Theorem fneq12 6453

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 486	. 2 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → 𝐹 = 𝐺)
2		simpr 488	. 2 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → 𝐴 = 𝐵)
3	1, 2	fneq12d 6452	1 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399   = wceq 1543   Fn wfn 6353

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-3an 1091  df-tru 1546  df-fal 1556  df-ex 1788  df-sb 2073  df-clab 2715  df-cleq 2728  df-clel 2809  df-rab 3060  df-v 3400  df-dif 3856  df-un 3858  df-in 3860  df-ss 3870  df-nul 4224  df-if 4426  df-sn 4528  df-pr 4530  df-op 4534  df-br 5040  df-opab 5102  df-rel 5543  df-cnv 5544  df-co 5545  df-dm 5546  df-fun 6360  df-fn 6361

This theorem is referenced by:  fprlem1  8019  tfrlem3a  8091  hashresfn  13871  frrlem15  33505