
Assume H3: x ∈ X.
We will prove R x (f x).
An exact proof term for the current goal is (H1 x H3 R H2).
∎
Theorem. (In_rec_i_G_inv)
∀X : set, ∀Y : set, In_rec_i_G X Y → ∃f : set → set, (∀x ∈ X, In_rec_i_G x (f x)) ∧ Y = F X f
In Proofgold the corresponding term root is ba1774... and proposition id is 5cfac2...
Proof:
Let X and Y be given.
Assume H1: In_rec_i_G X Y.
Set R to be the term (λX : set ⇒ λY : set ⇒ ∃f : set → set, (∀x ∈ X, In_rec_i_G x (f x)) ∧ Y = F X f).
Apply (H1 R) to the current goal.
We will prove ∀Z : set, ∀g : set → set, (∀z ∈ Z, R z (g z)) → R Z (F Z g).
Let Z and g be given.
Assume IH: ∀z ∈ Z, ∃f : set → set, (∀x ∈ z, In_rec_i_G x (f x)) ∧ g z = F z f.
We will prove ∃f : set → set, (∀x ∈ Z, In_rec_i_G x (f x)) ∧ F Z g = F Z f.
We use g to witness the existential quantifier.
Apply andI to the current goal.
Let z be given.
Assume H2: z ∈ Z.
Apply (exandE_ii (λf ⇒ ∀x ∈ z, In_rec_i_G x (f x)) (λf ⇒ g z = F z f) (IH z H2)) to the current goal.
Let f of type set → set be given.
Assume H3: ∀x ∈ z, In_rec_i_G x (f x).
Assume H4: g z = F z f.
We will prove In_rec_i_G z (g z).
rewrite the current goal using H4 (from left to right).
We will prove In_rec_i_G z (F z f).
An exact proof term for the current goal is (In_rec_i_G_c z f H3).
Use reflexivity.
∎
Hypothesis Fr : ∀X : set, ∀g h : set → set, (∀x ∈ X, g x = h x) → F X g = F X h
Theorem. (In_rec_i_G_f)
∀X : set, ∀Y Z : set, In_rec_i_G X Y → In_rec_i_G X Z → Y = Z
In Proofgold the corresponding term root is 25f7ff... and proposition id is 02b4e4...
Proof:
Apply In_ind to the current goal.
Let X be given.
Assume IH: ∀x ∈ X, ∀y z : set, In_rec_i_G x y → In_rec_i_G x z → y = z.
Let Y and Z be given.
Assume H1: In_rec_i_G X Y.
Assume H2: In_rec_i_G X Z.
We will prove Y = Z.
We prove the intermediate claim L1: ∃f : set → set, (∀x ∈ X, In_rec_i_G x (f x)) ∧ Y = F X f.
An exact proof term for the current goal is (In_rec_i_G_inv X Y H1).
We prove the intermediate claim L2: ∃f : set → set, (∀x ∈ X, In_rec_i_G x (f x)) ∧ Z = F X f.
An exact proof term for the current goal is (In_rec_i_G_inv X Z H2).
Apply (exandE_ii (λf ⇒ ∀x ∈ X, In_rec_i_G x (f x)) (λf ⇒ Y = F X f) L1) to the current goal.
Let g be given.
Assume H3: ∀x ∈ X, In_rec_i_G x (g x).
Assume H4: Y = F X g.
Apply (exandE_ii (λf ⇒ ∀x ∈ X, In_rec_i_G x (f x)) (λf ⇒ Z = F X f) L2) to the current goal.
Let h be given.
Assume H5: ∀x ∈ X, In_rec_i_G x (h x).
Assume H6: Z = F X h.
We will prove Y = Z.
rewrite the current goal using H4 (from left to right).
rewrite the current goal using H6 (from left to right).
We will prove F X g = F X h.
Apply Fr to the current goal.
We will prove ∀x ∈ X, g x = h x.
Let x be given.
Assume H7: x ∈ X.
An exact proof term for the current goal is (IH x H7 (g x) (h x) (H3 x H7) (H5 x H7)).
∎
Theorem. (In_rec_i_G_In_rec_i)
∀X : set, In_rec_i_G X (In_rec_i X)
In Proofgold the corresponding term root is 095407... and proposition id is dc2359...
Proof:
Apply In_ind to the current goal.
Let X be given.
Assume IH: ∀x ∈ X, In_rec_i_G x (In_rec_i x).
We will prove In_rec_i_G X (In_rec_i X).
We will prove In_rec_i_G X (Eps_i (In_rec_i_G X)).
Apply (Eps_i_ax (In_rec_i_G X) (F X In_rec_i)) to the current goal.
We will prove In_rec_i_G X (F X In_rec_i).
An exact proof term for the current goal is (In_rec_i_G_c X In_rec_i IH).
∎
Theorem. (In_rec_i_G_In_rec_i_d)
∀X : set, In_rec_i_G X (F X In_rec_i)
In Proofgold the corresponding term root is b480fe... and proposition id is 5b92ff...
Proof:
Let X and R be given.
Assume H1: ∀X : set, ∀f : set → set, (∀x ∈ X, R x (f x)) → R X (F X f).
Apply (H1 X In_rec_i) to the current goal.
Let x be given.
Assume _.
An exact proof term for the current goal is (In_rec_i_G_In_rec_i x R H1).
∎
Theorem. (In_rec_i_eq)
∀X : set, In_rec_i X = F X In_rec_i
In Proofgold the corresponding term root is 110653... and proposition id is e8653b...
Proof:
An exact proof term for the current goal is (λX : set ⇒ In_rec_i_G_f X (In_rec_i X) (F X In_rec_i) (In_rec_i_G_In_rec_i X) (In_rec_i_G_In_rec_i_d X)).
∎
End of Section EpsilonRec_i
Beginning of Section EpsilonRec_ii
Variable F : set → (set → (set → set)) → (set → set)
Definition. We define In_rec_G_ii to be λX Y ⇒ ∀R : set → (set → set) → prop, (∀X : set, ∀f : set → (set → set), (∀x ∈ X, R x (f x)) → R X (F X f)) → R X Y of type set → (set → set) → prop.
In Proofgold the corresponding term root is 6edba9... and object id is b31dbf...
Definition. We define In_rec_ii to be λX ⇒ Descr_ii (In_rec_G_ii X) of type set → (set → set).
In Proofgold the corresponding term root is f3c9ab... and object id is 3841bd...
Theorem. (In_rec_G_ii_c)
∀X : set, ∀f : set → (set → set), (∀x ∈ X, In_rec_G_ii x (f x)) → In_rec_G_ii X (F X f)
In Proofgold the corresponding term root is 381613... and proposition id is 49fbeb...
Proof:
Let X of type set be given.
Let f of type set → (set → set) be given.
Assume H1: ∀x ∈ X, In_rec_G_ii x (f x).
We will prove In_rec_G_ii X (F X f).
Let R of type set → (set → set) → prop be given.
Assume H2: ∀X : set, ∀f : set → (set → set), (∀x ∈ X, R x (f x)) → R X (F X f).
We will prove R X (F X f).
Apply (H2 X f) to the current goal.
We will prove ∀x ∈ X, R x (f x).
Let x of type set be given.
Assume H3: x ∈ X.
We will prove R x (f x).
An exact proof term for the current goal is (H1 x H3 R H2).
∎
Theorem. (In_rec_G_ii_inv)
∀X : set, ∀Y : (set → set), In_rec_G_ii X Y → ∃f : set → (set → set), (∀x ∈ X, In_rec_G_ii x (f x)) ∧ Y = F X f
In Proofgold the corresponding term root is 336cfe... and proposition id is b4a180...
Proof:
Let X and Y be given.
Assume H1: In_rec_G_ii X Y.
Set R to be the term (λX : set ⇒ λY : (set → set) ⇒ ∃f : set → (set → set), (∀x ∈ X, In_rec_G_ii x (f x)) ∧ Y = F X f).
Apply (H1 R) to the current goal.
We will prove ∀Z : set, ∀g : set → (set → set), (∀z ∈ Z, R z (g z)) → R Z (F Z g).
Let Z and g be given.
Assume IH: ∀z ∈ Z, ∃f : set → (set → set), (∀x ∈ z, In_rec_G_ii x (f x)) ∧ g z = F z f.
We will prove ∃f : set → (set → set), (∀x ∈ Z, In_rec_G_ii x (f x)) ∧ F Z g = F Z f.
We use g to witness the existential quantifier.
Apply andI to the current goal.
Let z be given.
Assume H2: z ∈ Z.
Apply (exandE_iii (λf ⇒ ∀x ∈ z, In_rec_G_ii x (f x)) (λf ⇒ g z = F z f) (IH z H2)) to the current goal.
Let f of type set → (set → set) be given.
Assume H3: ∀x ∈ z, In_rec_G_ii x (f x).
Assume H4: g z = F z f.
We will prove In_rec_G_ii z (g z).
rewrite the current goal using H4 (from left to right).
We will prove In_rec_G_ii z (F z f).
An exact proof term for the current goal is (In_rec_G_ii_c z f H3).
An exact proof term for the current goal is (λq H ⇒ H).
∎
Hypothesis Fr : ∀X : set, ∀g h : set → (set → set), (∀x ∈ X, g x = h x) → F X g = F X h
Theorem. (In_rec_G_ii_f)
∀X : set, ∀Y Z : (set → set), In_rec_G_ii X Y → In_rec_G_ii X Z → Y = Z
In Proofgold the corresponding term root is 64b6e8... and proposition id is a3826b...
Proof:
Apply In_ind to the current goal.
Let X be given.
Assume IH: ∀x ∈ X, ∀y z : (set → set), In_rec_G_ii x y → In_rec_G_ii x z → y = z.
Let Y and Z be given.
Assume H1: In_rec_G_ii X Y.
Assume H2: In_rec_G_ii X Z.
We will prove Y = Z.
We prove the intermediate claim L1: ∃f : set → (set → set), (∀x ∈ X, In_rec_G_ii x (f x)) ∧ Y = F X f.
An exact proof term for the current goal is (In_rec_G_ii_inv X Y H1).
We prove the intermediate claim L2: ∃f : set → (set → set), (∀x ∈ X, In_rec_G_ii x (f x)) ∧ Z = F X f.
An exact proof term for the current goal is (In_rec_G_ii_inv X Z H2).
Apply (exandE_iii (λf ⇒ ∀x ∈ X, In_rec_G_ii x (f x)) (λf ⇒ Y = F X f) L1) to the current goal.
Let g be given.
Assume H3: ∀x ∈ X, In_rec_G_ii x (g x).
Assume H4: Y = F X g.
Apply (exandE_iii (λf ⇒ ∀x ∈ X, In_rec_G_ii x (f x)) (λf ⇒ Z = F X f) L2) to the current goal.
Let h be given.
Assume H5: ∀x ∈ X, In_rec_G_ii x (h x).
Assume H6: Z = F X h.
We will prove Y = Z.
rewrite the current goal using H4 (from left to right).
rewrite the current goal using H6 (from left to right).
We will prove F X g = F X h.
Apply Fr to the current goal.
We will prove ∀x ∈ X, g x = h x.
Let x be given.
Assume H7: x ∈ X.
An exact proof term for the current goal is (IH x H7 (g x) (h x) (H3 x H7) (H5 x H7)).
∎
Theorem. (In_rec_G_ii_In_rec_ii)
∀X : set, In_rec_G_ii X (In_rec_ii X)
In Proofgold the corresponding term root is b2f81f... and proposition id is 2893f4...
Proof:
Apply In_ind to the current goal.
Let X be given.
Assume IH: ∀x ∈ X, In_rec_G_ii x (In_rec_ii x).
We will prove In_rec_G_ii X (In_rec_ii X).
We will prove In_rec_G_ii X (Descr_ii (In_rec_G_ii X)).
Apply (Descr_ii_prop (In_rec_G_ii X)) to the current goal.
We use (F X In_rec_ii) to witness the existential quantifier.
We will prove In_rec_G_ii X (F X In_rec_ii).
An exact proof term for the current goal is (In_rec_G_ii_c X In_rec_ii IH).
An exact proof term for the current goal is In_rec_G_ii_f X.
∎
Theorem. (In_rec_G_ii_In_rec_ii_d)
∀X : set, In_rec_G_ii X (F X In_rec_ii)
In Proofgold the corresponding term root is 2ff1ca... and proposition id is f12072...
Proof:
Let X and R be given.
Assume H1: ∀X : set, ∀f : set → (set → set), (∀x ∈ X, R x (f x)) → R X (F X f).
Apply (H1 X In_rec_ii) to the current goal.
Let x be given.
Assume _.
An exact proof term for the current goal is (In_rec_G_ii_In_rec_ii x R H1).
