Theorem feq2 5515

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 5468	. . 3 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))
2	1	anbi1d 469	. 2 ⊢ (𝐴 = 𝐵 → ((𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶) ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹 ⊆ 𝐶)))
3		df-f 5379	. 2 ⊢ (𝐹:𝐴⟶𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶))
4		df-f 5379	. 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 1402   ⊆ wss 3220  ran crn 4773   Fn wfn 5370  ⟶wf 5371

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 1500  ax-gen 1502  ax-4 1563  ax-17 1579  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-cleq 2231  df-fn 5378  df-f 5379

This theorem is referenced by:  feq23  5517  feq2d  5519  feq2i  5525  f00  5582  f0dom0  5584  f1eq2  5592  fressnfv  5896  tfrcllemsucfn  6618  tfrcllemsucaccv  6619  tfrcllembxssdm  6621  tfrcllembfn  6622  tfrcllemaccex  6626  tfrcllemres  6627  tfrcldm  6628  tfrcl  6629  mapvalg  6926  map0g  6963  ac6sfi  7196  isomni  7470  ismkv  7487  iswomni  7499  isghm  14029