Theorem fneq1 5007

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 4941	. . 3 ⊢ (𝐹 = 𝐺 → (Fun 𝐹 ↔ Fun 𝐺))
2		dmeq 4553	. . . 4 ⊢ (𝐹 = 𝐺 → dom 𝐹 = dom 𝐺)
3	2	eqeq1d 2089	. . 3 ⊢ (𝐹 = 𝐺 → (dom 𝐹 = 𝐴 ↔ dom 𝐺 = 𝐴))
4	1, 3	anbi12d 456	. 2 ⊢ (𝐹 = 𝐺 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐺 ∧ dom 𝐺 = 𝐴)))
5		df-fn 4925	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
6		df-fn 4925	. 2 ⊢ (𝐺 Fn 𝐴 ↔ (Fun 𝐺 ∧ dom 𝐺 = 𝐴))
7	4, 5, 6	3bitr4g 221	1 ⊢ (𝐹 = 𝐺 → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1284  dom cdm 4363  Fun wfun 4916   Fn wfn 4917

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-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-3an 921  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-un 2977  df-in 2979  df-ss 2986  df-sn 3404  df-pr 3405  df-op 3407  df-br 3786  df-opab 3840  df-rel 4370  df-cnv 4371  df-co 4372  df-dm 4373  df-fun 4924  df-fn 4925

This theorem is referenced by:  fneq1d  5009  fneq1i  5013  fn0  5038  feq1  5050  foeq1  5122  f1ocnv  5159  mpteqb  5282  eufnfv  5410  tfr0  5960  tfrlemiex  5968