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Theorem qdassr 4264
 Description: Two ways to write an unordered quadruple. (Contributed by Mario Carneiro, 5-Jan-2016.)
Assertion
Ref Expression
qdassr ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})

Proof of Theorem qdassr
StepHypRef Expression
1 unass 3753 . 2 (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2 df-pr 4156 . . 3 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
32uneq1i 3746 . 2 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4 tpass 4262 . . 3 {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
54uneq2i 3747 . 2 ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
61, 3, 53eqtr4i 2653 1 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})
 Colors of variables: wff setvar class Syntax hints:   = wceq 1480   ∪ cun 3557  {csn 4153  {cpr 4155  {ctp 4157 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3or 1037  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-v 3191  df-un 3564  df-sn 4154  df-pr 4156  df-tp 4158 This theorem is referenced by:  en4  8150  ex-pw  27157
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