Theorem feq2 5351

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 5307	. . 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 5222	. 2 ⊢ (𝐹:𝐴⟶𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶))
4		df-f 5222	. 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 1353   ⊆ wss 3131  ran crn 4629   Fn wfn 5213  ⟶wf 5214

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 1447  ax-gen 1449  ax-4 1510  ax-17 1526  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-cleq 2170  df-fn 5221  df-f 5222

This theorem is referenced by:  feq23  5353  feq2d  5355  feq2i  5361  f00  5409  f0dom0  5411  f1eq2  5419  fressnfv  5705  tfrcllemsucfn  6356  tfrcllemsucaccv  6357  tfrcllembxssdm  6359  tfrcllembfn  6360  tfrcllemaccex  6364  tfrcllemres  6365  tfrcldm  6366  tfrcl  6367  mapvalg  6660  map0g  6690  ac6sfi  6900  isomni  7136  ismkv  7153  iswomni  7165