Theorem diftpsn3 4827

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 4739	. . . . 5 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵} ∩ {𝐶}) = ∅)
2		disj3 4477	. . . . 5 ⊢ (({𝐴, 𝐵} ∩ {𝐶}) = ∅ ↔ {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
3	1, 2	sylib 218	. . . 4 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
4	3	eqcomd 2746	. . 3 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵} ∖ {𝐶}) = {𝐴, 𝐵})
5		difid 4398	. . . 4 ⊢ ({𝐶} ∖ {𝐶}) = ∅
6	5	a1i 11	. . 3 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐶} ∖ {𝐶}) = ∅)
7	4, 6	uneq12d 4192	. 2 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵} ∪ ∅))
8		df-tp 4653	. . . 4 ⊢ {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
9	8	difeq1i 4145	. . 3 ⊢ ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶})
10		difundir 4310	. . 3 ⊢ (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶}) = (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶}))
11	9, 10	eqtr2i 2769	. 2 ⊢ (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵, 𝐶} ∖ {𝐶})
12		un0 4417	. 2 ⊢ ({𝐴, 𝐵} ∪ ∅) = {𝐴, 𝐵}
13	7, 11, 12	3eqtr3g 2803	1 ⊢ ((𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1537   ≠ wne 2946   ∖ cdif 3973   ∪ cun 3974   ∩ cin 3975  ∅c0 4352  {csn 4648  {cpr 4650  {ctp 4652

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ne 2947  df-ral 3068  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-in 3983  df-ss 3993  df-nul 4353  df-sn 4649  df-pr 4651  df-tp 4653

This theorem is referenced by:  f13dfv  7310  nb3grprlem2  29416  cplgr3v  29470  frgr3v  30307  3vfriswmgr  30310  signswch  34538  signstfvcl  34550