Theorem fneq1i 5374

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 5368	. 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 5272

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 3172  df-in 3174  df-ss 3181  df-sn 3641  df-pr 3642  df-op 3644  df-br 4049  df-opab 4111  df-rel 4687  df-cnv 4688  df-co 4689  df-dm 4690  df-fun 5279  df-fn 5280

This theorem is referenced by:  fnunsn  5389  fnopabg  5406  f1oun  5551  f1oi  5570  f1osn  5572  ovid  6072  tfri1d  6431  frec2uzrand  10563  frec2uzf1od  10564  frecfzennn  10584  xnn0nnen  10595  prdsinvlem  13490  dfrelog  15382  edgstruct  15710  nninfsellemeqinf  16068