Theorem fneq2d 5365

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 5363	. 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 1373   Fn wfn 5266

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 1470  ax-gen 1472  ax-4 1533  ax-17 1549  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-cleq 2198  df-fn 5274

This theorem is referenced by:  fneq12d  5366  acfun  7319  ccfunen  7376  ccatlid  11062  ccatrid  11063  ccatass  11064  ccatswrd  11123  swrdccat2  11124  seq3shft  11149  ptex  13096  srg1zr  13749