Theorem feq23 6690

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 6688	. 2 ⊢ (𝐴 = 𝐶 → (𝐹:𝐴⟶𝐵 ↔ 𝐹:𝐶⟶𝐵))
2		feq3 6689	. 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 1570  ⟶wf 6536

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2757  df-ss 3923  df-fn 6543  df-f 6544

This theorem is used by:  feq23i  6703  ismgmOLD  38534  ismndo2  38558  rngomndo  38619  seff  45052