Theorem fneq1i 5449

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 5443	. 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 5346

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 2214

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-v 2814  df-un 3214  df-in 3216  df-ss 3223  df-sn 3694  df-pr 3695  df-op 3697  df-br 4109  df-opab 4171  df-rel 4755  df-cnv 4756  df-co 4757  df-dm 4758  df-fun 5353  df-fn 5354

This theorem is referenced by:  fnunsn  5464  fnopabg  5481  f1oun  5633  f1oi  5653  f1osn  5655  ovid  6169  tfri1d  6565  frec2uzrand  10766  frec2uzf1od  10767  frecfzennn  10787  xnn0nnen  10798  prdsinvlem  13813  dfrelog  15717  edgstruct  16051  nninfsellemeqinf  16786