Theorem fntp 6629

Description: A function with a domain of three elements. (Contributed by NM, 14-Sep-2011.) (Revised by Mario Carneiro, 26-Apr-2015.)

Hypotheses

Ref	Expression
fntp.1	⊢ 𝐴 ∈ V
fntp.2	⊢ 𝐵 ∈ V
fntp.3	⊢ 𝐶 ∈ V
fntp.4	⊢ 𝐷 ∈ V
fntp.5	⊢ 𝐸 ∈ V
fntp.6	⊢ 𝐹 ∈ V

Assertion

Ref	Expression
fntp	⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} Fn {𝐴, 𝐵, 𝐶})

Proof of Theorem fntp

Step	Hyp	Ref	Expression
1		fntp.1	. . 3 ⊢ 𝐴 ∈ V
2		fntp.2	. . 3 ⊢ 𝐵 ∈ V
3		fntp.3	. . 3 ⊢ 𝐶 ∈ V
4		fntp.4	. . 3 ⊢ 𝐷 ∈ V
5		fntp.5	. . 3 ⊢ 𝐸 ∈ V
6		fntp.6	. . 3 ⊢ 𝐹 ∈ V
7	1, 2, 3, 4, 5, 6	funtp 6625	. 2 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → Fun {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩})
8	4, 5, 6	dmtpop 6240	. 2 ⊢ dom {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} = {𝐴, 𝐵, 𝐶}
9		df-fn 6566	. 2 ⊢ ({⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} Fn {𝐴, 𝐵, 𝐶} ↔ (Fun {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} ∧ dom {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} = {𝐴, 𝐵, 𝐶}))
10	7, 8, 9	sylanblrc 590	1 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} Fn {𝐴, 𝐵, 𝐶})

Syntax hints:   → wi 4   ∧ w3a 1086   = wceq 1537   ∈ wcel 2106   ≠ wne 2938  Vcvv 3478  {ctp 4635  ⟨cop 4637  dom cdm 5689  Fun wfun 6557   Fn wfn 6558

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-10 2139  ax-12 2175  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pr 5438

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-nf 1781  df-sb 2063  df-mo 2538  df-clab 2713  df-cleq 2727  df-clel 2814  df-ne 2939  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-tp 4636  df-op 4638  df-br 5149  df-opab 5211  df-id 5583  df-xp 5695  df-rel 5696  df-cnv 5697  df-co 5698  df-dm 5699  df-fun 6565  df-fn 6566

This theorem is referenced by:  fntpb  7229  rabren3dioph  42803