Theorem fneq2 5008

Description: Equality theorem for function predicate with domain. (Contributed by NM, 1-Aug-1994.)

Assertion

Ref	Expression
fneq2	|- ( A = B -> ( F Fn A <-> F Fn B ) )

Proof of Theorem fneq2

Step	Hyp	Ref	Expression
1		eqeq2 2090	. . 3 |- ( A = B -> ( dom F = A <-> dom F = B ) )
2	1	anbi2d 451	. 2 |- ( A = B -> ( ( Fun F /\ dom F = A ) <-> ( Fun F /\ dom F = B ) ) )
3		df-fn 4925	. 2 |- ( F Fn A <-> ( Fun F /\ dom F = A ) )
4		df-fn 4925	. 2 |- ( F Fn B <-> ( Fun F /\ dom F = B ) )
5	2, 3, 4	3bitr4g 221	1 |- ( A = B -> ( F Fn A <-> F Fn B ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103   = wceq 1284   dom cdm 4363   Fun wfun 4916   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:  fneq2d  5010  fneq2i  5014  feq2  5051  foeq2  5123  f1o00  5181  eqfnfv2  5287  tfr0  5960  tfrlemisucaccv  5962  tfrlemi1  5969  tfrlemi14d  5970  tfrexlem  5971  0fz1  9064