Theorem fneq1 4930

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 4864	. . 3 ⊢ (𝐹 = 𝐺 → (Fun 𝐹 ↔ Fun 𝐺))
2		dmeq 4478	. . . 4 ⊢ (𝐹 = 𝐺 → dom 𝐹 = dom 𝐺)
3	2	eqeq1d 2045	. . 3 ⊢ (𝐹 = 𝐺 → (dom 𝐹 = A ↔ dom 𝐺 = A))
4	1, 3	anbi12d 442	. 2 ⊢ (𝐹 = 𝐺 → ((Fun 𝐹 ∧ dom 𝐹 = A) ↔ (Fun 𝐺 ∧ dom 𝐺 = A)))
5		df-fn 4848	. 2 ⊢ (𝐹 Fn A ↔ (Fun 𝐹 ∧ dom 𝐹 = A))
6		df-fn 4848	. 2 ⊢ (𝐺 Fn A ↔ (Fun 𝐺 ∧ dom 𝐺 = A))
7	4, 5, 6	3bitr4g 212	1 ⊢ (𝐹 = 𝐺 → (𝐹 Fn A ↔ 𝐺 Fn A))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1242  dom cdm 4288  Fun wfun 4839   Fn wfn 4840

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-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-3an 886  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-un 2916  df-in 2918  df-ss 2925  df-sn 3373  df-pr 3374  df-op 3376  df-br 3756  df-opab 3810  df-rel 4295  df-cnv 4296  df-co 4297  df-dm 4298  df-fun 4847  df-fn 4848

This theorem is referenced by:  fneq1d  4932  fneq1i  4936  fn0  4961  feq1  4973  foeq1  5045  f1ocnv  5082  mpteqb  5204  eufnfv  5332  tfr0  5878  tfrlemiex  5886