Theorem fneq12 6675

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 6674	1 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1537   Fn wfn 6568

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-br 5167  df-opab 5229  df-rel 5707  df-cnv 5708  df-co 5709  df-dm 5710  df-fun 6575  df-fn 6576

This theorem is referenced by:  fprlem1  8341  tfrlem3a  8433  frrlem15  9826  hashresfn  14389