∎
Theorem. (In_rec_ii_eq)
∀X : set, In_rec_ii X = F X In_rec_ii
In Proofgold the corresponding term root is f7aff6... and proposition id is 7bb1e9...
Proof:
An exact proof term for the current goal is (λX : set ⇒ In_rec_G_ii_f X (In_rec_ii X) (F X In_rec_ii) (In_rec_G_ii_In_rec_ii X) (In_rec_G_ii_In_rec_ii_d X)).
∎
End of Section EpsilonRec_ii
Beginning of Section EpsilonRec_iii
Variable F : set → (set → (set → set → set)) → (set → set → set)
Definition. We define In_rec_G_iii to be λX Y ⇒ ∀R : set → (set → set → set) → prop, (∀X : set, ∀f : set → (set → set → set), (∀x ∈ X, R x (f x)) → R X (F X f)) → R X Y of type set → (set → set → set) → prop.
In Proofgold the corresponding term root is 533cc7... and object id is 7a2f6b...
Definition. We define In_rec_iii to be λX ⇒ Descr_iii (In_rec_G_iii X) of type set → (set → set → set).
In Proofgold the corresponding term root is 9b3a85... and object id is 651077...
Theorem. (In_rec_G_iii_c)
∀X : set, ∀f : set → (set → set → set), (∀x ∈ X, In_rec_G_iii x (f x)) → In_rec_G_iii X (F X f)
In Proofgold the corresponding term root is 4ea572... and proposition id is a7d90f...
Proof:
Let X of type set be given.
Let f of type set → (set → set → set) be given.
Assume H1: ∀x ∈ X, In_rec_G_iii x (f x).
We will prove In_rec_G_iii X (F X f).
Let R of type set → (set → set → set) → prop be given.
Assume H2: ∀X : set, ∀f : set → (set → set → set), (∀x ∈ X, R x (f x)) → R X (F X f).
We will prove R X (F X f).
Apply (H2 X f) to the current goal.
We will prove ∀x ∈ X, R x (f x).
Let x of type set be given.
Assume H3: x ∈ X.
We will prove R x (f x).
An exact proof term for the current goal is (H1 x H3 R H2).
∎
Theorem. (In_rec_G_iii_inv)
∀X : set, ∀Y : (set → set → set), In_rec_G_iii X Y → ∃f : set → (set → set → set), (∀x ∈ X, In_rec_G_iii x (f x)) ∧ Y = F X f
In Proofgold the corresponding term root is 5f6e0c... and proposition id is f47dae...
Proof:
Let X and Y be given.
Assume H1: In_rec_G_iii X Y.
Set R to be the term (λX : set ⇒ λY : (set → set → set) ⇒ ∃f : set → (set → set → set), (∀x ∈ X, In_rec_G_iii x (f x)) ∧ Y = F X f).
Apply (H1 R) to the current goal.
We will prove ∀Z : set, ∀g : set → (set → set → set), (∀z ∈ Z, R z (g z)) → R Z (F Z g).
Let Z and g be given.
Assume IH: ∀z ∈ Z, ∃f : set → (set → set → set), (∀x ∈ z, In_rec_G_iii x (f x)) ∧ g z = F z f.
We will prove ∃f : set → (set → set → set), (∀x ∈ Z, In_rec_G_iii x (f x)) ∧ F Z g = F Z f.
We use g to witness the existential quantifier.
Apply andI to the current goal.
Let z be given.
Assume H2: z ∈ Z.
Apply (exandE_iiii (λf ⇒ ∀x ∈ z, In_rec_G_iii x (f x)) (λf ⇒ g z = F z f) (IH z H2)) to the current goal.
Let f of type set → (set → set → set) be given.
Assume H3: ∀x ∈ z, In_rec_G_iii x (f x).
Assume H4: g z = F z f.
We will prove In_rec_G_iii z (g z).
rewrite the current goal using H4 (from left to right).
We will prove In_rec_G_iii z (F z f).
An exact proof term for the current goal is (In_rec_G_iii_c z f H3).
An exact proof term for the current goal is (λq H ⇒ H).
∎
Hypothesis Fr : ∀X : set, ∀g h : set → (set → set → set), (∀x ∈ X, g x = h x) → F X g = F X h
Theorem. (In_rec_G_iii_f)
∀X : set, ∀Y Z : (set → set → set), In_rec_G_iii X Y → In_rec_G_iii X Z → Y = Z
In Proofgold the corresponding term root is 574d71... and proposition id is 40978a...
Proof:
Apply In_ind to the current goal.
Let X be given.
Assume IH: ∀x ∈ X, ∀y z : (set → set → set), In_rec_G_iii x y → In_rec_G_iii x z → y = z.
Let Y and Z be given.
Assume H1: In_rec_G_iii X Y.
Assume H2: In_rec_G_iii X Z.
We will prove Y = Z.
We prove the intermediate claim L1: ∃f : set → (set → set → set), (∀x ∈ X, In_rec_G_iii x (f x)) ∧ Y = F X f.
An exact proof term for the current goal is (In_rec_G_iii_inv X Y H1).
We prove the intermediate claim L2: ∃f : set → (set → set → set), (∀x ∈ X, In_rec_G_iii x (f x)) ∧ Z = F X f.
An exact proof term for the current goal is (In_rec_G_iii_inv X Z H2).
Apply (exandE_iiii (λf ⇒ ∀x ∈ X, In_rec_G_iii x (f x)) (λf ⇒ Y = F X f) L1) to the current goal.
Let g be given.
Assume H3: ∀x ∈ X, In_rec_G_iii x (g x).
Assume H4: Y = F X g.
Apply (exandE_iiii (λf ⇒ ∀x ∈ X, In_rec_G_iii x (f x)) (λf ⇒ Z = F X f) L2) to the current goal.
Let h be given.
Assume H5: ∀x ∈ X, In_rec_G_iii x (h x).
Assume H6: Z = F X h.
We will prove Y = Z.
rewrite the current goal using H4 (from left to right).
rewrite the current goal using H6 (from left to right).
We will prove F X g = F X h.
Apply Fr to the current goal.
We will prove ∀x ∈ X, g x = h x.
Let x be given.
Assume H7: x ∈ X.
An exact proof term for the current goal is (IH x H7 (g x) (h x) (H3 x H7) (H5 x H7)).
∎
Theorem. (In_rec_G_iii_In_rec_iii)
∀X : set, In_rec_G_iii X (In_rec_iii X)
In Proofgold the corresponding term root is d8c3ba... and proposition id is 37f9a7...
Proof:
Apply In_ind to the current goal.
Let X be given.
Assume IH: ∀x ∈ X, In_rec_G_iii x (In_rec_iii x).
We will prove In_rec_G_iii X (In_rec_iii X).
We will prove In_rec_G_iii X (Descr_iii (In_rec_G_iii X)).
Apply (Descr_iii_prop (In_rec_G_iii X)) to the current goal.
We use (F X In_rec_iii) to witness the existential quantifier.
We will prove In_rec_G_iii X (F X In_rec_iii).
An exact proof term for the current goal is (In_rec_G_iii_c X In_rec_iii IH).
An exact proof term for the current goal is In_rec_G_iii_f X.
∎
Theorem. (In_rec_G_iii_In_rec_iii_d)
∀X : set, In_rec_G_iii X (F X In_rec_iii)
In Proofgold the corresponding term root is 60c6d7... and proposition id is c05f77...
Proof:
Let X and R be given.
Assume H1: ∀X : set, ∀f : set → (set → set → set), (∀x ∈ X, R x (f x)) → R X (F X f).
Apply (H1 X In_rec_iii) to the current goal.
Let x be given.
Assume _.
An exact proof term for the current goal is (In_rec_G_iii_In_rec_iii x R H1).
∎
Theorem. (In_rec_iii_eq)
∀X : set, In_rec_iii X = F X In_rec_iii
In Proofgold the corresponding term root is 5f0574... and proposition id is a0709d...
Proof:
An exact proof term for the current goal is (λX : set ⇒ In_rec_G_iii_f X (In_rec_iii X) (F X In_rec_iii) (In_rec_G_iii_In_rec_iii X) (In_rec_G_iii_In_rec_iii_d X)).
∎
End of Section EpsilonRec_iii
Beginning of Section NatRec
Variable z : set
Variable f : set → set → set
Let F : set → (set → set) → set ≝ λn g ⇒ if ⋃ n ∈ n then f (⋃ n) (g (⋃ n)) else z
Definition. We define nat_primrec to be In_rec_i F of type set → set.
In Proofgold the corresponding term root is 3be1c5... and object id is 1a64d5...
Theorem. (nat_primrec_r)
∀X : set, ∀g h : set → set, (∀x ∈ X, g x = h x) → F X g = F X h
In Proofgold the corresponding term root is c955cd... and proposition id is 9ba887...
Proof:
Let X, g and h be given.
Assume H1: ∀x ∈ X, g x = h x.
We will prove F X g = F X h.
Apply xm (⋃ X ∈ X) to the current goal.
Assume H2: (⋃ X ∈ X).
We will prove (if ⋃ X ∈ X then f (⋃ X) (g (⋃ X)) else z) = (if ⋃ X ∈ X then f (⋃ X) (h (⋃ X)) else z).
rewrite the current goal using (H1 (⋃ X) H2) (from left to right).
We will prove (if ⋃ X ∈ X then f (⋃ X) (h (⋃ X)) else z) = (if ⋃ X ∈ X then f (⋃ X) (h (⋃ X)) else z).
An exact proof term for the current goal is (λq H ⇒ H).
Assume H2: ¬ (⋃ X ∈ X).
We will prove (if ⋃ X ∈ X then f (⋃ X) (g (⋃ X)) else z) = (if ⋃ X ∈ X then f (⋃ X) (h (⋃ X)) else z).
We prove the intermediate claim L1: (if ⋃ X ∈ X then f (⋃ X) (g (⋃ X)) else z) = z.
An exact proof term for the current goal is (If_i_0 (⋃ X ∈ X) (f (⋃ X) (g (⋃ X))) z H2).
We prove the intermediate claim L2: (if ⋃ X ∈ X then f (⋃ X) (h (⋃ X)) else z) = z.
An exact proof term for the current goal is (If_i_0 (⋃ X ∈ X) (f (⋃ X) (h (⋃ X))) z H2).
rewrite the current goal using L2 (from left to right).
An exact proof term for the current goal is L1.
∎
Theorem. (nat_primrec_0)
nat_primrec 0 = z
In Proofgold the corresponding term root is 416eaf... and proposition id is be11a2...
Proof:
We will prove (In_rec_i F 0 = z).
rewrite the current goal using (In_rec_i_eq F nat_primrec_r) (from left to right).
We will prove F 0 (In_rec_i F) = z.
We will prove (if ⋃ 0 ∈ 0 then f (⋃ 0) (In_rec_i F (⋃ 0)) else z) = z.
An exact proof term for the current goal is (If_i_0 (⋃ 0 ∈ 0) (f (⋃ 0) (In_rec_i F (⋃ 0))) z (EmptyE (⋃ 0))).
∎
Theorem. (nat_primrec_S)
∀n : set, nat_p n → nat_primrec (ordsucc n) = f n (nat_primrec n)
In Proofgold the corresponding term root is 797185... and proposition id is fe4c88...
Proof:
Let n be given.
Assume Hn: nat_p n.
We will prove (In_rec_i F (ordsucc n) = f n (In_rec_i F n)).
rewrite the current goal using (In_rec_i_eq F nat_primrec_r) (from left to right) at position 1.
We will prove F (ordsucc n) (In_rec_i F) = f n (In_rec_i F n).
We will prove (if ⋃ (ordsucc n) ∈ ordsucc n then f (⋃ (ordsucc n)) (In_rec_i F (⋃ (ordsucc n))) else z) = f n (In_rec_i F n).
rewrite the current goal using (Union_ordsucc_eq n Hn) (from left to right).
We will prove (if n ∈ ordsucc n then f n (In_rec_i F n) else z) = f n (In_rec_i F n).
An exact proof term for the current goal is (If_i_1 (n ∈ ordsucc n) (f n (In_rec_i F n)) z (ordsuccI2 n)).
∎
End of Section NatRec
Beginning of Section NatArith
Definition. We define add_nat to be λn m : set ⇒ nat_primrec n (λ_ r ⇒ ordsucc r) m of type set → set → set.
In Proofgold the corresponding term root is afa8ae... and object id is 29768b...
Notation. We use + as an infix operator with priority 360 and which associates to the right corresponding to applying term add_nat.
Theorem. (add_nat_0R)
∀n : set, n + 0 = n
In Proofgold the corresponding term root is 402bcb... and proposition id is 23277f...
Proof:
Let n be given.
An exact proof term for the current goal is (nat_primrec_0 n (λ_ r ⇒ ordsucc r)).
