Theorem fneq2d 5384

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 5382	. 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 5285

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 1471  ax-gen 1473  ax-4 1534  ax-17 1550  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-cleq 2200  df-fn 5293

This theorem is referenced by:  fneq12d  5385  acfun  7350  ccfunen  7411  ccatlid  11100  ccatrid  11101  ccatass  11102  ccatswrd  11161  swrdccat2  11162  ccatpfx  11192  swrdswrd  11196  swrdccatin2  11220  pfxccatin12  11224  seq3shft  11264  ptex  13211  srg1zr  13864