Metamath Proof Explorer


Theorem en2prd

Description: Two proper unordered pairs are equinumerous. (Contributed by BTernaryTau, 23-Dec-2024)

Ref Expression
Hypotheses en2prd.1 ⊢ φ → A ∈ V
en2prd.2 ⊢ φ → B ∈ W
en2prd.3 ⊢ φ → C ∈ X
en2prd.4 ⊢ φ → D ∈ Y
en2prd.5 ⊢ φ → A ≠ B
en2prd.6 ⊢ φ → C ≠ D
Assertion en2prd ⊢ φ → A B ≈ C D

Proof

Step Hyp Ref Expression
1 en2prd.1 ⊢ φ → A ∈ V
2 en2prd.2 ⊢ φ → B ∈ W
3 en2prd.3 ⊢ φ → C ∈ X
4 en2prd.4 ⊢ φ → D ∈ Y
5 en2prd.5 ⊢ φ → A ≠ B
6 en2prd.6 ⊢ φ → C ≠ D
7 prex ⊢ A C B D ∈ V
8 f1oprg ⊢ A ∈ V ∧ C ∈ X ∧ B ∈ W ∧ D ∈ Y → A ≠ B ∧ C ≠ D → A C B D : A B ⟶ 1-1 onto C D
9 1 3 2 4 8 syl22anc ⊢ φ → A ≠ B ∧ C ≠ D → A C B D : A B ⟶ 1-1 onto C D
10 5 6 9 mp2and ⊢ φ → A C B D : A B ⟶ 1-1 onto C D
11 f1oeq1 ⊢ f = A C B D → f : A B ⟶ 1-1 onto C D ↔ A C B D : A B ⟶ 1-1 onto C D
12 11 spcegv ⊢ A C B D ∈ V → A C B D : A B ⟶ 1-1 onto C D → ∃ f f : A B ⟶ 1-1 onto C D
13 7 10 12 mpsyl ⊢ φ → ∃ f f : A B ⟶ 1-1 onto C D
14 prex ⊢ A B ∈ V
15 prex ⊢ C D ∈ V
16 breng ⊢ A B ∈ V ∧ C D ∈ V → A B ≈ C D ↔ ∃ f f : A B ⟶ 1-1 onto C D
17 14 15 16 mp2an ⊢ A B ≈ C D ↔ ∃ f f : A B ⟶ 1-1 onto C D
18 13 17 sylibr ⊢ φ → A B ≈ C D