Theorem fneq2 5089

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 2097	. . 3 |- ( A = B -> ( dom F = A <-> dom F = B ) )
2	1	anbi2d 452	. 2 |- ( A = B -> ( ( Fun F /\ dom F = A ) <-> ( Fun F /\ dom F = B ) ) )
3		df-fn 5005	. 2 |- ( F Fn A <-> ( Fun F /\ dom F = A ) )
4		df-fn 5005	. 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 1289   dom cdm 4428   Fun wfun 4996   Fn wfn 4997

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 1381  ax-gen 1383  ax-4 1445  ax-17 1464  ax-ext 2070

This theorem depends on definitions:  df-bi 115  df-cleq 2081  df-fn 5005

This theorem is referenced by:  fneq2d  5091  fneq2i  5095  feq2  5132  foeq2  5214  f1o00  5272  eqfnfv2  5382  tfr0dm  6069  tfrlemisucaccv  6072  tfrlemi1  6079  tfrlemi14d  6080  tfrexlem  6081  tfr1onlemsucfn  6087  tfr1onlemsucaccv  6088  tfr1onlembxssdm  6090  tfr1onlembfn  6091  tfr1onlemaccex  6095  tfr1onlemres  6096  0fz1  9428