Theorem feq3 5052

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

Assertion

Ref	Expression
feq3	⊢ (𝐴 = 𝐵 → (𝐹:𝐶⟶𝐴 ↔ 𝐹:𝐶⟶𝐵))

Proof of Theorem feq3

Step	Hyp	Ref	Expression
1		sseq2 3021	. . 3 ⊢ (𝐴 = 𝐵 → (ran 𝐹 ⊆ 𝐴 ↔ ran 𝐹 ⊆ 𝐵))
2	1	anbi2d 451	. 2 ⊢ (𝐴 = 𝐵 → ((𝐹 Fn 𝐶 ∧ ran 𝐹 ⊆ 𝐴) ↔ (𝐹 Fn 𝐶 ∧ ran 𝐹 ⊆ 𝐵)))
3		df-f 4926	. 2 ⊢ (𝐹:𝐶⟶𝐴 ↔ (𝐹 Fn 𝐶 ∧ ran 𝐹 ⊆ 𝐴))
4		df-f 4926	. 2 ⊢ (𝐹:𝐶⟶𝐵 ↔ (𝐹 Fn 𝐶 ∧ ran 𝐹 ⊆ 𝐵))
5	2, 3, 4	3bitr4g 221	1 ⊢ (𝐴 = 𝐵 → (𝐹:𝐶⟶𝐴 ↔ 𝐹:𝐶⟶𝐵))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1284   ⊆ wss 2973  ran crn 4364   Fn wfn 4917  ⟶wf 4918

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-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  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-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-in 2979  df-ss 2986  df-f 4926

This theorem is referenced by:  feq23  5053  feq123d  5057  fun2  5084  fconstg  5103  f1eq3  5109  fsng  5357  fsn2  5358  fsnunf  5383