Metamath Proof Explorer


Theorem elcnvlem

Description: Two ways to say a set is a member of the converse of a class. (Contributed by RP, 19-Aug-2020)

Ref Expression
Hypothesis elcnvlem.f ⊢ F = x ∈ V × V ⟼ 2 nd ⁡ x 1 st ⁡ x
Assertion elcnvlem ⊢ A ∈ B -1 ↔ A ∈ V × V ∧ F ⁡ A ∈ B

Proof

Step Hyp Ref Expression
1 elcnvlem.f ⊢ F = x ∈ V × V ⟼ 2 nd ⁡ x 1 st ⁡ x
2 elcnv2 ⊢ A ∈ B -1 ↔ ∃ u ∃ v A = u v ∧ v u ∈ B
3 fveq2 ⊢ A = u v → F ⁡ A = F ⁡ u v
4 vex ⊢ u ∈ V
5 vex ⊢ v ∈ V
6 4 5 opelvv ⊢ u v ∈ V × V
7 4 5 op2ndd ⊢ x = u v → 2 nd ⁡ x = v
8 4 5 op1std ⊢ x = u v → 1 st ⁡ x = u
9 7 8 opeq12d ⊢ x = u v → 2 nd ⁡ x 1 st ⁡ x = v u
10 opex ⊢ v u ∈ V
11 9 1 10 fvmpt ⊢ u v ∈ V × V → F ⁡ u v = v u
12 6 11 ax-mp ⊢ F ⁡ u v = v u
13 3 12 eqtrdi ⊢ A = u v → F ⁡ A = v u
14 13 eleq1d ⊢ A = u v → F ⁡ A ∈ B ↔ v u ∈ B
15 14 copsex2gb ⊢ ∃ u ∃ v A = u v ∧ v u ∈ B ↔ A ∈ V × V ∧ F ⁡ A ∈ B
16 2 15 bitri ⊢ A ∈ B -1 ↔ A ∈ V × V ∧ F ⁡ A ∈ B