Metamath Proof Explorer


Theorem endisj

Description: Any two sets are equinumerous to two disjoint sets. Exercise 4.39 of Mendelson p. 255. (Contributed by NM, 16-Apr-2004)

Ref Expression
Hypotheses endisj.1 ⊢ A ∈ V
endisj.2 ⊢ B ∈ V
Assertion endisj ⊢ ∃ x ∃ y x ≈ A ∧ y ≈ B ∧ x ∩ y = ∅

Proof

Step Hyp Ref Expression
1 endisj.1 ⊢ A ∈ V
2 endisj.2 ⊢ B ∈ V
3 0ex ⊢ ∅ ∈ V
4 1 3 xpsnen ⊢ A × ∅ ≈ A
5 1oex ⊢ 1 𝑜 ∈ V
6 2 5 xpsnen ⊢ B × 1 𝑜 ≈ B
7 4 6 pm3.2i ⊢ A × ∅ ≈ A ∧ B × 1 𝑜 ≈ B
8 xp01disj ⊢ A × ∅ ∩ B × 1 𝑜 = ∅
9 p0ex ⊢ ∅ ∈ V
10 1 9 xpex ⊢ A × ∅ ∈ V
11 snex ⊢ 1 𝑜 ∈ V
12 2 11 xpex ⊢ B × 1 𝑜 ∈ V
13 breq1 ⊢ x = A × ∅ → x ≈ A ↔ A × ∅ ≈ A
14 breq1 ⊢ y = B × 1 𝑜 → y ≈ B ↔ B × 1 𝑜 ≈ B
15 13 14 bi2anan9 ⊢ x = A × ∅ ∧ y = B × 1 𝑜 → x ≈ A ∧ y ≈ B ↔ A × ∅ ≈ A ∧ B × 1 𝑜 ≈ B
16 ineq12 ⊢ x = A × ∅ ∧ y = B × 1 𝑜 → x ∩ y = A × ∅ ∩ B × 1 𝑜
17 16 eqeq1d ⊢ x = A × ∅ ∧ y = B × 1 𝑜 → x ∩ y = ∅ ↔ A × ∅ ∩ B × 1 𝑜 = ∅
18 15 17 anbi12d ⊢ x = A × ∅ ∧ y = B × 1 𝑜 → x ≈ A ∧ y ≈ B ∧ x ∩ y = ∅ ↔ A × ∅ ≈ A ∧ B × 1 𝑜 ≈ B ∧ A × ∅ ∩ B × 1 𝑜 = ∅
19 10 12 18 spc2ev ⊢ A × ∅ ≈ A ∧ B × 1 𝑜 ≈ B ∧ A × ∅ ∩ B × 1 𝑜 = ∅ → ∃ x ∃ y x ≈ A ∧ y ≈ B ∧ x ∩ y = ∅
20 7 8 19 mp2an ⊢ ∃ x ∃ y x ≈ A ∧ y ≈ B ∧ x ∩ y = ∅