∎
Theorem. (add_nat_SR)
∀n m : set, nat_p m → n + ordsucc m = ordsucc (n + m)
In Proofgold the corresponding term root is a53523... and proposition id is e61e18...
Proof:
Let n and m be given.
Assume Hm.
An exact proof term for the current goal is (nat_primrec_S n (λ_ r ⇒ ordsucc r) m Hm).
∎
Theorem. (add_nat_p)
∀n : set, nat_p n → ∀m : set, nat_p m → nat_p (n + m)
In Proofgold the corresponding term root is a68eaa... and proposition id is 4c1d5d...
Proof:
Let n be given.
Assume Hn: nat_p n.
Apply nat_ind to the current goal.
We will prove nat_p (n + 0).
rewrite the current goal using (add_nat_0R n) (from left to right).
We will prove nat_p n.
An exact proof term for the current goal is Hn.
Let m be given.
Assume Hm: nat_p m.
Assume IHm: nat_p (n + m).
We will prove nat_p (n + ordsucc m).
rewrite the current goal using (add_nat_SR n m Hm) (from left to right).
We will prove nat_p (ordsucc (n + m)).
Apply nat_ordsucc to the current goal.
We will prove nat_p (n + m).
An exact proof term for the current goal is IHm.
∎
Theorem. (add_nat_1_1_2)
1 + 1 = 2
In Proofgold the corresponding term root is 174b87... and proposition id is 77f23f...
Proof:
Use symmetry.
We will prove ordsucc 1 = 1 + ordsucc 0.
rewrite the current goal using add_nat_SR 1 0 nat_0 (from left to right).
We will prove ordsucc 1 = ordsucc (1 + 0).
rewrite the current goal using add_nat_0R 1 (from left to right).
An exact proof term for the current goal is (λq H ⇒ H).
∎
Theorem. (add_nat_0L)
∀m : set, nat_p m → 0 + m = m
In Proofgold the corresponding term root is 8ceeb9... and proposition id is 84fbe4...
Proof:
Apply nat_ind to the current goal.
We will prove 0 + 0 = 0.
Apply add_nat_0R to the current goal.
Let m be given.
Assume Hm: nat_p m.
Assume IHm: 0 + m = m.
We will prove 0 + ordsucc m = ordsucc m.
rewrite the current goal using (add_nat_SR 0 m Hm) (from left to right).
We will prove ordsucc (0 + m) = ordsucc m.
rewrite the current goal using IHm (from left to right).
Use reflexivity.
∎
Theorem. (add_nat_SL)
∀n : set, nat_p n → ∀m : set, nat_p m → ordsucc n + m = ordsucc (n + m)
In Proofgold the corresponding term root is 9c442a... and proposition id is e6efb7...
Proof:
Let n be given.
Assume Hn: nat_p n.
Apply nat_ind to the current goal.
We will prove ordsucc n + 0 = ordsucc (n + 0).
rewrite the current goal using add_nat_0R (from left to right).
rewrite the current goal using add_nat_0R (from left to right).
Use reflexivity.
Let m be given.
Assume Hm: nat_p m.
Assume IHm: ordsucc n + m = ordsucc (n + m).
We will prove ordsucc n + ordsucc m = ordsucc (n + ordsucc m).
rewrite the current goal using (add_nat_SR (ordsucc n) m Hm) (from left to right).
We will prove ordsucc (ordsucc n + m) = ordsucc (n + ordsucc m).
rewrite the current goal using (add_nat_SR n m Hm) (from left to right).
We will prove ordsucc (ordsucc n + m) = ordsucc (ordsucc (n + m)).
rewrite the current goal using IHm (from left to right).
Use reflexivity.
∎
Theorem. (add_nat_com)
∀n : set, nat_p n → ∀m : set, nat_p m → n + m = m + n
In Proofgold the corresponding term root is 40e952... and proposition id is 4ab179...
Proof:
Let n be given.
Assume Hn: nat_p n.
Apply nat_ind to the current goal.
We will prove n + 0 = 0 + n.
rewrite the current goal using (add_nat_0L n Hn) (from left to right).
An exact proof term for the current goal is (add_nat_0R n).
Let m be given.
Assume Hm: nat_p m.
Assume IHm: n + m = m + n.
We will prove n + ordsucc m = ordsucc m + n.
rewrite the current goal using (add_nat_SL m Hm n Hn) (from left to right).
We will prove n + ordsucc m = ordsucc (m + n).
rewrite the current goal using IHm (from right to left).
We will prove n + ordsucc m = ordsucc (n + m).
An exact proof term for the current goal is (add_nat_SR n m Hm).
∎
Theorem. (nat_Subq_add_ex)
∀n, nat_p n → ∀m, nat_p m → n ⊆ m → ∃k, nat_p k ∧ m = k + n
In Proofgold the corresponding term root is a1e406... and proposition id is 2c7fa4...
Proof:
Apply nat_ind to the current goal.
Let m be given.
Assume Hm Hnm.
We use m to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is Hm.
We will prove m = m + 0.
Use symmetry.
An exact proof term for the current goal is add_nat_0R m.
Let n be given.
Assume Hn.
Assume IH: ∀m, nat_p m → n ⊆ m → ∃k, nat_p k ∧ m = k + n.
Apply nat_inv_impred to the current goal.
Assume Hnm: ordsucc n ⊆ 0.
We will prove False.
Apply EmptyE n to the current goal.
We will prove n ∈ 0.
Apply Hnm to the current goal.
Apply ordsuccI2 to the current goal.
Let m be given.
Assume Hm: nat_p m.
Assume Hnm: ordsucc n ⊆ ordsucc m.
We prove the intermediate claim Lnm: n ⊆ m.
Apply ordinal_In_Or_Subq m n (nat_p_ordinal m Hm) (nat_p_ordinal n Hn) to the current goal.
Assume H1: m ∈ n.
We will prove False.
Apply In_irref (ordsucc m) to the current goal.
Apply Hnm to the current goal.
We will prove ordsucc m ∈ ordsucc n.
An exact proof term for the current goal is nat_ordsucc_in_ordsucc n Hn m H1.
Assume H1: n ⊆ m.
An exact proof term for the current goal is H1.
Apply IH m Hm Lnm to the current goal.
Let k be given.
Assume H.
Apply H to the current goal.
Assume Hk: nat_p k.
Assume H1: m = k + n.
We will prove ∃k, nat_p k ∧ ordsucc m = k + ordsucc n.
We use k to witness the existential quantifier.
Apply andI to the current goal.
An exact proof term for the current goal is Hk.
We will prove ordsucc m = k + ordsucc n.
rewrite the current goal using add_nat_SR k n Hn (from left to right).
We will prove ordsucc m = ordsucc (k + n).
Use f_equal.
An exact proof term for the current goal is H1.
∎
Definition. We define mul_nat to be λn m : set ⇒ nat_primrec 0 (λ_ r ⇒ n + r) m of type set → set → set.
In Proofgold the corresponding term root is 35a1ef... and object id is 80ccc6...
Notation. We use * as an infix operator with priority 355 and which associates to the right corresponding to applying term mul_nat.
Theorem. (mul_nat_0R)
∀n : set, n * 0 = 0
In Proofgold the corresponding term root is b2f56c... and proposition id is 1e5b2c...
Proof:
Let n be given.
An exact proof term for the current goal is (nat_primrec_0 0 (λ_ r ⇒ n + r)).
∎
Theorem. (mul_nat_SR)
∀n m : set, nat_p m → n * ordsucc m = n + n * m
In Proofgold the corresponding term root is 37df54... and proposition id is 6dd6fc...
Proof:
Let n and m be given.
Assume Hm.
An exact proof term for the current goal is (nat_primrec_S 0 (λ_ r ⇒ n + r) m Hm).
∎
Theorem. (mul_nat_p)
∀n : set, nat_p n → ∀m : set, nat_p m → nat_p (n * m)
In Proofgold the corresponding term root is d63438... and proposition id is e910cb...
Proof:
Let n be given.
Assume Hn: nat_p n.
Apply nat_ind to the current goal.
We will prove nat_p (n * 0).
rewrite the current goal using (mul_nat_0R n) (from left to right).
We will prove nat_p 0.
An exact proof term for the current goal is nat_0.
Let m be given.
Assume Hm: nat_p m.
Assume IHm: nat_p (n * m).
We will prove nat_p (n * ordsucc m).
rewrite the current goal using (mul_nat_SR n m Hm) (from left to right).
We will prove nat_p (n + n * m).
An exact proof term for the current goal is (add_nat_p n Hn (n * m) IHm).
∎
End of Section NatArith
Definition. We define Inj1 to be In_rec_i (λX f ⇒ {0} ∪ {f x|x ∈ X}) of type set → set.
In Proofgold the corresponding term root is 8f0026... and object id is 7736ca...
Theorem. (Inj1_eq)
∀X : set, Inj1 X = {0} ∪ {Inj1 x|x ∈ X}
In Proofgold the corresponding term root is 73d189... and proposition id is b80457...
Proof:
We prove the intermediate claim L1: ∀X : set, ∀g h : set → set, (∀x ∈ X, g x = h x) → {0} ∪ {g x|x ∈ X} = {0} ∪ {h x|x ∈ X}.
Let X, g and h be given.
Assume H: ∀x ∈ X, g x = h x.
We prove the intermediate claim L1a: {g x|x ∈ X} = {h x|x ∈ X}.
An exact proof term for the current goal is (ReplEq_ext X g h H).
We will prove {0} ∪ {g x|x ∈ X} = {0} ∪ {h x|x ∈ X}.
rewrite the current goal using L1a (from left to right).
Use reflexivity.
An exact proof term for the current goal is (In_rec_i_eq (λX f ⇒ {0} ∪ {f x|x ∈ X}) L1).
∎
Theorem. (Inj1I1)
∀X : set, 0 ∈ Inj1 X
In Proofgold the corresponding term root is 3c2cc4... and proposition id is 4ccac3...
Proof:
Let X be given.
rewrite the current goal using (Inj1_eq X) (from left to right).
We will prove 0 ∈ {0} ∪ {Inj1 x|x ∈ X}.
Apply binunionI1 to the current goal.
Apply SingI to the current goal.
∎
Theorem. (Inj1I2)
∀X x : set, x ∈ X → Inj1 x ∈ Inj1 X
In Proofgold the corresponding term root is b6f23b... and proposition id is 086475...
Proof:
Let X and x be given.
Assume H: x ∈ X.
rewrite the current goal using (Inj1_eq X) (from left to right).
We will prove Inj1 x ∈ {0} ∪ {Inj1 x|x ∈ X}.
Apply binunionI2 to the current goal.
An exact proof term for the current goal is (ReplI X Inj1 x H).
∎
Theorem. (Inj1E)
∀X y : set, y ∈ Inj1 X → y = 0 ∨ ∃x ∈ X, y = Inj1 x
In Proofgold the corresponding term root is d4177b... and proposition id is 259217...
Proof:
Let X and y be given.
rewrite the current goal using (Inj1_eq X) (from left to right).
Assume H1: y ∈ {0} ∪ {Inj1 x|x ∈ X}.
We will prove y = 0 ∨ ∃x ∈ X, y = Inj1 x.
Apply (binunionE {0} {Inj1 x|x ∈ X} y H1) to the current goal.
Assume H2: y ∈ {0}.
Apply orIL to the current goal.
An exact proof term for the current goal is (SingE 0 y H2).
Assume H2: y ∈ {Inj1 x|x ∈ X}.
Apply orIR to the current goal.
We will prove ∃x ∈ X, y = Inj1 x.
An exact proof term for the current goal is (ReplE X Inj1 y H2).
∎
Theorem. (Inj1NE1)
∀x : set, Inj1 x ≠ 0
In Proofgold the corresponding term root is 5e3197... and proposition id is 09bd10...
Proof:
Let x be given.
Assume H1: Inj1 x = 0.
Apply (EmptyE 0) to the current goal.
We will prove 0 ∈ 0.
rewrite the current goal using H1 (from right to left) at position 2.
We will prove 0 ∈ Inj1 x.
An exact proof term for the current goal is (Inj1I1 x).
∎
Theorem. (Inj1NE2)
∀x : set, Inj1 x ∉ {0}
In Proofgold the corresponding term root is 9e94b1... and proposition id is 3c4dd0...
Proof:
Let x be given.
Assume H1: Inj1 x ∈ {0}.
An exact proof term for the current goal is (Inj1NE1 x (SingE 0 (Inj1 x) H1)).
∎
Definition. We define Inj0 to be λX ⇒ {Inj1 x|x ∈ X} of type set → set.
In Proofgold the corresponding term root is 814321... and object id is 7d54be...
Theorem. (Inj0I)
∀X x : set, x ∈ X → Inj1 x ∈ Inj0 X
In Proofgold the corresponding term root is 7fc2e9... and proposition id is 565378...
Proof:
An exact proof term for the current goal is (λX x H ⇒ ReplI X Inj1 x H).
