Theorem feq2 5099

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

Assertion

Ref	Expression
feq2	⊢ (𝐴 = 𝐵 → (𝐹:𝐴⟶𝐶 ↔ 𝐹:𝐵⟶𝐶))

Proof of Theorem feq2

Step	Hyp	Ref	Expression
1		fneq2 5056	. . 3 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))
2	1	anbi1d 453	. 2 ⊢ (𝐴 = 𝐵 → ((𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶) ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹 ⊆ 𝐶)))
3		df-f 4973	. 2 ⊢ (𝐹:𝐴⟶𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶))
4		df-f 4973	. 2 ⊢ (𝐹:𝐵⟶𝐶 ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹 ⊆ 𝐶))
5	2, 3, 4	3bitr4g 221	1 ⊢ (𝐴 = 𝐵 → (𝐹:𝐴⟶𝐶 ↔ 𝐹:𝐵⟶𝐶))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1285   ⊆ wss 2984  ran crn 4402   Fn wfn 4964  ⟶wf 4965

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 1377  ax-gen 1379  ax-4 1441  ax-17 1460  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-cleq 2076  df-fn 4972  df-f 4973

This theorem is referenced by:  feq23  5101  feq2d  5103  feq2i  5108  f00  5150  f0dom0  5152  f1eq2  5160  fressnfv  5426  tfrcllemsucfn  6050  tfrcllemsucaccv  6051  tfrcllembxssdm  6053  tfrcllembfn  6054  tfrcllemaccex  6058  tfrcllemres  6059  tfrcldm  6060  tfrcl  6061  mapvalg  6345  map0g  6375  ac6sfi  6544  isomni  6697