Theorem fntp 6599

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 6595	. 2 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → Fun {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩})
8	4, 5, 6	dmtpop 6221	. 2 ⊢ dom {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} = {𝐴, 𝐵, 𝐶}
9		df-fn 6541	. 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 601	1 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} Fn {𝐴, 𝐵, 𝐶})

Syntax hints:   → wi 4   ∧ w3a 1103   = wceq 1570   ∈ wcel 2143   ≠ wne 2958  Vcvv 3455  {ctp 4594  ⟨cop 4596  dom cdm 5663  Fun wfun 6532   Fn wfn 6533

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-12 2213  ax-ext 2735  ax-sep 5258  ax-pr 5406

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-mo 2567  df-clab 2742  df-cleq 2755  df-clel 2838  df-ne 2959  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4288  df-if 4489  df-sn 4591  df-pr 4593  df-tp 4595  df-op 4597  df-br 5111  df-opab 5175  df-id 5558  df-xp 5669  df-rel 5670  df-cnv 5671  df-co 5672  df-dm 5673  df-fun 6540  df-fn 6541

This theorem is referenced by:  fntpb  7209  rabren3dioph  43525