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Theorem diftpsn3 4735
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 4646 . . . . 5 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∩ {𝐶}) = ∅)
2 disj3 4382 . . . . 5 (({𝐴, 𝐵} ∩ {𝐶}) = ∅ ↔ {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
31, 2sylib 219 . . . 4 ((𝐴𝐶𝐵𝐶) → {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
43eqcomd 2745 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∖ {𝐶}) = {𝐴, 𝐵})
5 difid 4304 . . . 4 ({𝐶} ∖ {𝐶}) = ∅
65a1i 11 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐶} ∖ {𝐶}) = ∅)
74, 6uneq12d 4099 . 2 ((𝐴𝐶𝐵𝐶) → (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵} ∪ ∅))
8 df-tp 4560 . . . 4 {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
98difeq1i 4053 . . 3 ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶})
10 difundir 4219 . . 3 (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶}) = (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶}))
119, 10eqtr2i 2763 . 2 (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵, 𝐶} ∖ {𝐶})
12 un0 4322 . 2 ({𝐴, 𝐵} ∪ ∅) = {𝐴, 𝐵}
137, 11, 123eqtr3g 2797 1 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 396   = wceq 1547  wne 2934  cdif 3880  cun 3881  cin 3882  c0 4261  {csn 4555  {cpr 4557  {ctp 4559
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-ne 2935  df-ral 3054  df-rab 3392  df-v 3433  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4262  df-sn 4556  df-pr 4558  df-tp 4560
This theorem is referenced by:  f13dfv  7218  ex-chn2  18595  nb3grprlem2  29468  cplgr3v  29522  frgr3v  30363  3vfriswmgr  30366  signswch  34745  signstfvcl  34757
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