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Theorem qdass 3458
 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 3094 . 2 (({A, B} ∪ {𝐶}) ∪ {𝐷}) = ({A, B} ∪ ({𝐶} ∪ {𝐷}))
2 df-tp 3375 . . 3 {A, B, 𝐶} = ({A, B} ∪ {𝐶})
32uneq1i 3087 . 2 ({A, B, 𝐶} ∪ {𝐷}) = (({A, B} ∪ {𝐶}) ∪ {𝐷})
4 df-pr 3374 . . 3 {𝐶, 𝐷} = ({𝐶} ∪ {𝐷})
54uneq2i 3088 . 2 ({A, B} ∪ {𝐶, 𝐷}) = ({A, B} ∪ ({𝐶} ∪ {𝐷}))
61, 3, 53eqtr4ri 2068 1 ({A, B} ∪ {𝐶, 𝐷}) = ({A, B, 𝐶} ∪ {𝐷})
 Colors of variables: wff set class Syntax hints:   = wceq 1242   ∪ cun 2909  {csn 3367  {cpr 3368  {ctp 3369 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019 This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-un 2916  df-pr 3374  df-tp 3375 This theorem is referenced by: (None)
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