Theorem fneq2d 5475

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 5473	. 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 5375

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 5383

This theorem is used by:  fneq12d  5476  fncofn  5896  acfun  7569  ccfunen  7636  ccatlid  11395  ccatrid  11396  ccatass  11397  ccatswrd  11463  swrdccat2  11464  ccatpfx  11494  swrdswrd  11498  swrdccatin2  11522  pfxccatin12  11526  seq3shft  11624  ptex  13676  rng1zrlem  14364