ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  qdassr GIF version

Theorem qdassr 3679
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 3284 . 2 (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2 df-pr 3588 . . 3 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
32uneq1i 3277 . 2 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4 tpass 3677 . . 3 {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
54uneq2i 3278 . 2 ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
61, 3, 53eqtr4i 2201 1 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})
Colors of variables: wff set class
Syntax hints:   = wceq 1348  cun 3119  {csn 3581  {cpr 3582  {ctp 3583
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-3or 974  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-un 3125  df-sn 3587  df-pr 3588  df-tp 3589
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator