Theorem fneq12 6229

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 476	. 2 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → 𝐹 = 𝐺)
2		simpr 479	. 2 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → 𝐴 = 𝐵)
3	1, 2	fneq12d 6228	1 ⊢ ((𝐹 = 𝐺 ∧ 𝐴 = 𝐵) → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 386   = wceq 1601   Fn wfn 6130

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-9 2116  ax-10 2135  ax-11 2150  ax-12 2163  ax-13 2334  ax-ext 2754

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-3an 1073  df-tru 1605  df-ex 1824  df-nf 1828  df-sb 2012  df-clab 2764  df-cleq 2770  df-clel 2774  df-nfc 2921  df-rab 3099  df-v 3400  df-dif 3795  df-un 3797  df-in 3799  df-ss 3806  df-nul 4142  df-if 4308  df-sn 4399  df-pr 4401  df-op 4405  df-br 4887  df-opab 4949  df-rel 5362  df-cnv 5363  df-co 5364  df-dm 5365  df-fun 6137  df-fn 6138

This theorem is referenced by:  tfrlem3a  7756  hashresfn  13445