∎
Theorem. (Inj0E)
∀X y : set, y ∈ Inj0 X → ∃x : set, x ∈ X ∧ y = Inj1 x
In Proofgold the corresponding term root is 416684... and proposition id is 855213...
Proof:
An exact proof term for the current goal is (λX y H ⇒ ReplE X Inj1 y H).
∎
Definition. We define Unj to be In_rec_i (λX f ⇒ {f x|x ∈ X ∖ {0}}) of type set → set.
In Proofgold the corresponding term root is f202c1... and object id is 2e1675...
Theorem. (Unj_eq)
∀X : set, Unj X = {Unj x|x ∈ X ∖ {0}}
In Proofgold the corresponding term root is 3b1aa6... and proposition id is b4c852...
Proof:
We prove the intermediate claim L1: ∀X : set, ∀g h : set → set, (∀x ∈ X, g x = h x) → {g x|x ∈ X ∖ {0}} = {h x|x ∈ X ∖ {0}}.
Let X, g and h be given.
Assume H1: ∀x ∈ X, g x = h x.
We will prove {g x|x ∈ X ∖ {0}} = {h x|x ∈ X ∖ {0}}.
Apply (ReplEq_ext (X ∖ {0}) g h) to the current goal.
Let x be given.
Assume H2: x ∈ X ∖ {0}.
We will prove g x = h x.
Apply H1 to the current goal.
We will prove x ∈ X.
An exact proof term for the current goal is (setminusE1 X {0} x H2).
An exact proof term for the current goal is (In_rec_i_eq (λX f ⇒ {f x|x ∈ X ∖ {0}}) L1).
∎
Theorem. (Unj_Inj1_eq)
∀X : set, Unj (Inj1 X) = X
In Proofgold the corresponding term root is e2cdfc... and proposition id is ae00e1...
Proof:
Apply In_ind to the current goal.
Let X be given.
Assume IH: ∀x ∈ X, Unj (Inj1 x) = x.
We will prove Unj (Inj1 X) = X.
rewrite the current goal using Unj_eq (from left to right).
We will prove {Unj x|x ∈ Inj1 X ∖ {0}} = X.
Apply set_ext to the current goal.
We will prove {Unj x|x ∈ Inj1 X ∖ {0}} ⊆ X.
Let x be given.
Assume H1: x ∈ {Unj x|x ∈ Inj1 X ∖ {0}}.
We will prove x ∈ X.
Apply (ReplE_impred (Inj1 X ∖ {0}) Unj x H1) to the current goal.
Let y be given.
Assume H2: y ∈ Inj1 X ∖ {0}.
Assume H3: x = Unj y.
rewrite the current goal using H3 (from left to right).
We will prove Unj y ∈ X.
Apply (setminusE (Inj1 X) {0} y H2) to the current goal.
Assume H4: y ∈ Inj1 X.
Assume H5: y ∉ {0}.
Apply (Inj1E X y H4) to the current goal.
Assume H6: y = 0.
We will prove False.
Apply H5 to the current goal.
rewrite the current goal using H6 (from left to right).
We will prove 0 ∈ {0}.
Apply SingI to the current goal.
Assume H6: ∃x ∈ X, y = Inj1 x.
Apply (exandE_i (λx ⇒ x ∈ X) (λx ⇒ y = Inj1 x) H6) to the current goal.
Let z be given.
Assume H7: z ∈ X.
Assume H8: y = Inj1 z.
rewrite the current goal using H8 (from left to right).
We will prove Unj (Inj1 z) ∈ X.
rewrite the current goal using (IH z H7) (from left to right).
We will prove z ∈ X.
An exact proof term for the current goal is H7.
We will prove X ⊆ {Unj x|x ∈ Inj1 X ∖ {0}}.
Let x be given.
Assume H1: x ∈ X.
We will prove x ∈ {Unj x|x ∈ Inj1 X ∖ {0}}.
rewrite the current goal using (IH x H1) (from right to left).
We will prove Unj (Inj1 x) ∈ {Unj x|x ∈ Inj1 X ∖ {0}}.
Apply (ReplI (Inj1 X ∖ {0}) Unj) to the current goal.
We will prove Inj1 x ∈ Inj1 X ∖ {0}.
Apply setminusI to the current goal.
An exact proof term for the current goal is (Inj1I2 X x H1).
We will prove Inj1 x ∉ {0}.
An exact proof term for the current goal is (Inj1NE2 x).
∎
Theorem. (Inj1_inj)
∀X Y : set, Inj1 X = Inj1 Y → X = Y
In Proofgold the corresponding term root is ce8b66... and proposition id is 3e918a...
Proof:
Let X and Y be given.
Assume H1: Inj1 X = Inj1 Y.
We will prove X = Y.
rewrite the current goal using (Unj_Inj1_eq X) (from right to left).
rewrite the current goal using (Unj_Inj1_eq Y) (from right to left).
rewrite the current goal using H1 (from left to right).
Use reflexivity.
∎
Theorem. (Unj_Inj0_eq)
∀X : set, Unj (Inj0 X) = X
In Proofgold the corresponding term root is f3e39c... and proposition id is 3808ad...
Proof:
Let X be given.
rewrite the current goal using (Unj_eq (Inj0 X)) (from left to right).
We will prove {Unj x|x ∈ Inj0 X ∖ {0}} = X.
Apply set_ext to the current goal.
We will prove {Unj x|x ∈ Inj0 X ∖ {0}} ⊆ X.
Let x be given.
Assume H1: x ∈ {Unj x|x ∈ Inj0 X ∖ {0}}.
We will prove x ∈ X.
Apply (ReplE_impred (Inj0 X ∖ {0}) Unj x H1) to the current goal.
Let y be given.
Assume H2: y ∈ Inj0 X ∖ {0}.
Assume H3: x = Unj y.
Apply (setminusE (Inj0 X) {0} y H2) to the current goal.
Assume H4: y ∈ {Inj1 x|x ∈ X}.
Assume H5: y ∉ {0}.
Apply (ReplE_impred X Inj1 y H4) to the current goal.
Let z be given.
Assume H6: z ∈ X.
Assume H7: y = Inj1 z.
We prove the intermediate claim L1: x = z.
rewrite the current goal using H3 (from left to right).
We will prove Unj y = z.
rewrite the current goal using H7 (from left to right).
We will prove Unj (Inj1 z) = z.
An exact proof term for the current goal is (Unj_Inj1_eq z).
We will prove x ∈ X.
rewrite the current goal using L1 (from left to right).
We will prove z ∈ X.
An exact proof term for the current goal is H6.
We will prove X ⊆ {Unj x|x ∈ Inj0 X ∖ {0}}.
Let x be given.
Assume H1: x ∈ X.
We will prove x ∈ {Unj x|x ∈ Inj0 X ∖ {0}}.
rewrite the current goal using (Unj_Inj1_eq x) (from right to left).
We will prove Unj (Inj1 x) ∈ {Unj x|x ∈ Inj0 X ∖ {0}}.
Apply (ReplI (Inj0 X ∖ {0}) Unj) to the current goal.
We will prove Inj1 x ∈ Inj0 X ∖ {0}.
Apply setminusI to the current goal.
We will prove Inj1 x ∈ {Inj1 x|x ∈ X}.
Apply ReplI to the current goal.
An exact proof term for the current goal is H1.
We will prove Inj1 x ∉ {0}.
An exact proof term for the current goal is (Inj1NE2 x).
∎
Theorem. (Inj0_inj)
∀X Y : set, Inj0 X = Inj0 Y → X = Y
In Proofgold the corresponding term root is 662f53... and proposition id is 29f945...
Proof:
Let X and Y be given.
Assume H1: Inj0 X = Inj0 Y.
We will prove X = Y.
rewrite the current goal using (Unj_Inj0_eq X) (from right to left).
rewrite the current goal using (Unj_Inj0_eq Y) (from right to left).
rewrite the current goal using H1 (from left to right).
Use reflexivity.
∎
Theorem. (Inj0_0)
Inj0 0 = 0
In Proofgold the corresponding term root is 171f47... and proposition id is 57f5af...
Proof:
An exact proof term for the current goal is (Repl_Empty Inj1).
∎
Theorem. (Inj0_Inj1_neq)
∀X Y : set, Inj0 X ≠ Inj1 Y
In Proofgold the corresponding term root is 21bb03... and proposition id is 2b14d8...
Proof:
Let X and Y be given.
Assume H1: Inj0 X = Inj1 Y.
We prove the intermediate claim L1: 0 ∈ Inj1 Y.
An exact proof term for the current goal is (Inj1I1 Y).
We prove the intermediate claim L2: 0 ∈ Inj0 X.
rewrite the current goal using H1 (from left to right).
An exact proof term for the current goal is L1.
Apply (Inj0E X 0 L2) to the current goal.
Let x be given.
Assume H2: x ∈ X ∧ 0 = Inj1 x.
Apply Inj1NE1 x to the current goal.
Use symmetry.
An exact proof term for the current goal is andER (x ∈ X) (0 = Inj1 x) H2.
∎
Definition. We define setsum to be λX Y ⇒ {Inj0 x|x ∈ X} ∪ {Inj1 y|y ∈ Y} of type set → set → set.
In Proofgold the corresponding term root is afe37e... and object id is 5b7e09...
Notation. We use + as an infix operator with priority 450 and which associates to the left corresponding to applying term setsum.
Theorem. (Inj0_setsum)
∀X Y x : set, x ∈ X → Inj0 x ∈ X + Y
In Proofgold the corresponding term root is 1c6408... and proposition id is 246ac5...
Proof:
Let X, Y and x be given.
Assume H: x ∈ X.
We will prove Inj0 x ∈ {Inj0 x|x ∈ X} ∪ {Inj1 y|y ∈ Y}.
Apply binunionI1 to the current goal.
We will prove Inj0 x ∈ {Inj0 x|x ∈ X}.
Apply ReplI to the current goal.
An exact proof term for the current goal is H.
∎
Theorem. (Inj1_setsum)
∀X Y y : set, y ∈ Y → Inj1 y ∈ X + Y
In Proofgold the corresponding term root is d47a0f... and proposition id is 2b4f94...
Proof:
Let X, Y and y be given.
Assume H: y ∈ Y.
We will prove Inj1 y ∈ {Inj0 x|x ∈ X} ∪ {Inj1 y|y ∈ Y}.
Apply binunionI2 to the current goal.
We will prove Inj1 y ∈ {Inj1 y|y ∈ Y}.
Apply ReplI to the current goal.
An exact proof term for the current goal is H.
∎
Theorem. (setsum_Inj_inv)
∀X Y z : set, z ∈ X + Y → (∃x ∈ X, z = Inj0 x) ∨ (∃y ∈ Y, z = Inj1 y)
In Proofgold the corresponding term root is 645a6c... and proposition id is a36400...
Proof:
Let X, Y and z be given.
Assume H1: z ∈ {Inj0 x|x ∈ X} ∪ {Inj1 y|y ∈ Y}.
Apply (binunionE {Inj0 x|x ∈ X} {Inj1 y|y ∈ Y} z H1) to the current goal.
Assume H2: z ∈ {Inj0 x|x ∈ X}.
Apply orIL to the current goal.
An exact proof term for the current goal is (ReplE X Inj0 z H2).
Assume H2: z ∈ {Inj1 y|y ∈ Y}.
Apply orIR to the current goal.
An exact proof term for the current goal is (ReplE Y Inj1 z H2).
∎
Theorem. (Inj0_setsum_0L)
∀X : set, 0 + X = Inj0 X
In Proofgold the corresponding term root is 0be1e7... and proposition id is 36fabd...
Proof:
Let X be given.
Apply set_ext to the current goal.
Let z be given.
Assume H1: z ∈ 0 + X.
We will prove z ∈ Inj0 X.
Apply (setsum_Inj_inv 0 X z H1) to the current goal.
Assume H2: ∃x ∈ 0, z = Inj0 x.
Apply (exandE_i (λx ⇒ x ∈ 0) (λx ⇒ z = Inj0 x) H2) to the current goal.
Let x be given.
Assume H3: x ∈ 0.
We will prove False.
An exact proof term for the current goal is (EmptyE x H3).
Assume H2: ∃x ∈ X, z = Inj1 x.
Apply (exandE_i (λx ⇒ x ∈ X) (λx ⇒ z = Inj1 x) H2) to the current goal.
Let x be given.
Assume H3: x ∈ X.
Assume H4: z = Inj1 x.
rewrite the current goal using H4 (from left to right).
We will prove Inj1 x ∈ Inj0 X.
An exact proof term for the current goal is (Inj0I X x H3).
Let z be given.
Assume H1: z ∈ Inj0 X.
We will prove z ∈ 0 + X.
