Theorem fneq2 5363

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 2215	. . 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 5274	. 2 |- ( F Fn A <-> ( Fun F /\ dom F = A ) )
4		df-fn 5274	. 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 1373   dom cdm 4675   Fun wfun 5265   Fn wfn 5266

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 1470  ax-gen 1472  ax-4 1533  ax-17 1549  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-cleq 2198  df-fn 5274

This theorem is referenced by:  fneq2d  5365  fneq2i  5369  feq2  5409  foeq2  5495  f1o00  5557  eqfnfv2  5678  tfr0dm  6408  tfrlemisucaccv  6411  tfrlemi1  6418  tfrlemi14d  6419  tfrexlem  6420  tfr1onlemsucfn  6426  tfr1onlemsucaccv  6427  tfr1onlembxssdm  6429  tfr1onlembfn  6430  tfr1onlemaccex  6434  tfr1onlemres  6435  ixpeq1  6796  0fz1  10167