Theorem fneq2d 5418

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 5416	. 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 1395   Fn wfn 5319

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 1493  ax-gen 1495  ax-4 1556  ax-17 1572  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-cleq 2222  df-fn 5327

This theorem is referenced by:  fneq12d  5419  fncofn  5827  acfun  7412  ccfunen  7473  ccatlid  11173  ccatrid  11174  ccatass  11175  ccatswrd  11241  swrdccat2  11242  ccatpfx  11272  swrdswrd  11276  swrdccatin2  11300  pfxccatin12  11304  seq3shft  11389  ptex  13337  srg1zr  13990