Metamath Proof Explorer


Theorem fneq12d

Description: Equality deduction for function predicate with domain. (Contributed by NM, 26-Jun-2011)

Ref Expression
Hypotheses fneq12d.1 ⊢ φ → F = G
fneq12d.2 ⊢ φ → A = B
Assertion fneq12d ⊢ φ → F Fn A ↔ G Fn B

Proof

Step Hyp Ref Expression
1 fneq12d.1 ⊢ φ → F = G
2 fneq12d.2 ⊢ φ → A = B
3 1 fneq1d ⊢ φ → F Fn A ↔ G Fn A
4 2 fneq2d ⊢ φ → G Fn A ↔ G Fn B
5 3 4 bitrd ⊢ φ → F Fn A ↔ G Fn B