Apply (exandE_i (λx ⇒ x ∈ X) (λx ⇒ z = Inj1 x) (Inj0E X z H1)) to the current goal.
Let x be given.
Assume H2: x ∈ X.
Assume H3: z = Inj1 x.
rewrite the current goal using H3 (from left to right).
We will prove Inj1 x ∈ 0 + X.
Apply Inj1_setsum to the current goal.
An exact proof term for the current goal is H2.
∎
Theorem. (Subq_1_Sing0)
1 ⊆ {0}
In Proofgold the corresponding term root is aeae9a... and proposition id is abab2d...
Proof:
Let x be given.
Assume H1: x ∈ 1.
We will prove x ∈ {0}.
Apply ordsuccE 0 x H1 to the current goal.
Assume H2: x ∈ 0.
We will prove False.
An exact proof term for the current goal is (EmptyE x H2).
Assume H2: x = 0.
rewrite the current goal using H2 (from left to right).
We will prove 0 ∈ {0}.
An exact proof term for the current goal is (SingI 0).
∎
Theorem. (Subq_Sing0_1)
{0} ⊆ 1
In Proofgold the corresponding term root is cd7864... and proposition id is 1806bf...
Proof:
An exact proof term for the current goal is (λx H1 ⇒ SingE 0 x H1 (λs _ : set ⇒ x ∈ ordsucc s) (ordsuccI2 x)).
∎
Theorem. (eq_1_Sing0)
1 = {0}
In Proofgold the corresponding term root is 5e5996... and proposition id is 264f38...
Proof:
An exact proof term for the current goal is (set_ext 1 (Sing 0) Subq_1_Sing0 Subq_Sing0_1).
∎
Theorem. (Inj1_setsum_1L)
∀X : set, 1 + X = Inj1 X
In Proofgold the corresponding term root is f8d721... and proposition id is 45b1c8...
Proof:
Let X be given.
Apply set_ext to the current goal.
Let z be given.
Assume H1: z ∈ 1 + X.
We will prove z ∈ Inj1 X.
Apply (setsum_Inj_inv 1 X z H1) to the current goal.
Assume H2: ∃x ∈ 1, z = Inj0 x.
Apply (exandE_i (λx ⇒ x ∈ 1) (λx ⇒ z = Inj0 x) H2) to the current goal.
Let x be given.
Assume H3: x ∈ 1.
Assume H4: z = Inj0 x.
rewrite the current goal using H4 (from left to right).
We will prove Inj0 x ∈ Inj1 X.
We prove the intermediate claim L1: x = 0.
An exact proof term for the current goal is (SingE 0 x (Subq_1_Sing0 x H3)).
rewrite the current goal using L1 (from left to right).
We will prove Inj0 0 ∈ Inj1 X.
rewrite the current goal using Inj0_0 (from left to right).
We will prove 0 ∈ Inj1 X.
An exact proof term for the current goal is (Inj1I1 X).
Assume H2: ∃x ∈ X, z = Inj1 x.
Apply (exandE_i (λx ⇒ x ∈ X) (λx ⇒ z = Inj1 x) H2) to the current goal.
Let x be given.
Assume H3: x ∈ X.
Assume H4: z = Inj1 x.
rewrite the current goal using H4 (from left to right).
We will prove Inj1 x ∈ Inj1 X.
An exact proof term for the current goal is (Inj1I2 X x H3).
Let z be given.
Assume H1: z ∈ Inj1 X.
We will prove z ∈ 1 + X.
Apply (Inj1E X z H1) to the current goal.
Assume H2: z = 0.
rewrite the current goal using H2 (from left to right).
We will prove 0 ∈ 1 + X.
rewrite the current goal using Inj0_0 (from right to left) at position 1.
We will prove Inj0 0 ∈ 1 + X.
Apply Inj0_setsum to the current goal.
We will prove 0 ∈ 1.
An exact proof term for the current goal is In_0_1.
Assume H2: ∃x ∈ X, z = Inj1 x.
Apply (exandE_i (λx ⇒ x ∈ X) (λx ⇒ z = Inj1 x) H2) to the current goal.
Let x be given.
Assume H2: x ∈ X.
Assume H3: z = Inj1 x.
rewrite the current goal using H3 (from left to right).
We will prove Inj1 x ∈ 1 + X.
Apply Inj1_setsum to the current goal.
An exact proof term for the current goal is H2.
∎
Theorem. (nat_setsum1_ordsucc)
∀n : set, nat_p n → 1 + n = ordsucc n
In Proofgold the corresponding term root is 3be3cc... and proposition id is 8fbdda...
Proof:
We prove the intermediate claim L: ∀n : set, nat_p n → Inj1 n = ordsucc n.
Apply nat_complete_ind to the current goal.
Let n be given.
Assume Hn: nat_p n.
Assume IHn: ∀m ∈ n, Inj1 m = ordsucc m.
We will prove Inj1 n = ordsucc n.
Apply set_ext to the current goal.
We will prove Inj1 n ⊆ ordsucc n.
Let z be given.
Assume H1: z ∈ Inj1 n.
We will prove z ∈ ordsucc n.
Apply (Inj1E n z H1) to the current goal.
Assume H2: z = 0.
rewrite the current goal using H2 (from left to right).
We will prove 0 ∈ ordsucc n.
An exact proof term for the current goal is (nat_0_in_ordsucc n Hn).
Assume H2: ∃m ∈ n, z = Inj1 m.
Apply (exandE_i (λm ⇒ m ∈ n) (λm ⇒ z = Inj1 m) H2) to the current goal.
Let m be given.
Assume H3: m ∈ n.
Assume H4: z = Inj1 m.
rewrite the current goal using H4 (from left to right).
We will prove Inj1 m ∈ ordsucc n.
rewrite the current goal using (IHn m H3) (from left to right).
We will prove ordsucc m ∈ ordsucc n.
An exact proof term for the current goal is (nat_ordsucc_in_ordsucc n Hn m H3).
We will prove ordsucc n ⊆ Inj1 n.
Let m be given.
Assume H1: m ∈ ordsucc n.
We will prove m ∈ Inj1 n.
Apply (ordsuccE n m H1) to the current goal.
Assume H2: m ∈ n.
Apply (nat_inv m (nat_p_trans n Hn m H2)) to the current goal.
Assume H3: m = 0.
rewrite the current goal using H3 (from left to right).
We will prove 0 ∈ Inj1 n.
An exact proof term for the current goal is (Inj1I1 n).
Assume H3: ∃k, nat_p k ∧ m = ordsucc k.
Apply (exandE_i nat_p (λk ⇒ m = ordsucc k) H3) to the current goal.
Let k be given.
Assume H4: nat_p k.
Assume H5: m = ordsucc k.
rewrite the current goal using H5 (from left to right).
We will prove ordsucc k ∈ Inj1 n.
We prove the intermediate claim L1: k ∈ m.
rewrite the current goal using H5 (from left to right).
An exact proof term for the current goal is (ordsuccI2 k).
We prove the intermediate claim L2: k ∈ n.
An exact proof term for the current goal is (nat_trans n Hn m H2 k L1).
rewrite the current goal using (IHn k L2) (from right to left).
We will prove Inj1 k ∈ Inj1 n.
An exact proof term for the current goal is (Inj1I2 n k L2).
Assume H2: m = n.
rewrite the current goal using H2 (from left to right).
We will prove n ∈ Inj1 n.
Apply (nat_inv n Hn) to the current goal.
Assume H3: n = 0.
rewrite the current goal using H3 (from left to right) at position 1.
We will prove 0 ∈ Inj1 n.
An exact proof term for the current goal is (Inj1I1 n).
Assume H3: ∃k, nat_p k ∧ n = ordsucc k.
Apply (exandE_i nat_p (λk ⇒ n = ordsucc k) H3) to the current goal.
Let k be given.
Assume H4: nat_p k.
Assume H5: n = ordsucc k.
rewrite the current goal using H5 (from left to right) at position 1.
We will prove ordsucc k ∈ Inj1 n.
We prove the intermediate claim L1: k ∈ n.
rewrite the current goal using H5 (from left to right).
An exact proof term for the current goal is (ordsuccI2 k).
rewrite the current goal using (IHn k L1) (from right to left).
We will prove Inj1 k ∈ Inj1 n.
An exact proof term for the current goal is (Inj1I2 n k L1).
Let n be given.
Assume Hn: nat_p n.
We will prove 1 + n = ordsucc n.
rewrite the current goal using Inj1_setsum_1L (from left to right).
We will prove Inj1 n = ordsucc n.
An exact proof term for the current goal is (L n Hn).
∎
Theorem. (setsum_0_0)
0 + 0 = 0
In Proofgold the corresponding term root is 5dcaeb... and proposition id is 4f3651...
Proof:
rewrite the current goal using (Inj0_setsum_0L 0) (from left to right).
We will prove Inj0 0 = 0.
An exact proof term for the current goal is Inj0_0.
∎
Theorem. (setsum_1_0_1)
1 + 0 = 1
In Proofgold the corresponding term root is c2107f... and proposition id is e8609a...
Proof:
An exact proof term for the current goal is (nat_setsum1_ordsucc 0 nat_0).
∎
Theorem. (setsum_1_1_2)
1 + 1 = 2
In Proofgold the corresponding term root is 9c89c0... and proposition id is ef3f41...
Proof:
An exact proof term for the current goal is (nat_setsum1_ordsucc 1 nat_1).
∎
Beginning of Section pair_setsum
Let pair ≝ setsum
Definition. We define proj0 to be λZ ⇒ {Unj z|z ∈ Z, ∃x : set, Inj0 x = z} of type set → set.
In Proofgold the corresponding term root is 21a488... and object id is 9c5e4e...
Definition. We define proj1 to be λZ ⇒ {Unj z|z ∈ Z, ∃y : set, Inj1 y = z} of type set → set.
In Proofgold the corresponding term root is 7aba21... and object id is 7fd81e...
Theorem. (Inj0_pair_0_eq)
Inj0 = pair 0
In Proofgold the corresponding term root is 0fe601... and proposition id is 490c2a...
Proof:
Apply (func_ext set set) to the current goal.
Let x be given.
Use symmetry.
We will prove 0 + x = Inj0 x.
An exact proof term for the current goal is (Inj0_setsum_0L x).
∎
Theorem. (Inj1_pair_1_eq)
Inj1 = pair 1
In Proofgold the corresponding term root is 0bbde8... and proposition id is 8272d7...
Proof:
Apply (func_ext set set) to the current goal.
Let x be given.
Use symmetry.
We will prove 1 + x = Inj1 x.
An exact proof term for the current goal is (Inj1_setsum_1L x).
∎
Theorem. (pairI0)
∀X Y x, x ∈ X → pair 0 x ∈ pair X Y
In Proofgold the corresponding term root is 21a414... and proposition id is 34813b...
Proof:
rewrite the current goal using Inj0_pair_0_eq (from right to left).
An exact proof term for the current goal is Inj0_setsum.
∎
Theorem. (pairI1)
∀X Y y, y ∈ Y → pair 1 y ∈ pair X Y
In Proofgold the corresponding term root is bdaab5... and proposition id is b5248b...
Proof:
rewrite the current goal using Inj1_pair_1_eq (from right to left).
An exact proof term for the current goal is Inj1_setsum.
∎
Theorem. (pairE)
∀X Y z, z ∈ pair X Y → (∃x ∈ X, z = pair 0 x) ∨ (∃y ∈ Y, z = pair 1 y)
In Proofgold the corresponding term root is aac610... and proposition id is 0afdeb...
Proof:
rewrite the current goal using Inj0_pair_0_eq (from right to left).
rewrite the current goal using Inj1_pair_1_eq (from right to left).
An exact proof term for the current goal is setsum_Inj_inv.
∎
Theorem. (pairE0)
∀X Y x, pair 0 x ∈ pair X Y → x ∈ X
In Proofgold the corresponding term root is 042f30... and proposition id is f50b25...
Proof:
Let X, Y and x be given.
Assume H1: pair 0 x ∈ pair X Y.
We will prove x ∈ X.
Apply (pairE X Y (pair 0 x) H1) to the current goal.
rewrite the current goal using Inj0_pair_0_eq (from right to left).
Assume H2: ∃x' ∈ X, Inj0 x = Inj0 x'.
Apply (exandE_i (λx' ⇒ x' ∈ X) (λx' ⇒ Inj0 x = Inj0 x') H2) to the current goal.
Let x' be given.
Assume H3: x' ∈ X.
Assume H4: Inj0 x = Inj0 x'.
We will prove x ∈ X.
rewrite the current goal using (Inj0_inj x x' H4) (from left to right).
We will prove x' ∈ X.
An exact proof term for the current goal is H3.
rewrite the current goal using Inj0_pair_0_eq (from right to left).
rewrite the current goal using Inj1_pair_1_eq (from right to left).
