Theorem fneq1 5469

Description: Equality theorem for function predicate with domain. (Contributed by NM, 1-Aug-1994.)

Assertion

Ref	Expression
fneq1	|- ( F = G -> ( F Fn A <-> G Fn A ) )

Proof of Theorem fneq1

Step	Hyp	Ref	Expression
1		funeq 5397	. . 3 |- ( F = G -> ( Fun F <-> Fun G ) )
2		dmeq 4981	. . . 4 |- ( F = G -> dom F = dom G )
3	2	eqeq1d 2247	. . 3 |- ( F = G -> ( dom F = A <-> dom G = A ) )
4	1, 3	anbi12d 477	. 2 |- ( F = G -> ( ( Fun F /\ dom F = A ) <-> ( Fun G /\ dom G = A ) ) )
5		df-fn 5380	. 2 |- ( F Fn A <-> ( Fun F /\ dom F = A ) )
6		df-fn 5380	. 2 |- ( G Fn A <-> ( Fun G /\ dom G = A ) )
7	4, 5, 6	3bitr4g 223	1 |- ( F = G -> ( F Fn A <-> G Fn A ) )

This proof depends on syntax axioms:   -> wi 4   /\ wa 104   <-> wb 105   = wceq 1402   dom cdm 4774   Fun wfun 5371   Fn wfn 5372

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-un 3224  df-in 3226  df-ss 3233  df-sn 3715  df-pr 3716  df-op 3718  df-br 4131  df-opab 4193  df-rel 4781  df-cnv 4782  df-co 4783  df-dm 4784  df-fun 5379  df-fn 5380

This theorem is used by:  fneq1d  5471  fneq1i  5475  fn0  5503  feq1  5516  foeq1  5611  f1ocnv  5652  mpteqb  5796  eufnfv  5949  uchoice  6371  tfr0dm  6593  tfrlemiex  6602  tfr1onlemsucfn  6611  tfr1onlemsucaccv  6612  tfr1onlembxssdm  6614  tfr1onlembfn  6615  tfr1onlemex  6618  tfr1onlemaccex  6619  tfr1onlemres  6620  fsetdmprc0  6950  mapval2  6959  elixp2  6984  ixpfn  6986  elixpsn  7017  cc2lem  7632  cc3  7634  lmodfopnelem1  14661