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Theorem qdass 4718
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 4124 . 2 (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
2 df-tp 4593 . . 3 {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
32uneq1i 4117 . 2 ({𝐴, 𝐵, 𝐶} ∪ {𝐷}) = (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷})
4 df-pr 4591 . . 3 {𝐶, 𝐷} = ({𝐶} ∪ {𝐷})
54uneq2i 4118 . 2 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
61, 3, 53eqtr4ri 2795 1 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵, 𝐶} ∪ {𝐷})
Colors of variables: wff setvar class
Syntax hints:   = wceq 1568  cun 3902  {csn 4588  {cpr 4590  {ctp 4592
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2143  ax-9 2151  ax-ext 2733
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1571  df-ex 1808  df-sb 2095  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3455  df-un 3909  df-pr 4591  df-tp 4593
This theorem is referenced by:  cnlmodlem1  25274  cnlmodlem2  25275  cnlmodlem3  25276  cnlmod4  25277  cnstrcvs  25279  ex-pw  30746
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