Theorem diftpsn3 4770

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 4680	. . . . 5 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵} ∩ {𝐶}) = ∅)
2		disj3 4414	. . . . 5 ⊢ (({𝐴, 𝐵} ∩ {𝐶}) = ∅ ↔ {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
3	1, 2	sylib 221	. . . 4 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
4	3	eqcomd 2769	. . 3 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵} ∖ {𝐶}) = {𝐴, 𝐵})
5		difid 4332	. . . 4 ⊢ ({𝐶} ∖ {𝐶}) = ∅
6	5	a1i 11	. . 3 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐶} ∖ {𝐶}) = ∅)
7	4, 6	uneq12d 4123	. 2 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵} ∪ ∅))
8		df-tp 4594	. . . 4 ⊢ {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
9	8	difeq1i 4077	. . 3 ⊢ ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶})
10		difundir 4244	. . 3 ⊢ (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶}) = (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶}))
11	9, 10	eqtr2i 2787	. 2 ⊢ (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵, 𝐶} ∖ {𝐶})
12		un0 4351	. 2 ⊢ ({𝐴, 𝐵} ∪ ∅) = {𝐴, 𝐵}
13	7, 11, 12	3eqtr3g 2821	1 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})

Syntax hints:   → wi 4   ∧ wa 400   = wceq 1570   ≠ wne 2958   ∖ cdif 3902   ∪ cun 3903   ∩ cin 3904  ∅c0 4286  {csn 4589  {cpr 4591  {ctp 4593

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-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ne 2959  df-ral 3080  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-sn 4590  df-pr 4592  df-tp 4594

This theorem is referenced by:  f13dfv  7272  ex-chn2  18689  nb3grprlem2  29731  cplgr3v  29785  frgr3v  30626  3vfriswmgr  30629  signswch  34948  signstfvcl  34960