Theorem feq2 5214

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 5170	. . 3 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))
2	1	anbi1d 458	. 2 ⊢ (𝐴 = 𝐵 → ((𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶) ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹 ⊆ 𝐶)))
3		df-f 5085	. 2 ⊢ (𝐹:𝐴⟶𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶))
4		df-f 5085	. 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 1314   ⊆ wss 3037  ran crn 4500   Fn wfn 5076  ⟶wf 5077

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1406  ax-gen 1408  ax-4 1470  ax-17 1489  ax-ext 2097

This theorem depends on definitions:  df-bi 116  df-cleq 2108  df-fn 5084  df-f 5085

This theorem is referenced by:  feq23  5216  feq2d  5218  feq2i  5224  f00  5272  f0dom0  5274  f1eq2  5282  fressnfv  5561  tfrcllemsucfn  6204  tfrcllemsucaccv  6205  tfrcllembxssdm  6207  tfrcllembfn  6208  tfrcllemaccex  6212  tfrcllemres  6213  tfrcldm  6214  tfrcl  6215  mapvalg  6506  map0g  6536  ac6sfi  6745  isomni  6958  ismkv  6977