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Theorem diftpsn3 4804
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 4717 . . . . 5 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∩ {𝐶}) = ∅)
2 disj3 4452 . . . . 5 (({𝐴, 𝐵} ∩ {𝐶}) = ∅ ↔ {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
31, 2sylib 217 . . . 4 ((𝐴𝐶𝐵𝐶) → {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
43eqcomd 2736 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∖ {𝐶}) = {𝐴, 𝐵})
5 difid 4369 . . . 4 ({𝐶} ∖ {𝐶}) = ∅
65a1i 11 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐶} ∖ {𝐶}) = ∅)
74, 6uneq12d 4163 . 2 ((𝐴𝐶𝐵𝐶) → (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵} ∪ ∅))
8 df-tp 4632 . . . 4 {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
98difeq1i 4117 . . 3 ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶})
10 difundir 4279 . . 3 (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶}) = (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶}))
119, 10eqtr2i 2759 . 2 (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵, 𝐶} ∖ {𝐶})
12 un0 4389 . 2 ({𝐴, 𝐵} ∪ ∅) = {𝐴, 𝐵}
137, 11, 123eqtr3g 2793 1 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 394   = wceq 1539  wne 2938  cdif 3944  cun 3945  cin 3946  c0 4321  {csn 4627  {cpr 4629  {ctp 4631
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2701
This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-tru 1542  df-fal 1552  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-ne 2939  df-ral 3060  df-rab 3431  df-v 3474  df-dif 3950  df-un 3952  df-in 3954  df-ss 3964  df-nul 4322  df-sn 4628  df-pr 4630  df-tp 4632
This theorem is referenced by:  f13dfv  7274  nb3grprlem2  28905  cplgr3v  28959  frgr3v  29795  3vfriswmgr  29798  signswch  33870  signstfvcl  33882
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