Theorem feq23 6693

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 6691	. 2 ⊢ (𝐴 = 𝐶 → (𝐹:𝐴⟶𝐵 ↔ 𝐹:𝐶⟶𝐵))
2		feq3 6692	. 2 ⊢ (𝐵 = 𝐷 → (𝐹:𝐶⟶𝐵 ↔ 𝐹:𝐶⟶𝐷))
3	1, 2	sylan9bb 519	1 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐹:𝐴⟶𝐵 ↔ 𝐹:𝐶⟶𝐷))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569  ⟶wf 6537

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-cleq 2752  df-ss 3897  df-fn 6544  df-f 6545

This theorem is used by:  feq23i  6706  ismgmOLD  38891  ismndo2  38915  rngomndo  38976  seff  45400