Theorem fneq2 5473

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 2248	. . 3 |- ( A = B -> ( dom F = A <-> dom F = B ) )
2	1	anbi2d 468	. 2 |- ( A = B -> ( ( Fun F /\ dom F = A ) <-> ( Fun F /\ dom F = B ) ) )
3		df-fn 5383	. 2 |- ( F Fn A <-> ( Fun F /\ dom F = A ) )
4		df-fn 5383	. 2 |- ( F Fn B <-> ( Fun F /\ dom F = B ) )
5	2, 3, 4	3bitr4g 223	1 |- ( A = B -> ( F Fn A <-> F Fn B ) )

This proof depends on syntax axioms:   -> wi 4   /\ wa 104   <-> wb 105   = wceq 1402   dom cdm 4776   Fun wfun 5374   Fn wfn 5375

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-4 1563  ax-17 1579  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-cleq 2231  df-fn 5383

This theorem is used by:  fneq2d  5475  fneq2i  5479  feq2  5520  foeq2  5615  f1o00  5679  eqfnfv2  5810  tfr0dm  6597  tfrlemisucaccv  6600  tfrlemi1  6607  tfrlemi14d  6608  tfrexlem  6609  tfr1onlemsucfn  6615  tfr1onlemsucaccv  6616  tfr1onlembxssdm  6618  tfr1onlembfn  6619  tfr1onlemaccex  6623  tfr1onlemres  6624  ixpeq1  6995  0fz1  10466