Theorem feq2 5132

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 5089	. . 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 5006	. 2 ⊢ (𝐹:𝐴⟶𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶))
4		df-f 5006	. 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 1289   ⊆ wss 2997  ran crn 4429   Fn wfn 4997  ⟶wf 4998

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 1381  ax-gen 1383  ax-4 1445  ax-17 1464  ax-ext 2070

This theorem depends on definitions:  df-bi 115  df-cleq 2081  df-fn 5005  df-f 5006

This theorem is referenced by:  feq23  5134  feq2d  5136  feq2i  5141  f00  5186  f0dom0  5188  f1eq2  5196  fressnfv  5468  tfrcllemsucfn  6100  tfrcllemsucaccv  6101  tfrcllembxssdm  6103  tfrcllembfn  6104  tfrcllemaccex  6108  tfrcllemres  6109  tfrcldm  6110  tfrcl  6111  mapvalg  6395  map0g  6425  ac6sfi  6594  isomni  6771