Theorem feq2 5453

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 5406	. . 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 5318	. 2 ⊢ (𝐹:𝐴⟶𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶))
4		df-f 5318	. 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 1395   ⊆ wss 3197  ran crn 4717   Fn wfn 5309  ⟶wf 5310

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 1493  ax-gen 1495  ax-4 1556  ax-17 1572  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-cleq 2222  df-fn 5317  df-f 5318

This theorem is referenced by:  feq23  5455  feq2d  5457  feq2i  5463  f00  5513  f0dom0  5515  f1eq2  5523  fressnfv  5819  tfrcllemsucfn  6489  tfrcllemsucaccv  6490  tfrcllembxssdm  6492  tfrcllembfn  6493  tfrcllemaccex  6497  tfrcllemres  6498  tfrcldm  6499  tfrcl  6500  mapvalg  6795  map0g  6825  ac6sfi  7048  isomni  7291  ismkv  7308  iswomni  7320  isghm  13766