Theorem fneq12 6513

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 6512	1 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1539   Fn wfn 6413

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-br 5071  df-opab 5133  df-rel 5587  df-cnv 5588  df-co 5589  df-dm 5590  df-fun 6420  df-fn 6421

This theorem is referenced by:  fprlem1  8087  tfrlem3a  8179  frrlem15  9446  hashresfn  13982