Assume H2: ∃y ∈ Y, Inj0 x = Inj1 y.
We will prove False.
Apply (exandE_i (λy ⇒ y ∈ Y) (λy ⇒ Inj0 x = Inj1 y) H2) to the current goal.
Let y be given.
Assume _.
Assume H3: Inj0 x = Inj1 y.
An exact proof term for the current goal is (Inj0_Inj1_neq x y H3).
∎
Theorem. (pairE1)
∀X Y y, pair 1 y ∈ pair X Y → y ∈ Y
In Proofgold the corresponding term root is 505cfc... and proposition id is 30a181...
Proof:
Let X, Y and y be given.
Assume H1: pair 1 y ∈ pair X Y.
We will prove y ∈ Y.
Apply (pairE X Y (pair 1 y) H1) to the current goal.
rewrite the current goal using Inj0_pair_0_eq (from right to left).
rewrite the current goal using Inj1_pair_1_eq (from right to left).
Assume H2: ∃x ∈ X, Inj1 y = Inj0 x.
We will prove False.
Apply (exandE_i (λx ⇒ x ∈ X) (λx ⇒ Inj1 y = Inj0 x) H2) to the current goal.
Let x be given.
Assume _.
Assume H3: Inj1 y = Inj0 x.
Apply (Inj0_Inj1_neq x y) to the current goal.
Use symmetry.
An exact proof term for the current goal is H3.
rewrite the current goal using Inj1_pair_1_eq (from right to left).
Assume H2: ∃y' ∈ Y, Inj1 y = Inj1 y'.
Apply (exandE_i (λy' ⇒ y' ∈ Y) (λy' ⇒ Inj1 y = Inj1 y') H2) to the current goal.
Let y' be given.
Assume H3: y' ∈ Y.
Assume H4: Inj1 y = Inj1 y'.
We will prove y ∈ Y.
rewrite the current goal using (Inj1_inj y y' H4) (from left to right).
We will prove y' ∈ Y.
An exact proof term for the current goal is H3.
∎
Theorem. (proj0I)
∀w u : set, pair 0 u ∈ w → u ∈ proj0 w
In Proofgold the corresponding term root is be55b9... and proposition id is 840c8d...
Proof:
rewrite the current goal using Inj0_pair_0_eq (from right to left).
Let w and u be given.
Assume H1: Inj0 u ∈ w.
We will prove u ∈ {Unj z|z ∈ w, ∃x : set, Inj0 x = z}.
rewrite the current goal using (Unj_Inj0_eq u) (from right to left).
We will prove Unj (Inj0 u) ∈ {Unj z|z ∈ w, ∃x : set, Inj0 x = z}.
Apply (ReplSepI w (λz ⇒ ∃x : set, Inj0 x = z) Unj (Inj0 u)) to the current goal.
We will prove Inj0 u ∈ w.
An exact proof term for the current goal is H1.
We will prove ∃x, Inj0 x = Inj0 u.
We use u to witness the existential quantifier.
Use reflexivity.
∎
Theorem. (proj0E)
∀w u : set, u ∈ proj0 w → pair 0 u ∈ w
In Proofgold the corresponding term root is b9afa9... and proposition id is 5b7f58...
Proof:
Let w and u be given.
Assume H1: u ∈ {Unj z|z ∈ w, ∃x : set, Inj0 x = z}.
rewrite the current goal using Inj0_pair_0_eq (from right to left).
We will prove Inj0 u ∈ w.
Apply (ReplSepE_impred w (λz ⇒ ∃x : set, Inj0 x = z) Unj u H1) to the current goal.
Let z be given.
Assume H2: z ∈ w.
Assume H3: ∃x, Inj0 x = z.
Assume H4: u = Unj z.
Apply H3 to the current goal.
Let x be given.
Assume H5: Inj0 x = z.
We will prove Inj0 u ∈ w.
rewrite the current goal using H4 (from left to right).
We will prove Inj0 (Unj z) ∈ w.
rewrite the current goal using H5 (from right to left).
We will prove Inj0 (Unj (Inj0 x)) ∈ w.
rewrite the current goal using Unj_Inj0_eq (from left to right).
We will prove Inj0 x ∈ w.
rewrite the current goal using H5 (from left to right).
An exact proof term for the current goal is H2.
∎
Theorem. (proj1I)
∀w u : set, pair 1 u ∈ w → u ∈ proj1 w
In Proofgold the corresponding term root is fbddb7... and proposition id is 595b3c...
Proof:
rewrite the current goal using Inj1_pair_1_eq (from right to left).
Let w and u be given.
Assume H1: Inj1 u ∈ w.
We will prove u ∈ {Unj z|z ∈ w, ∃y : set, Inj1 y = z}.
rewrite the current goal using (Unj_Inj1_eq u) (from right to left).
We will prove Unj (Inj1 u) ∈ {Unj z|z ∈ w, ∃y : set, Inj1 y = z}.
Apply (ReplSepI w (λz ⇒ ∃y : set, Inj1 y = z) Unj (Inj1 u)) to the current goal.
We will prove Inj1 u ∈ w.
An exact proof term for the current goal is H1.
We will prove ∃y, Inj1 y = Inj1 u.
We use u to witness the existential quantifier.
Use reflexivity.
∎
Theorem. (proj1E)
∀w u : set, u ∈ proj1 w → pair 1 u ∈ w
In Proofgold the corresponding term root is 06e32e... and proposition id is c3c3e6...
Proof:
Let w and u be given.
Assume H1: u ∈ {Unj z|z ∈ w, ∃y : set, Inj1 y = z}.
rewrite the current goal using Inj1_pair_1_eq (from right to left).
We will prove Inj1 u ∈ w.
Apply (ReplSepE_impred w (λz ⇒ ∃y : set, Inj1 y = z) Unj u H1) to the current goal.
Let z be given.
Assume H2: z ∈ w.
Assume H3: ∃y, Inj1 y = z.
Assume H4: u = Unj z.
Apply H3 to the current goal.
Let y be given.
Assume H5: Inj1 y = z.
We will prove Inj1 u ∈ w.
rewrite the current goal using H4 (from left to right).
We will prove Inj1 (Unj z) ∈ w.
rewrite the current goal using H5 (from right to left).
We will prove Inj1 (Unj (Inj1 y)) ∈ w.
rewrite the current goal using Unj_Inj1_eq (from left to right).
We will prove Inj1 y ∈ w.
rewrite the current goal using H5 (from left to right).
An exact proof term for the current goal is H2.
∎
Theorem. (proj0_pair_eq)
∀X Y : set, proj0 (pair X Y) = X
In Proofgold the corresponding term root is d546f3... and proposition id is 0c12fb...
Proof:
Let X and Y be given.
Apply set_ext to the current goal.
We will prove proj0 (pair X Y) ⊆ X.
Let u be given.
Assume H1: u ∈ proj0 (pair X Y).
We will prove u ∈ X.
Apply (pairE0 X Y u) to the current goal.
We will prove pair 0 u ∈ pair X Y.
An exact proof term for the current goal is (proj0E (pair X Y) u H1).
We will prove X ⊆ proj0 (pair X Y).
Let u be given.
Assume H1: u ∈ X.
We will prove u ∈ proj0 (pair X Y).
Apply proj0I to the current goal.
We will prove pair 0 u ∈ pair X Y.
Apply pairI0 to the current goal.
We will prove u ∈ X.
An exact proof term for the current goal is H1.
∎
Theorem. (proj1_pair_eq)
∀X Y : set, proj1 (pair X Y) = Y
In Proofgold the corresponding term root is 9f98a8... and proposition id is e0cb0f...
Proof:
Let X and Y be given.
Apply set_ext to the current goal.
We will prove proj1 (pair X Y) ⊆ Y.
Let u be given.
Assume H1: u ∈ proj1 (pair X Y).
We will prove u ∈ Y.
Apply (pairE1 X Y u) to the current goal.
We will prove pair 1 u ∈ pair X Y.
An exact proof term for the current goal is (proj1E (pair X Y) u H1).
We will prove Y ⊆ proj1 (pair X Y).
Let u be given.
Assume H1: u ∈ Y.
We will prove u ∈ proj1 (pair X Y).
Apply proj1I to the current goal.
We will prove pair 1 u ∈ pair X Y.
Apply pairI1 to the current goal.
We will prove u ∈ Y.
An exact proof term for the current goal is H1.
∎
Definition. We define Sigma to be λX Y ⇒ ⋃x ∈ X{pair x y|y ∈ Y x} of type set → (set → set) → set.
In Proofgold the corresponding term root is 8acbb2... and object id is d64bcb...
Notation. We use ∑ x...y [possibly with ascriptions] , B as a binder notation corresponding to a term constructed using Sigma.
Theorem. (pair_Sigma)
∀X : set, ∀Y : set → set, ∀x ∈ X, ∀y ∈ Y x, pair x y ∈ ∑x ∈ X, Y x
In Proofgold the corresponding term root is 0d8645... and proposition id is 1456cb...
Proof:
Let X, Y and x be given.
Assume Hx: x ∈ X.
Let y be given.
Assume Hy: y ∈ Y x.
We will prove pair x y ∈ ⋃x ∈ X{pair x y|y ∈ Y x}.
Apply (famunionI X (λx ⇒ {pair x y|y ∈ Y x}) x (pair x y)) to the current goal.
We will prove x ∈ X.
An exact proof term for the current goal is Hx.
We will prove pair x y ∈ {pair x y|y ∈ Y x}.
Apply ReplI to the current goal.
We will prove y ∈ Y x.
An exact proof term for the current goal is Hy.
∎
Theorem. (Sigma_eta_proj0_proj1)
∀X : set, ∀Y : set → set, ∀z ∈ (∑x ∈ X, Y x), pair (proj0 z) (proj1 z) = z ∧ proj0 z ∈ X ∧ proj1 z ∈ Y (proj0 z)
In Proofgold the corresponding term root is afe962... and proposition id is 77ba8e...
Proof:
Let X, Y and z be given.
Assume H1: z ∈ ∑x ∈ X, Y x.
We prove the intermediate claim L1: ∃x ∈ X, z ∈ {pair x y|y ∈ Y x}.
An exact proof term for the current goal is (famunionE X (λx ⇒ {pair x y|y ∈ Y x}) z H1).
Apply (exandE_i (λx ⇒ x ∈ X) (λx ⇒ z ∈ {pair x y|y ∈ Y x}) L1) to the current goal.
Let x be given.
Assume H2: x ∈ X.
Assume H3: z ∈ {pair x y|y ∈ Y x}.
Apply (ReplE_impred (Y x) (pair x) z H3) to the current goal.
Let y be given.
Assume H4: y ∈ Y x.
Assume H5: z = pair x y.
rewrite the current goal using H5 (from left to right).
We will prove pair (proj0 (pair x y)) (proj1 (pair x y)) = pair x y ∧ proj0 (pair x y) ∈ X ∧ proj1 (pair x y) ∈ Y (proj0 (pair x y)).
rewrite the current goal using proj0_pair_eq (from left to right).
We will prove pair x (proj1 (pair x y)) = pair x y ∧ x ∈ X ∧ proj1 (pair x y) ∈ Y x.
rewrite the current goal using proj1_pair_eq (from left to right).
We will prove pair x y = pair x y ∧ x ∈ X ∧ y ∈ Y x.
Apply and3I to the current goal.
We will prove pair x y = pair x y.
Use reflexivity.
We will prove x ∈ X.
An exact proof term for the current goal is H2.
We will prove y ∈ Y x.
An exact proof term for the current goal is H4.
∎
Theorem. (proj_Sigma_eta)
∀X : set, ∀Y : set → set, ∀z ∈ (∑x ∈ X, Y x), pair (proj0 z) (proj1 z) = z
In Proofgold the corresponding term root is 8b31cd... and proposition id is f30f76...
Proof:
Let X, Y and z be given.
Assume H1: z ∈ ∑x ∈ X, Y x.
Apply (and3E (pair (proj0 z) (proj1 z) = z) (proj0 z ∈ X) (proj1 z ∈ Y (proj0 z)) (Sigma_eta_proj0_proj1 X Y z H1)) to the current goal.
Assume H2 _ _.
An exact proof term for the current goal is H2.
∎
Theorem. (proj0_Sigma)
∀X : set, ∀Y : set → set, ∀z : set, z ∈ (∑x ∈ X, Y x) → proj0 z ∈ X
In Proofgold the corresponding term root is ae12a7... and proposition id is 506869...
Proof:
Let X, Y and z be given.
Assume H1: z ∈ ∑x ∈ X, Y x.
Apply (and3E (pair (proj0 z) (proj1 z) = z) (proj0 z ∈ X) (proj1 z ∈ Y (proj0 z)) (Sigma_eta_proj0_proj1 X Y z H1)) to the current goal.
Assume _ H2 _.
An exact proof term for the current goal is H2.
