Theorem fneq2d 5472

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 5470	. 2 |- ( A = B -> ( F Fn A <-> F Fn B ) )
3	1, 2	syl 14	1 |- ( ph -> ( F Fn A <-> F Fn B ) )

This proof depends on syntax axioms:   -> wi 4   <-> wb 105   = wceq 1402   Fn wfn 5372

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-4 1563  ax-17 1579  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-cleq 2231  df-fn 5380

This theorem is used by:  fneq12d  5473  fncofn  5893  acfun  7563  ccfunen  7630  ccatlid  11388  ccatrid  11389  ccatass  11390  ccatswrd  11456  swrdccat2  11457  ccatpfx  11487  swrdswrd  11491  swrdccatin2  11515  pfxccatin12  11519  seq3shft  11617  ptex  13667  rng1zrlem  14307