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Theorem qdass 4753
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 4172 . 2 (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
2 df-tp 4631 . . 3 {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
32uneq1i 4164 . 2 ({𝐴, 𝐵, 𝐶} ∪ {𝐷}) = (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷})
4 df-pr 4629 . . 3 {𝐶, 𝐷} = ({𝐶} ∪ {𝐷})
54uneq2i 4165 . 2 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
61, 3, 53eqtr4ri 2776 1 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵, 𝐶} ∪ {𝐷})
Colors of variables: wff setvar class
Syntax hints:   = wceq 1540  cun 3949  {csn 4626  {cpr 4628  {ctp 4630
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 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-v 3482  df-un 3956  df-pr 4629  df-tp 4631
This theorem is referenced by:  cnlmodlem1  25169  cnlmodlem2  25170  cnlmodlem3  25171  cnlmod4  25172  cnstrcvs  25174  ex-pw  30448
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