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Theorem diftpsn3 4754
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 4667 . . . . 5 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∩ {𝐶}) = ∅)
2 disj3 4404 . . . . 5 (({𝐴, 𝐵} ∩ {𝐶}) = ∅ ↔ {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
31, 2sylib 218 . . . 4 ((𝐴𝐶𝐵𝐶) → {𝐴, 𝐵} = ({𝐴, 𝐵} ∖ {𝐶}))
43eqcomd 2737 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵} ∖ {𝐶}) = {𝐴, 𝐵})
5 difid 4326 . . . 4 ({𝐶} ∖ {𝐶}) = ∅
65a1i 11 . . 3 ((𝐴𝐶𝐵𝐶) → ({𝐶} ∖ {𝐶}) = ∅)
74, 6uneq12d 4119 . 2 ((𝐴𝐶𝐵𝐶) → (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵} ∪ ∅))
8 df-tp 4581 . . . 4 {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
98difeq1i 4072 . . 3 ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶})
10 difundir 4241 . . 3 (({𝐴, 𝐵} ∪ {𝐶}) ∖ {𝐶}) = (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶}))
119, 10eqtr2i 2755 . 2 (({𝐴, 𝐵} ∖ {𝐶}) ∪ ({𝐶} ∖ {𝐶})) = ({𝐴, 𝐵, 𝐶} ∖ {𝐶})
12 un0 4344 . 2 ({𝐴, 𝐵} ∪ ∅) = {𝐴, 𝐵}
137, 11, 123eqtr3g 2789 1 ((𝐴𝐶𝐵𝐶) → ({𝐴, 𝐵, 𝐶} ∖ {𝐶}) = {𝐴, 𝐵})
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1541  wne 2928  cdif 3899  cun 3900  cin 3901  c0 4283  {csn 4576  {cpr 4578  {ctp 4580
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ne 2929  df-ral 3048  df-rab 3396  df-v 3438  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4284  df-sn 4577  df-pr 4579  df-tp 4581
This theorem is referenced by:  f13dfv  7208  ex-chn2  18544  nb3grprlem2  29360  cplgr3v  29414  frgr3v  30253  3vfriswmgr  30256  signswch  34572  signstfvcl  34584
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