∎
Theorem. (proj1_Sigma)
∀X : set, ∀Y : set → set, ∀z : set, z ∈ (∑x ∈ X, Y x) → proj1 z ∈ Y (proj0 z)
In Proofgold the corresponding term root is 553db6... and proposition id is d79917...
Proof:
Let X, Y and z be given.
Assume H1: z ∈ ∑x ∈ X, Y x.
Apply (and3E (pair (proj0 z) (proj1 z) = z) (proj0 z ∈ X) (proj1 z ∈ Y (proj0 z)) (Sigma_eta_proj0_proj1 X Y z H1)) to the current goal.
Assume _ _ H2.
An exact proof term for the current goal is H2.
∎
Theorem. (pair_Sigma_E1)
∀X : set, ∀Y : set → set, ∀x y : set, pair x y ∈ (∑x ∈ X, Y x) → y ∈ Y x
In Proofgold the corresponding term root is c294a7... and proposition id is 81d8b0...
Proof:
Let X, Y, x and y be given.
Assume H1: pair x y ∈ ∑x ∈ X, Y x.
We will prove y ∈ Y x.
rewrite the current goal using (proj0_pair_eq x y) (from right to left).
We will prove y ∈ Y (proj0 (pair x y)).
rewrite the current goal using (proj1_pair_eq x y) (from right to left) at position 1.
We will prove proj1 (pair x y) ∈ Y (proj0 (pair x y)).
An exact proof term for the current goal is (proj1_Sigma X Y (pair x y) H1).
∎
Theorem. (Sigma_E)
∀X : set, ∀Y : set → set, ∀z : set, z ∈ (∑x ∈ X, Y x) → ∃x ∈ X, ∃y ∈ Y x, z = pair x y
In Proofgold the corresponding term root is 5b1bd2... and proposition id is 8ecfab...
Proof:
Let X, Y and z be given.
Assume Hz: z ∈ ∑x ∈ X, Y x.
Apply (and3E (pair (proj0 z) (proj1 z) = z) (proj0 z ∈ X) (proj1 z ∈ Y (proj0 z)) (Sigma_eta_proj0_proj1 X Y z Hz)) to the current goal.
Assume H1: pair (proj0 z) (proj1 z) = z.
Assume H2: proj0 z ∈ X.
Assume H3: proj1 z ∈ Y (proj0 z).
We use (proj0 z) to witness the existential quantifier.
Apply andI to the current goal.
We will prove proj0 z ∈ X.
An exact proof term for the current goal is H2.
We use (proj1 z) to witness the existential quantifier.
Apply andI to the current goal.
We will prove proj1 z ∈ Y (proj0 z).
An exact proof term for the current goal is H3.
We will prove z = pair (proj0 z) (proj1 z).
Use symmetry.
An exact proof term for the current goal is H1.
∎
Definition. We define setprod to be λX Y : set ⇒ ∑x ∈ X, Y of type set → set → set.
In Proofgold the corresponding term root is fc0b60... and object id is 107b6b...
Notation. We use ⨯ as an infix operator with priority 440 and which associates to the left corresponding to applying term setprod.
Let lam : set → (set → set) → set ≝ Sigma
Definition. We define ap to be λf x ⇒ {proj1 z|z ∈ f, ∃y : set, z = pair x y} of type set → set → set.
In Proofgold the corresponding term root is c7aaa6... and object id is a4d82c...
Notation. When x is a set, a term x y is notation for ap x y.
Notation. λ x ∈ A ⇒ B is notation for the set Sigma A (λ x : set ⇒ B).
Notation. We now use n-tuple notation (a0,...,an-1) for n ≥ 2 for λ i ∈ n . if i = 0 then a0 else ... if i = n-2 then an-2 else an-1.
Theorem. (lamI)
∀X : set, ∀F : set → set, ∀x ∈ X, ∀y ∈ F x, pair x y ∈ λx ∈ X ⇒ F x
In Proofgold the corresponding term root is 0d8645... and proposition id is 1456cb...
Proof:
An exact proof term for the current goal is pair_Sigma.
∎
Theorem. (lamE)
∀X : set, ∀F : set → set, ∀z : set, z ∈ (λx ∈ X ⇒ F x) → ∃x ∈ X, ∃y ∈ F x, z = pair x y
In Proofgold the corresponding term root is 5b1bd2... and proposition id is 8ecfab...
Proof:
An exact proof term for the current goal is Sigma_E.
∎
Theorem. (apI)
∀f x y, pair x y ∈ f → y ∈ f x
In Proofgold the corresponding term root is 64667c... and proposition id is 770816...
Proof:
Let f, x and y be given.
Assume H1: pair x y ∈ f.
We will prove y ∈ {proj1 z|z ∈ f, ∃y : set, z = pair x y}.
rewrite the current goal using (proj1_pair_eq x y) (from right to left).
We will prove proj1 (pair x y) ∈ {proj1 z|z ∈ f, ∃y : set, z = pair x y}.
Apply (ReplSepI f (λz ⇒ ∃y : set, z = pair x y) proj1 (pair x y) H1) to the current goal.
We will prove ∃y' : set, pair x y = pair x y'.
We use y to witness the existential quantifier.
Use reflexivity.
∎
Theorem. (apE)
∀f x y, y ∈ f x → pair x y ∈ f
In Proofgold the corresponding term root is 85693a... and proposition id is c08565...
Proof:
Let f, x and y be given.
Assume H1: y ∈ {proj1 z|z ∈ f, ∃y : set, z = pair x y}.
We will prove pair x y ∈ f.
Apply (ReplSepE_impred f (λz ⇒ ∃y : set, z = pair x y) proj1 y H1) to the current goal.
Let z be given.
Assume Hz: z ∈ f.
Assume H1: ∃y : set, z = pair x y.
Assume H2: y = proj1 z.
Apply H1 to the current goal.
Let v be given.
Assume H3: z = pair x v.
We prove the intermediate claim L1: y = v.
rewrite the current goal using H2 (from left to right).
rewrite the current goal using H3 (from left to right).
We will prove proj1 (pair x v) = v.
Apply proj1_pair_eq to the current goal.
We prove the intermediate claim L2: z = pair x y.
rewrite the current goal using L1 (from left to right).
An exact proof term for the current goal is H3.
rewrite the current goal using L2 (from right to left).
An exact proof term for the current goal is Hz.
∎
Theorem. (beta)
∀X : set, ∀F : set → set, ∀x : set, x ∈ X → (λx ∈ X ⇒ F x) x = F x
In Proofgold the corresponding term root is 95991d... and proposition id is b5182d...
Proof:
Let X, F and x be given.
Assume Hx: x ∈ X.
Apply set_ext to the current goal.
Let w be given.
Assume Hw: w ∈ (λx ∈ X ⇒ F x) x.
We prove the intermediate claim L1: pair x w ∈ (λx ∈ X ⇒ F x).
An exact proof term for the current goal is (apE (λx ∈ X ⇒ F x) x w Hw).
An exact proof term for the current goal is (pair_Sigma_E1 X F x w L1).
Let w be given.
Assume Hw: w ∈ F x.
We will prove w ∈ (λx ∈ X ⇒ F x) x.
Apply apI to the current goal.
We will prove pair x w ∈ λx ∈ X ⇒ F x.
We will prove pair x w ∈ ∑x ∈ X, F x.
Apply pair_Sigma to the current goal.
An exact proof term for the current goal is Hx.
An exact proof term for the current goal is Hw.
∎
Theorem. (proj0_ap_0)
∀u, proj0 u = u 0
In Proofgold the corresponding term root is 588f37... and proposition id is d4f351...
Proof:
Let u be given.
Apply set_ext to the current goal.
Let w be given.
Assume H1: w ∈ proj0 u.
We will prove w ∈ u 0.
Apply apI to the current goal.
We will prove pair 0 w ∈ u.
Apply proj0E to the current goal.
We will prove w ∈ proj0 u.
An exact proof term for the current goal is H1.
Let w be given.
Assume H1: w ∈ u 0.
We will prove w ∈ proj0 u.
Apply proj0I to the current goal.
We will prove pair 0 w ∈ u.
Apply apE to the current goal.
We will prove w ∈ u 0.
An exact proof term for the current goal is H1.
∎
Theorem. (proj1_ap_1)
∀u, proj1 u = u 1
In Proofgold the corresponding term root is d116b3... and proposition id is 2328b0...
Proof:
Let u be given.
Apply set_ext to the current goal.
Let w be given.
Assume H1: w ∈ proj1 u.
We will prove w ∈ u 1.
Apply apI to the current goal.
We will prove pair 1 w ∈ u.
Apply proj1E to the current goal.
We will prove w ∈ proj1 u.
An exact proof term for the current goal is H1.
Let w be given.
Assume H1: w ∈ u 1.
We will prove w ∈ proj1 u.
Apply proj1I to the current goal.
We will prove pair 1 w ∈ u.
Apply apE to the current goal.
We will prove w ∈ u 1.
An exact proof term for the current goal is H1.
∎
Theorem. (pair_ap_0)
∀x y : set, (pair x y) 0 = x
In Proofgold the corresponding term root is ac34b9... and proposition id is bb0477...
Proof:
Let x and y be given.
rewrite the current goal using (proj0_ap_0 (pair x y)) (from right to left).
We will prove proj0 (pair x y) = x.
Apply proj0_pair_eq to the current goal.
∎
Theorem. (pair_ap_1)
∀x y : set, (pair x y) 1 = y
In Proofgold the corresponding term root is 2274fc... and proposition id is cce3d2...
Proof:
Let x and y be given.
rewrite the current goal using (proj1_ap_1 (pair x y)) (from right to left).
We will prove proj1 (pair x y) = y.
Apply proj1_pair_eq to the current goal.
∎
Theorem. (ap0_Sigma)
∀X : set, ∀Y : set → set, ∀z : set, z ∈ (∑x ∈ X, Y x) → (z 0) ∈ X
In Proofgold the corresponding term root is 861a79... and proposition id is 1affcc...
Proof:
Let X, Y and z be given.
rewrite the current goal using proj0_ap_0 (from right to left).
Apply proj0_Sigma to the current goal.
∎
Theorem. (ap1_Sigma)
∀X : set, ∀Y : set → set, ∀z : set, z ∈ (∑x ∈ X, Y x) → (z 1) ∈ (Y (z 0))
In Proofgold the corresponding term root is ad3bfe... and proposition id is 520af5...
Proof:
Let X, Y and z be given.
rewrite the current goal using proj0_ap_0 (from right to left).
rewrite the current goal using proj1_ap_1 (from right to left).
Apply proj1_Sigma to the current goal.
∎
Definition. We define pair_p to be λu : set ⇒ pair (u 0) (u 1) = u of type set → prop.
In Proofgold the corresponding term root is ade2bb... and object id is 5c8a35...
Theorem. (pair_p_I)
∀x y, pair_p (pair x y)
In Proofgold the corresponding term root is 06d9a2... and proposition id is 8f2fde...
Proof:
Let x and y be given.
We will prove pair (pair x y 0) (pair x y 1) = pair x y.
rewrite the current goal using pair_ap_0 (from left to right).
rewrite the current goal using pair_ap_1 (from left to right).
Use reflexivity.
∎
Theorem. (Subq_2_UPair01)
2 ⊆ {0,1}
In Proofgold the corresponding term root is d90884... and proposition id is 94203e...
Proof:
Let x be given.
Assume H1: x ∈ 2.
Apply ordsuccE 1 x H1 to the current goal.
Assume H2: x ∈ 1.
We prove the intermediate claim L1: x = 0.
An exact proof term for the current goal is (SingE 0 x (Subq_1_Sing0 x H2)).
We will prove x ∈ {0,1}.
rewrite the current goal using L1 (from left to right).
We will prove 0 ∈ {0,1}.
An exact proof term for the current goal is (UPairI1 0 1).
Assume H2: x = 1.
We will prove x ∈ {0,1}.
rewrite the current goal using H2 (from left to right).
We will prove 1 ∈ {0,1}.
An exact proof term for the current goal is (UPairI2 0 1).
∎
Theorem. (tuple_pair)
∀x y : set, pair x y = (x,y)
In Proofgold the corresponding term root is c122f6... and proposition id is 18198e...
Proof:
Let x and y be given.
Apply set_ext to the current goal.
Let z be given.
Assume Hz: z ∈ pair x y.
Apply (pairE x y z Hz) to the current goal.
Assume H1: ∃u ∈ x, z = pair 0 u.
Apply (exandE_i (λu ⇒ u ∈ x) (λu ⇒ z = pair 0 u) H1) to the current goal.
Let u be given.
Assume Hu: u ∈ x.
Assume H2: z = pair 0 u.
We will prove z ∈ (x,y).
We will prove z ∈ λi ∈ 2 ⇒ if i = 0 then x else y.
rewrite the current goal using H2 (from left to right).
