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Theorem qdassr 4653
 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 4096 . 2 (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2 df-pr 4531 . . 3 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
32uneq1i 4089 . 2 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4 tpass 4651 . . 3 {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
54uneq2i 4090 . 2 ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
61, 3, 53eqtr4i 2834 1 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})
 Colors of variables: wff setvar class Syntax hints:   = wceq 1538   ∪ cun 3882  {csn 4528  {cpr 4530  {ctp 4532 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2114  ax-9 2122  ax-ext 2773 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3or 1085  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2780  df-cleq 2794  df-clel 2873  df-v 3446  df-un 3889  df-sn 4529  df-pr 4531  df-tp 4533 This theorem is referenced by:  en4  8744  ex-pw  28217
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