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Theorem qdassr 4656
Description: Two ways to write an unordered quadruple. (Contributed by Mario Carneiro, 5-Jan-2016.)
Assertion
Ref Expression
qdassr ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})

Proof of Theorem qdassr
StepHypRef Expression
1 unass 4066 . 2 (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2 df-pr 4530 . . 3 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
32uneq1i 4059 . 2 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4 tpass 4654 . . 3 {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
54uneq2i 4060 . 2 ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
61, 3, 53eqtr4i 2769 1 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})
Colors of variables: wff setvar class
Syntax hints:   = wceq 1543  cun 3851  {csn 4527  {cpr 4529  {ctp 4531
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-ext 2708
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-3or 1090  df-tru 1546  df-ex 1788  df-sb 2073  df-clab 2715  df-cleq 2728  df-clel 2809  df-v 3400  df-un 3858  df-sn 4528  df-pr 4530  df-tp 4532
This theorem is referenced by:  en4  8890  ex-pw  28466
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