Theorem fneq2 4988

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 2049	. . 3 |- ( A = B -> ( dom F = A <-> dom F = B ) )
2	1	anbi2d 437	. 2 |- ( A = B -> ( ( Fun F /\ dom F = A ) <-> ( Fun F /\ dom F = B ) ) )
3		df-fn 4905	. 2 |- ( F Fn A <-> ( Fun F /\ dom F = A ) )
4		df-fn 4905	. 2 |- ( F Fn B <-> ( Fun F /\ dom F = B ) )
5	2, 3, 4	3bitr4g 212	1 |- ( A = B -> ( F Fn A <-> F Fn B ) )

Syntax hints:   -> wi 4   /\ wa 97   <-> wb 98   = wceq 1243   dom cdm 4345   Fun wfun 4896   Fn wfn 4897

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-4 1400  ax-17 1419  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-cleq 2033  df-fn 4905

This theorem is referenced by:  fneq2d  4990  fneq2i  4994  feq2  5031  foeq2  5103  f1o00  5161  eqfnfv2  5266  tfr0  5937  tfrlemisucaccv  5939  tfrlemi1  5946  tfrlemi14d  5947  tfrexlem  5948  0fz1  8909