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Theorem qdass 4688
Description: Two ways to write an unordered quadruple. (Contributed by Mario Carneiro, 5-Jan-2016.)
Assertion
Ref Expression
qdass ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵, 𝐶} ∪ {𝐷})

Proof of Theorem qdass
StepHypRef Expression
1 unass 4146 . 2 (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
2 df-tp 4569 . . 3 {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
32uneq1i 4139 . 2 ({𝐴, 𝐵, 𝐶} ∪ {𝐷}) = (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷})
4 df-pr 4567 . . 3 {𝐶, 𝐷} = ({𝐶} ∪ {𝐷})
54uneq2i 4140 . 2 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
61, 3, 53eqtr4ri 2860 1 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵, 𝐶} ∪ {𝐷})
Colors of variables: wff setvar class
Syntax hints:   = wceq 1530  cun 3938  {csn 4564  {cpr 4566  {ctp 4568
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1904  ax-6 1963  ax-7 2008  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2153  ax-12 2169  ax-ext 2798
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 844  df-tru 1533  df-ex 1774  df-nf 1778  df-sb 2063  df-clab 2805  df-cleq 2819  df-clel 2898  df-nfc 2968  df-v 3502  df-un 3945  df-pr 4567  df-tp 4569
This theorem is referenced by:  cnlmodlem1  23674  cnlmodlem2  23675  cnlmodlem3  23676  cnlmod4  23677  cnstrcvs  23679  ex-pw  28141
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