Theorem feq2 5497

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 5450	. . 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 5361	. 2 ⊢ (𝐹:𝐴⟶𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ 𝐶))
4		df-f 5361	. 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 1398   ⊆ wss 3214  ran crn 4755   Fn wfn 5352  ⟶wf 5353

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 1496  ax-gen 1498  ax-4 1559  ax-17 1575  ax-ext 2216

This theorem depends on definitions:  df-bi 117  df-cleq 2227  df-fn 5360  df-f 5361

This theorem is referenced by:  feq23  5499  feq2d  5501  feq2i  5507  f00  5564  f0dom0  5566  f1eq2  5574  fressnfv  5876  tfrcllemsucfn  6597  tfrcllemsucaccv  6598  tfrcllembxssdm  6600  tfrcllembfn  6601  tfrcllemaccex  6605  tfrcllemres  6606  tfrcldm  6607  tfrcl  6608  mapvalg  6905  map0g  6935  ac6sfi  7168  isomni  7440  ismkv  7457  iswomni  7469  isghm  14044