Theorem fneq1i 5382

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 5376	. 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 1373   Fn wfn 5280

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-un 3174  df-in 3176  df-ss 3183  df-sn 3644  df-pr 3645  df-op 3647  df-br 4055  df-opab 4117  df-rel 4695  df-cnv 4696  df-co 4697  df-dm 4698  df-fun 5287  df-fn 5288

This theorem is referenced by:  fnunsn  5397  fnopabg  5414  f1oun  5559  f1oi  5578  f1osn  5580  ovid  6080  tfri1d  6439  frec2uzrand  10582  frec2uzf1od  10583  frecfzennn  10603  xnn0nnen  10614  prdsinvlem  13525  dfrelog  15417  edgstruct  15745  nninfsellemeqinf  16125