Theorem feq2 5473

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 5426	. . 3 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))
2	1	anbi1d 465	. 2 ⊢ (𝐴 = 𝐵 → ((𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶) ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹 ⊆ 𝐶)))
3		df-f 5337	. 2 ⊢ (𝐹:𝐴⟶𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶))
4		df-f 5337	. 2 ⊢ (𝐹:𝐵⟶𝐶 ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹 ⊆ 𝐶))
5	2, 3, 4	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝐹:𝐴⟶𝐶 ↔ 𝐹:𝐵⟶𝐶))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1398   ⊆ wss 3201  ran crn 4732   Fn wfn 5328  ⟶wf 5329

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1496  ax-gen 1498  ax-4 1559  ax-17 1575  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-cleq 2224  df-fn 5336  df-f 5337

This theorem is referenced by:  feq23  5475  feq2d  5477  feq2i  5483  f00  5537  f0dom0  5539  f1eq2  5547  fressnfv  5849  tfrcllemsucfn  6562  tfrcllemsucaccv  6563  tfrcllembxssdm  6565  tfrcllembfn  6566  tfrcllemaccex  6570  tfrcllemres  6571  tfrcldm  6572  tfrcl  6573  mapvalg  6870  map0g  6900  ac6sfi  7130  isomni  7378  ismkv  7395  iswomni  7407  isghm  13891