Theorem feq2 5321

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 5277	. . 3 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))
2	1	anbi1d 461	. 2 ⊢ (𝐴 = 𝐵 → ((𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶) ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹 ⊆ 𝐶)))
3		df-f 5192	. 2 ⊢ (𝐹:𝐴⟶𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶))
4		df-f 5192	. 2 ⊢ (𝐹:𝐵⟶𝐶 ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹 ⊆ 𝐶))
5	2, 3, 4	3bitr4g 222	1 ⊢ (𝐴 = 𝐵 → (𝐹:𝐴⟶𝐶 ↔ 𝐹:𝐵⟶𝐶))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1343   ⊆ wss 3116  ran crn 4605   Fn wfn 5183  ⟶wf 5184

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-gen 1437  ax-4 1498  ax-17 1514  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-cleq 2158  df-fn 5191  df-f 5192

This theorem is referenced by:  feq23  5323  feq2d  5325  feq2i  5331  f00  5379  f0dom0  5381  f1eq2  5389  fressnfv  5672  tfrcllemsucfn  6321  tfrcllemsucaccv  6322  tfrcllembxssdm  6324  tfrcllembfn  6325  tfrcllemaccex  6329  tfrcllemres  6330  tfrcldm  6331  tfrcl  6332  mapvalg  6624  map0g  6654  ac6sfi  6864  isomni  7100  ismkv  7117  iswomni  7129