Theorem fneq2d 5299

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

Hypothesis

Ref	Expression
fneq2d.1	|- ( ph -> A = B )

Assertion

Ref	Expression
fneq2d	|- ( ph -> ( F Fn A <-> F Fn B ) )

Proof of Theorem fneq2d

Step	Hyp	Ref	Expression
1		fneq2d.1	. 2 |- ( ph -> A = B )
2		fneq2 5297	. 2 |- ( A = B -> ( F Fn A <-> F Fn B ) )
3	1, 2	syl 14	1 |- ( ph -> ( F Fn A <-> F Fn B ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1353   Fn wfn 5203

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1445  ax-gen 1447  ax-4 1508  ax-17 1524  ax-ext 2157

This theorem depends on definitions:  df-bi 117  df-cleq 2168  df-fn 5211

This theorem is referenced by:  fneq12d  5300  acfun  7196  ccfunen  7238  seq3shft  10815  srg1zr  12976