Theorem fneq1d

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

	Ref	Expression
Hypothesis	fneq1d.1	|- ( ph -> F = G )
Assertion	fneq1d	|- ( ph -> ( F Fn A <-> G Fn A ) )

Proof

Step	Hyp	Ref	Expression
1		fneq1d.1	|- ( ph -> F = G )
2		fneq1	|- ( F = G -> ( F Fn A <-> G Fn A ) )
3	1 2	syl	|- ( ph -> ( F Fn A <-> G Fn A ) )