Theorem fntp 6542

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 6538	. 2 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → Fun {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩})
8	4, 5, 6	dmtpop 6165	. 2 ⊢ dom {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} = {𝐴, 𝐵, 𝐶}
9		df-fn 6484	. 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 1541   ∈ wcel 2111   ≠ wne 2928  Vcvv 3436  {ctp 4577  ⟨cop 4579  dom cdm 5614  Fun wfun 6475   Fn wfn 6476

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-10 2144  ax-12 2180  ax-ext 2703  ax-sep 5232  ax-nul 5242  ax-pr 5368

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2068  df-mo 2535  df-clab 2710  df-cleq 2723  df-clel 2806  df-ne 2929  df-ral 3048  df-rex 3057  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-in 3904  df-ss 3914  df-nul 4281  df-if 4473  df-sn 4574  df-pr 4576  df-tp 4578  df-op 4580  df-br 5090  df-opab 5152  df-id 5509  df-xp 5620  df-rel 5621  df-cnv 5622  df-co 5623  df-dm 5624  df-fun 6483  df-fn 6484

This theorem is referenced by:  fntpb  7143  rabren3dioph  42918