Metamath Proof Explorer


Theorem fneq12d

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

Ref Expression
Hypotheses fneq12d.1
|- ( ph -> F = G )
fneq12d.2
|- ( ph -> A = B )
Assertion fneq12d
|- ( ph -> ( F Fn A <-> G Fn B ) )

Proof

Step Hyp Ref Expression
1 fneq12d.1
 |-  ( ph -> F = G )
2 fneq12d.2
 |-  ( ph -> A = B )
3 1 fneq1d
 |-  ( ph -> ( F Fn A <-> G Fn A ) )
4 2 fneq2d
 |-  ( ph -> ( G Fn A <-> G Fn B ) )
5 3 4 bitrd
 |-  ( ph -> ( F Fn A <-> G Fn B ) )