Theorem fneq12 6650

Description: Equality theorem for function predicate with domain. (Contributed by Thierry Arnoux, 31-Jan-2017.)

Assertion

Ref	Expression
fneq12	⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐵))

Proof of Theorem fneq12

Step	Hyp	Ref	Expression
1		simpl 482	. 2 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → 𝐹 = 𝐺)
2		simpr 484	. 2 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → 𝐴 = 𝐵)
3	1, 2	fneq12d 6649	1 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1534   Fn wfn 6543

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-3an 1087  df-tru 1537  df-fal 1547  df-ex 1775  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-rab 3430  df-v 3473  df-dif 3950  df-un 3952  df-in 3954  df-ss 3964  df-nul 4324  df-if 4530  df-sn 4630  df-pr 4632  df-op 4636  df-br 5149  df-opab 5211  df-rel 5685  df-cnv 5686  df-co 5687  df-dm 5688  df-fun 6550  df-fn 6551

This theorem is referenced by:  fprlem1  8306  tfrlem3a  8398  frrlem15  9781  hashresfn  14332