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Theorem qdass 3819
 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 3420 . 2 (({A, B} ∪ {C}) ∪ {D}) = ({A, B} ∪ ({C} ∪ {D}))
2 df-tp 3743 . . 3 {A, B, C} = ({A, B} ∪ {C})
32uneq1i 3414 . 2 ({A, B, C} ∪ {D}) = (({A, B} ∪ {C}) ∪ {D})
4 df-pr 3742 . . 3 {C, D} = ({C} ∪ {D})
54uneq2i 3415 . 2 ({A, B} ∪ {C, D}) = ({A, B} ∪ ({C} ∪ {D}))
61, 3, 53eqtr4ri 2384 1 ({A, B} ∪ {C, D}) = ({A, B, C} ∪ {D})
 Colors of variables: wff setvar class Syntax hints:   = wceq 1642   ∪ cun 3207  {csn 3737  {cpr 3738  {ctp 3739 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-nin 3211  df-compl 3212  df-un 3214  df-pr 3742  df-tp 3743 This theorem is referenced by: (None)
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