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Theorem diftpsn3 4273
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 4196 . . . . 5 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∩ {𝐶}) = ∅)
2 disj3 3973 . . . . 5 (({𝐴, 𝐵} ∩ {𝐶}) = ∅ ↔ {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
31, 2sylib 207 . . . 4 ((𝐴𝐶𝐵𝐶) → {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
43eqcomd 2616 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∖ {𝐶}) = {𝐴, 𝐵})
5 difid 3902 . . . 4 ({𝐶} ∖ {𝐶}) = ∅
65a1i 11 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐶} ∖ {𝐶}) = ∅)
74, 6uneq12d 3730 . 2 ((𝐴𝐶𝐵𝐶) → (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵} ∪ ∅))
8 df-tp 4130 . . . 4 {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
98difeq1i 3686 . . 3 ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶})
10 difundir 3839 . . 3 (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶}) = (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶}))
119, 10eqtr2i 2633 . 2 (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵, 𝐶} ∖ {𝐶})
12 un0 3919 . 2 ({𝐴, 𝐵} ∪ ∅) = {𝐴, 𝐵}
137, 11, 123eqtr3g 2667 1 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 383   = wceq 1475  wne 2780  cdif 3537  cun 3538  cin 3539  c0 3874  {csn 4125  {cpr 4127  {ctp 4129
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590
This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-ral 2901  df-rab 2905  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-sn 4126  df-pr 4128  df-tp 4130
This theorem is referenced by:  f13dfv  6430  nb3graprlem2  25981  cusgra3v  25993  frgra3v  26529  3vfriswmgra  26532  signswch  29964  signstfvcl  29976  nb3grprlem2  40609  cplgr3v  40657  frgr3v  41445  3vfriswmgr  41448
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