Theorem fntp 6559

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 6555	. 2 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → Fun {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩})
8	4, 5, 6	dmtpop 6182	. 2 ⊢ dom {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} = {𝐴, 𝐵, 𝐶}
9		df-fn 6501	. 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 591	1 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} Fn {𝐴, 𝐵, 𝐶})

Syntax hints:   → wi 4   ∧ w3a 1087   = wceq 1542   ∈ wcel 2114   ≠ wne 2932  Vcvv 3429  {ctp 4571  ⟨cop 4573  dom cdm 5631  Fun wfun 6492   Fn wfn 6493

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-12 2185  ax-ext 2708  ax-sep 5231  ax-pr 5375

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-mo 2539  df-clab 2715  df-cleq 2728  df-clel 2811  df-ne 2933  df-ral 3052  df-rex 3062  df-rab 3390  df-v 3431  df-dif 3892  df-un 3894  df-in 3896  df-ss 3906  df-nul 4274  df-if 4467  df-sn 4568  df-pr 4570  df-tp 4572  df-op 4574  df-br 5086  df-opab 5148  df-id 5526  df-xp 5637  df-rel 5638  df-cnv 5639  df-co 5640  df-dm 5641  df-fun 6500  df-fn 6501

This theorem is referenced by:  fntpb  7164  rabren3dioph  43243