Theorem fneq1i 5418

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 5412	. 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 1395   Fn wfn 5316

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-3an 1004  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-un 3201  df-in 3203  df-ss 3210  df-sn 3672  df-pr 3673  df-op 3675  df-br 4084  df-opab 4146  df-rel 4727  df-cnv 4728  df-co 4729  df-dm 4730  df-fun 5323  df-fn 5324

This theorem is referenced by:  fnunsn  5433  fnopabg  5450  f1oun  5597  f1oi  5616  f1osn  5618  ovid  6130  tfri1d  6492  frec2uzrand  10644  frec2uzf1od  10645  frecfzennn  10665  xnn0nnen  10676  prdsinvlem  13662  dfrelog  15555  edgstruct  15885  nninfsellemeqinf  16496