Theorem fntp 6600

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 6596	. 2 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → Fun {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩})
8	4, 5, 6	dmtpop 6208	. 2 ⊢ dom {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} = {𝐴, 𝐵, 𝐶}
9		df-fn 6537	. 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 589	1 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → {⟨𝐴, 𝐷⟩, ⟨𝐵, 𝐸⟩, ⟨𝐶, 𝐹⟩} Fn {𝐴, 𝐵, 𝐶})

Syntax hints:   → wi 4   ∧ w3a 1084   = wceq 1533   ∈ wcel 2098   ≠ wne 2932  Vcvv 3466  {ctp 4625  ⟨cop 4627  dom cdm 5667  Fun wfun 6528   Fn wfn 6529

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-12 2163  ax-ext 2695  ax-sep 5290  ax-nul 5297  ax-pr 5418

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-mo 2526  df-clab 2702  df-cleq 2716  df-clel 2802  df-ne 2933  df-ral 3054  df-rex 3063  df-rab 3425  df-v 3468  df-dif 3944  df-un 3946  df-in 3948  df-ss 3958  df-nul 4316  df-if 4522  df-sn 4622  df-pr 4624  df-tp 4626  df-op 4628  df-br 5140  df-opab 5202  df-id 5565  df-xp 5673  df-rel 5674  df-cnv 5675  df-co 5676  df-dm 5677  df-fun 6536  df-fn 6537

This theorem is referenced by:  fntpb  7203  rabren3dioph  42067