Theorem fneq1i 5455

Description: Equality inference for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011.)

Hypothesis

Ref	Expression
fneq1i.1	⊢ 𝐹 = 𝐺

Assertion

Ref	Expression
fneq1i	⊢ (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴)

Proof of Theorem fneq1i

Step	Hyp	Ref	Expression
1		fneq1i.1	. 2 ⊢ 𝐹 = 𝐺
2		fneq1 5449	. 2 ⊢ (𝐹 = 𝐺 → (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴))
3	1, 2	ax-mp 5	1 ⊢ (𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴)

Syntax hints:   ↔ wb 105   = wceq 1398   Fn wfn 5352

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2216

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-v 2817  df-un 3218  df-in 3220  df-ss 3227  df-sn 3700  df-pr 3701  df-op 3703  df-br 4115  df-opab 4177  df-rel 4761  df-cnv 4762  df-co 4763  df-dm 4764  df-fun 5359  df-fn 5360

This theorem is referenced by:  fnunsn  5470  fnopabg  5487  f1oun  5639  f1oi  5659  f1osn  5661  ovid  6178  tfri1d  6579  frec2uzrand  10791  frec2uzf1od  10792  frecfzennn  10812  xnn0nnen  10823  prdsinvlem  13905  dfrelog  15837  edgstruct  16171  nninfsellemeqinf  16906