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Theorem diftpsn3 4698
 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 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})

Proof of Theorem diftpsn3
StepHypRef Expression
1 disjprsn 4613 . . . . 5 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∩ {𝐶}) = ∅)
2 disj3 4364 . . . . 5 (({𝐴, 𝐵} ∩ {𝐶}) = ∅ ↔ {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
31, 2sylib 221 . . . 4 ((𝐴𝐶𝐵𝐶) → {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
43eqcomd 2804 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∖ {𝐶}) = {𝐴, 𝐵})
5 difid 4287 . . . 4 ({𝐶} ∖ {𝐶}) = ∅
65a1i 11 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐶} ∖ {𝐶}) = ∅)
74, 6uneq12d 4094 . 2 ((𝐴𝐶𝐵𝐶) → (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵} ∪ ∅))
8 df-tp 4533 . . . 4 {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
98difeq1i 4049 . . 3 ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶})
10 difundir 4210 . . 3 (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶}) = (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶}))
119, 10eqtr2i 2822 . 2 (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵, 𝐶} ∖ {𝐶})
12 un0 4301 . 2 ({𝐴, 𝐵} ∪ ∅) = {𝐴, 𝐵}
137, 11, 123eqtr3g 2856 1 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 399   = wceq 1538   ≠ wne 2987   ∖ cdif 3880   ∪ cun 3881   ∩ cin 3882  ∅c0 4246  {csn 4528  {cpr 4530  {ctp 4532 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 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-ne 2988  df-ral 3111  df-rab 3115  df-v 3444  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4247  df-sn 4529  df-pr 4531  df-tp 4533 This theorem is referenced by:  f13dfv  7019  nb3grprlem2  27215  cplgr3v  27269  frgr3v  28104  3vfriswmgr  28107  signswch  32007  signstfvcl  32019
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