Theorem feq23 5405

Description: Equality theorem for functions. (Contributed by FL, 14-Jul-2007.) (Proof shortened by Andrew Salmon, 17-Sep-2011.)

Assertion

Ref	Expression
feq23	⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐹:𝐴⟶𝐵 ↔ 𝐹:𝐶⟶𝐷))

Proof of Theorem feq23

Step	Hyp	Ref	Expression
1		feq2 5403	. 2 ⊢ (𝐴 = 𝐶 → (𝐹:𝐴⟶𝐵 ↔ 𝐹:𝐶⟶𝐵))
2		feq3 5404	. 2 ⊢ (𝐵 = 𝐷 → (𝐹:𝐶⟶𝐵 ↔ 𝐹:𝐶⟶𝐷))
3	1, 2	sylan9bb 462	1 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐹:𝐴⟶𝐵 ↔ 𝐹:𝐶⟶𝐷))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1372  ⟶wf 5264

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 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-11 1528  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-in 3171  df-ss 3178  df-fn 5271  df-f 5272

This theorem is referenced by:  feq23i  5414