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Theorem qdassr 4757
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 4165 . 2 (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2 df-pr 4630 . . 3 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
32uneq1i 4158 . 2 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4 tpass 4755 . . 3 {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
54uneq2i 4159 . 2 ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
61, 3, 53eqtr4i 2768 1 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})
Colors of variables: wff setvar class
Syntax hints:   = wceq 1539  cun 3945  {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-3or 1086  df-tru 1542  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-v 3474  df-un 3952  df-sn 4628  df-pr 4630  df-tp 4632
This theorem is referenced by:  en4  9285  ex-pw  29949
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