Theorem diftpsn3 4798

Description: Removal of a singleton from an unordered triple. (Contributed by Alexander van der Vekens, 5-Oct-2017.) (Proof shortened by JJ, 23-Jul-2021.)

Assertion

Ref	Expression
diftpsn3	⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})

Proof of Theorem diftpsn3

Step	Hyp	Ref	Expression
1		disjprsn 4711	. . . . 5 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵} ∩ {𝐶}) = ∅)
2		disj3 4449	. . . . 5 ⊢ (({𝐴, 𝐵} ∩ {𝐶}) = ∅ ↔ {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
3	1, 2	sylib 217	. . . 4 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
4	3	eqcomd 2737	. . 3 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵} ∖ {𝐶}) = {𝐴, 𝐵})
5		difid 4366	. . . 4 ⊢ ({𝐶} ∖ {𝐶}) = ∅
6	5	a1i 11	. . 3 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐶} ∖ {𝐶}) = ∅)
7	4, 6	uneq12d 4160	. 2 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵} ∪ ∅))
8		df-tp 4627	. . . 4 ⊢ {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
9	8	difeq1i 4114	. . 3 ⊢ ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶})
10		difundir 4276	. . 3 ⊢ (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶}) = (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶}))
11	9, 10	eqtr2i 2760	. 2 ⊢ (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵, 𝐶} ∖ {𝐶})
12		un0 4386	. 2 ⊢ ({𝐴, 𝐵} ∪ ∅) = {𝐴, 𝐵}
13	7, 11, 12	3eqtr3g 2794	1 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})

Syntax hints:   → wi 4   ∧ wa 396   = wceq 1541   ≠ wne 2939   ∖ cdif 3941   ∪ cun 3942   ∩ cin 3943  ∅c0 4318  {csn 4622  {cpr 4624  {ctp 4626

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 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-fal 1554  df-ex 1782  df-sb 2068  df-clab 2709  df-cleq 2723  df-clel 2809  df-ne 2940  df-ral 3061  df-rab 3432  df-v 3475  df-dif 3947  df-un 3949  df-in 3951  df-ss 3961  df-nul 4319  df-sn 4623  df-pr 4625  df-tp 4627

This theorem is referenced by:  f13dfv  7256  nb3grprlem2  28503  cplgr3v  28557  frgr3v  29393  3vfriswmgr  29396  signswch  33403  signstfvcl  33415