Theorem fneq2 5410

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 2239	. . 3 |- ( A = B -> ( dom F = A <-> dom F = B ) )
2	1	anbi2d 464	. 2 |- ( A = B -> ( ( Fun F /\ dom F = A ) <-> ( Fun F /\ dom F = B ) ) )
3		df-fn 5321	. 2 |- ( F Fn A <-> ( Fun F /\ dom F = A ) )
4		df-fn 5321	. 2 |- ( F Fn B <-> ( Fun F /\ dom F = B ) )
5	2, 3, 4	3bitr4g 223	1 |- ( A = B -> ( F Fn A <-> F Fn B ) )

Syntax hints:   -> wi 4   /\ wa 104   <-> wb 105   = wceq 1395   dom cdm 4719   Fun wfun 5312   Fn wfn 5313

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 5321

This theorem is referenced by:  fneq2d  5412  fneq2i  5416  feq2  5457  foeq2  5545  f1o00  5608  eqfnfv2  5733  tfr0dm  6468  tfrlemisucaccv  6471  tfrlemi1  6478  tfrlemi14d  6479  tfrexlem  6480  tfr1onlemsucfn  6486  tfr1onlemsucaccv  6487  tfr1onlembxssdm  6489  tfr1onlembfn  6490  tfr1onlemaccex  6494  tfr1onlemres  6495  ixpeq1  6856  0fz1  10241