Theorem fneq2d 5010

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 5008	. 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 103   = wceq 1284   Fn wfn 4917

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-4 1440  ax-17 1459  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-cleq 2074  df-fn 4925

This theorem is referenced by:  fneq12d  5011