Metamath Proof Explorer

Theorem fneq2d

Description: Equality deduction for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011)

Ref Expression
Hypothesis fneq2d.1 
|- ( ph -> A = B )
Assertion fneq2d 
|- ( ph -> ( F Fn A <-> F Fn B ) )

Proof

Step Hyp Ref Expression
1 fneq2d.1 
 |-  ( ph -> A = B )
2 fneq2 
 |-  ( A = B -> ( F Fn A <-> F Fn B ) )
3 1 2 syl 
 |-  ( ph -> ( F Fn A <-> F Fn B ) )