Theorem fneq12 5898

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 472	. 2 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → 𝐹 = 𝐺)
2		simpr 476	. 2 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → 𝐴 = 𝐵)
3	1, 2	fneq12d 5897	1 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383   = wceq 1475   Fn wfn 5799

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-rab 2905  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-if 4037  df-sn 4126  df-pr 4128  df-op 4132  df-br 4584  df-opab 4644  df-rel 5045  df-cnv 5046  df-co 5047  df-dm 5048  df-fun 5806  df-fn 5807

This theorem is referenced by:  tfrlem3a  7360  hashresfn  12990