We will prove pair 0 u ∈ λi ∈ 2 ⇒ if i = 0 then x else y.
Apply (lamI 2 (λi ⇒ if i = 0 then x else y) 0 In_0_2 u) to the current goal.
We will prove u ∈ if 0 = 0 then x else y.
rewrite the current goal using (If_i_1 (0 = 0) x y (λq H ⇒ H)) (from left to right).
We will prove u ∈ x.
An exact proof term for the current goal is Hu.
Assume H1: ∃u ∈ y, z = pair 1 u.
Apply (exandE_i (λu ⇒ u ∈ y) (λu ⇒ z = pair 1 u) H1) to the current goal.
Let u be given.
Assume Hu: u ∈ y.
Assume H2: z = pair 1 u.
We will prove z ∈ (x,y).
We will prove z ∈ λi ∈ 2 ⇒ if i = 0 then x else y.
rewrite the current goal using H2 (from left to right).
We will prove pair 1 u ∈ λi ∈ 2 ⇒ if i = 0 then x else y.
Apply (lamI 2 (λi ⇒ if i = 0 then x else y) 1 In_1_2 u) to the current goal.
We will prove u ∈ if 1 = 0 then x else y.
rewrite the current goal using (If_i_0 (1 = 0) x y neq_1_0) (from left to right).
We will prove u ∈ y.
An exact proof term for the current goal is Hu.
Let z be given.
Assume Hz: z ∈ (x,y).
We will prove z ∈ pair x y.
We prove the intermediate claim L1: ∃i ∈ 2, ∃w ∈ (if i = 0 then x else y), z = pair i w.
An exact proof term for the current goal is (lamE 2 (λi ⇒ if i = 0 then x else y) z Hz).
Apply (exandE_i (λi ⇒ i ∈ 2) (λi ⇒ ∃w ∈ (if i = 0 then x else y), z = pair i w) L1) to the current goal.
Let i be given.
Assume Hi: i ∈ 2.
Assume H1: ∃w ∈ (if i = 0 then x else y), z = pair i w.
Apply (exandE_i (λw ⇒ w ∈ if i = 0 then x else y) (λw ⇒ z = pair i w) H1) to the current goal.
Let w be given.
Assume Hw: w ∈ if i = 0 then x else y.
Assume H2: z = pair i w.
We will prove z ∈ pair x y.
rewrite the current goal using H2 (from left to right).
We will prove pair i w ∈ pair x y.
We prove the intermediate claim L2: i ∈ {0,1}.
An exact proof term for the current goal is (Subq_2_UPair01 i Hi).
Apply (UPairE i 0 1 L2) to the current goal.
Assume Hi0: i = 0.
rewrite the current goal using Hi0 (from left to right).
We will prove pair 0 w ∈ pair x y.
Apply pairI0 to the current goal.
We will prove w ∈ x.
We prove the intermediate claim L3: (if i = 0 then x else y) = x.
An exact proof term for the current goal is (If_i_1 (i = 0) x y Hi0).
rewrite the current goal using L3 (from right to left).
An exact proof term for the current goal is Hw.
Assume Hi1: i = 1.
rewrite the current goal using Hi1 (from left to right).
We will prove pair 1 w ∈ pair x y.
Apply pairI1 to the current goal.
We will prove w ∈ y.
We prove the intermediate claim L3: (if i = 0 then x else y) = y.
rewrite the current goal using Hi1 (from left to right).
An exact proof term for the current goal is (If_i_0 (1 = 0) x y neq_1_0).
rewrite the current goal using L3 (from right to left).
An exact proof term for the current goal is Hw.
∎
Definition. We define Pi to be λX Y ⇒ {f ∈ 𝒫 (∑x ∈ X, ⋃ (Y x))|∀x ∈ X, f x ∈ Y x} of type set → (set → set) → set.
In Proofgold the corresponding term root is 40f1c3... and object id is a7b1b0...
Notation. We use ∏ x...y [possibly with ascriptions] , B as a binder notation corresponding to a term constructed using Pi.
Theorem. (PiI)
∀X : set, ∀Y : set → set, ∀f : set, (∀u ∈ f, pair_p u ∧ u 0 ∈ X) → (∀x ∈ X, f x ∈ Y x) → f ∈ ∏x ∈ X, Y x
In Proofgold the corresponding term root is 2b7ef4... and proposition id is 1f8f6f...
Proof:
Let X, Y and f be given.
Assume H1: ∀u ∈ f, pair_p u ∧ u 0 ∈ X.
Assume H2: ∀x ∈ X, f x ∈ Y x.
We will prove f ∈ {f ∈ 𝒫 (∑x ∈ X, ⋃ (Y x))|∀x ∈ X, f x ∈ Y x}.
Apply SepI to the current goal.
We will prove f ∈ 𝒫 (∑x ∈ X, ⋃ (Y x)).
Apply PowerI to the current goal.
We will prove f ⊆ ∑x ∈ X, ⋃ (Y x).
Let z be given.
Assume Hz: z ∈ f.
We will prove z ∈ ∑x ∈ X, ⋃ (Y x).
Apply (H1 z Hz) to the current goal.
Assume H3: pair (z 0) (z 1) = z.
Assume H4: z 0 ∈ X.
rewrite the current goal using H3 (from right to left).
We will prove pair (z 0) (z 1) ∈ ∑x ∈ X, ⋃ (Y x).
Apply pair_Sigma to the current goal.
We will prove z 0 ∈ X.
An exact proof term for the current goal is H4.
We will prove z 1 ∈ ⋃ (Y (z 0)).
Apply (UnionI (Y (z 0)) (z 1) (f (z 0))) to the current goal.
We will prove z 1 ∈ f (z 0).
Apply apI to the current goal.
We will prove pair (z 0) (z 1) ∈ f.
rewrite the current goal using H3 (from left to right).
An exact proof term for the current goal is Hz.
We will prove f (z 0) ∈ Y (z 0).
An exact proof term for the current goal is (H2 (z 0) H4).
We will prove ∀x ∈ X, f x ∈ Y x.
An exact proof term for the current goal is H2.
∎
Theorem. (lam_Pi)
∀X : set, ∀Y : set → set, ∀F : set → set, (∀x ∈ X, F x ∈ Y x) → (λx ∈ X ⇒ F x) ∈ (∏x ∈ X, Y x)
In Proofgold the corresponding term root is c0b88f... and proposition id is 3733f8...
Proof:
Let X, Y and F be given.
Assume H1: ∀x ∈ X, F x ∈ Y x.
We will prove (λx ∈ X ⇒ F x) ∈ (∏x ∈ X, Y x).
Apply PiI to the current goal.
We will prove ∀u ∈ (λx ∈ X ⇒ F x), pair_p u ∧ u 0 ∈ X.
Let u be given.
Assume Hu: u ∈ λx ∈ X ⇒ F x.
We prove the intermediate claim L1: ∃x ∈ X, ∃y ∈ F x, u = pair x y.
An exact proof term for the current goal is (lamE X F u Hu).
Apply (exandE_i (λx ⇒ x ∈ X) (λx ⇒ ∃y ∈ F x, u = pair x y) L1) to the current goal.
Let x be given.
Assume Hx: x ∈ X.
Assume H2: ∃y ∈ F x, u = pair x y.
Apply (exandE_i (λy ⇒ y ∈ F x) (λy ⇒ u = pair x y) H2) to the current goal.
Let y be given.
Assume Hy: y ∈ F x.
Assume H3: u = pair x y.
Apply andI to the current goal.
We will prove pair_p u.
rewrite the current goal using H3 (from left to right).
Apply pair_p_I to the current goal.
We will prove u 0 ∈ X.
rewrite the current goal using H3 (from left to right).
We will prove pair x y 0 ∈ X.
rewrite the current goal using pair_ap_0 (from left to right).
We will prove x ∈ X.
An exact proof term for the current goal is Hx.
We will prove ∀x ∈ X, (λx ∈ X ⇒ F x) x ∈ Y x.
Let x be given.
Assume Hx: x ∈ X.
rewrite the current goal using (beta X F x Hx) (from left to right).
We will prove F x ∈ Y x.
An exact proof term for the current goal is (H1 x Hx).
∎
Theorem. (ap_Pi)
∀X : set, ∀Y : set → set, ∀f : set, ∀x : set, f ∈ (∏x ∈ X, Y x) → x ∈ X → f x ∈ Y x
In Proofgold the corresponding term root is 7c3617... and proposition id is cb962e...
Proof:
Let X, Y, f and x be given.
Assume Hf: f ∈ ∏x ∈ X, Y x.
An exact proof term for the current goal is (SepE2 (𝒫 (∑x ∈ X, ⋃ (Y x))) (λf ⇒ ∀x ∈ X, f x ∈ Y x) f Hf x).
∎
Definition. We define setexp to be λX Y : set ⇒ ∏y ∈ Y, X of type set → set → set.
In Proofgold the corresponding term root is 1de7fc... and object id is 021ae0...
Notation. We use :^: as an infix operator with priority 430 and which associates to the left corresponding to applying term setexp.
Theorem. (pair_tuple_fun)
pair = (λx y ⇒ (x,y))
In Proofgold the corresponding term root is b02dfa... and proposition id is ed37d0...
Proof:
Apply func_ext set (set → set) to the current goal.
Let x be given.
Apply func_ext set set to the current goal.
Let y be given.
We will prove pair x y = (x,y).
Apply tuple_pair to the current goal.
∎
Theorem. (lamI2)
∀X, ∀F : set → set, ∀x ∈ X, ∀y ∈ F x, (x,y) ∈ λx ∈ X ⇒ F x
In Proofgold the corresponding term root is d4a9d9... and proposition id is de381f...
Proof:
We will prove (∀X, ∀F : set → set, ∀x ∈ X, ∀y ∈ F x, ((λx y : set ⇒ (x,y)) x y) ∈ λx ∈ X ⇒ F x).
rewrite the current goal using pair_tuple_fun (from right to left).
An exact proof term for the current goal is lamI.
∎
Beginning of Section Tuples
Variable x0 x1 : set
Theorem. (tuple_2_0_eq)
(x0,x1) 0 = x0
In Proofgold the corresponding term root is 42d452... and proposition id is ad9037...
Proof:
rewrite the current goal using beta 2 (λi ⇒ if i = 0 then x0 else x1) 0 In_0_2 (from left to right).
Apply If_i_1 to the current goal.
Use reflexivity.
∎
Theorem. (tuple_2_1_eq)
(x0,x1) 1 = x1
In Proofgold the corresponding term root is 596569... and proposition id is 2c0ad1...
Proof:
rewrite the current goal using beta 2 (λi ⇒ if i = 0 then x0 else x1) 1 In_1_2 (from left to right).
Apply If_i_0 to the current goal.
Apply neq_1_0 to the current goal.
∎
End of Section Tuples
Theorem. (ReplEq_setprod_ext)
∀X Y, ∀F G : set → set → set, (∀x ∈ X, ∀y ∈ Y, F x y = G x y) → {F (w 0) (w 1)|w ∈ X ⨯ Y} = {G (w 0) (w 1)|w ∈ X ⨯ Y}
In Proofgold the corresponding term root is c486ab... and proposition id is f689a0...
Proof:
Let X, Y, F and G be given.
Assume H1: ∀x ∈ X, ∀y ∈ Y, F x y = G x y.
Apply ReplEq_ext (X ⨯ Y) (λw ⇒ F (w 0) (w 1)) (λw ⇒ G (w 0) (w 1)) to the current goal.
We will prove ∀w ∈ X ⨯ Y, F (w 0) (w 1) = G (w 0) (w 1).
Let w be given.
Assume Hw: w ∈ X ⨯ Y.
Apply H1 to the current goal.
We will prove w 0 ∈ X.
An exact proof term for the current goal is ap0_Sigma X (λ_ ⇒ Y) w Hw.
We will prove w 1 ∈ Y.
An exact proof term for the current goal is ap1_Sigma X (λ_ ⇒ Y) w Hw.
∎
Theorem. (tuple_2_Sigma)
∀X : set, ∀Y : set → set, ∀x ∈ X, ∀y ∈ Y x, (x,y) ∈ ∑x ∈ X, Y x
In Proofgold the corresponding term root is d4a9d9... and proposition id is de381f...
Proof:
An exact proof term for the current goal is lamI2.
∎
Theorem. (tuple_2_setprod)
∀X : set, ∀Y : set, ∀x ∈ X, ∀y ∈ Y, (x,y) ∈ X ⨯ Y
In Proofgold the corresponding term root is e907b8... and proposition id is 1f9d7f...
Proof:
An exact proof term for the current goal is λX Y x Hx y Hy ⇒ tuple_2_Sigma X (λ_ ⇒ Y) x Hx y Hy.
∎
End of Section pair_setsum