Theorem feq1 5110

Description: Equality theorem for functions. (Contributed by NM, 1-Aug-1994.)

Assertion

Ref	Expression
feq1	⊢ (𝐹 = 𝐺 → (𝐹:𝐴⟶𝐵 ↔ 𝐺:𝐴⟶𝐵))

Proof of Theorem feq1

Step	Hyp	Ref	Expression
1		fneq1 5067	. . 3 ⊢ (𝐹 = 𝐺 → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴))
2		rneq 4630	. . . 4 ⊢ (𝐹 = 𝐺 → ran 𝐹 = ran 𝐺)
3	2	sseq1d 3042	. . 3 ⊢ (𝐹 = 𝐺 → (ran 𝐹 ⊆ 𝐵 ↔ ran 𝐺 ⊆ 𝐵))
4	1, 3	anbi12d 457	. 2 ⊢ (𝐹 = 𝐺 → ((𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐵) ↔ (𝐺 Fn 𝐴 ∧ ran 𝐺 ⊆ 𝐵)))
5		df-f 4985	. 2 ⊢ (𝐹:𝐴⟶𝐵 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐵))
6		df-f 4985	. 2 ⊢ (𝐺:𝐴⟶𝐵 ↔ (𝐺 Fn 𝐴 ∧ ran 𝐺 ⊆ 𝐵))
7	4, 5, 6	3bitr4g 221	1 ⊢ (𝐹 = 𝐺 → (𝐹:𝐴⟶𝐵 ↔ 𝐺:𝐴⟶𝐵))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1287   ⊆ wss 2988  ran crn 4412   Fn wfn 4976  ⟶wf 4977

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 663  ax-5 1379  ax-7 1380  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-8 1438  ax-10 1439  ax-11 1440  ax-i12 1441  ax-bndl 1442  ax-4 1443  ax-17 1462  ax-i9 1466  ax-ial 1470  ax-i5r 1471  ax-ext 2067

This theorem depends on definitions:  df-bi 115  df-3an 924  df-tru 1290  df-nf 1393  df-sb 1690  df-clab 2072  df-cleq 2078  df-clel 2081  df-nfc 2214  df-v 2617  df-un 2992  df-in 2994  df-ss 3001  df-sn 3437  df-pr 3438  df-op 3440  df-br 3821  df-opab 3875  df-rel 4418  df-cnv 4419  df-co 4420  df-dm 4421  df-rn 4422  df-fun 4983  df-fn 4984  df-f 4985

This theorem is referenced by:  feq1d  5114  feq1i  5119  f00  5165  f0bi  5166  f0dom0  5167  fconstg  5170  f1eq1  5174  fconst2g  5473  tfrcllemsucfn  6072  tfrcllemsucaccv  6073  tfrcllembxssdm  6075  tfrcllembfn  6076  tfrcllemex  6079  tfrcllemaccex  6080  tfrcllemres  6081  tfrcl  6083  elmapg  6370  ac6sfi  6566  updjud  6717  finomni  6740  exmidomni  6742  1fv  9478