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