Theorem qdassr 3493

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

Step	Hyp	Ref	Expression
1		unass 3125	. 2 ⊢ (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2		df-pr 3407	. . 3 ⊢ {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
3	2	uneq1i 3118	. 2 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4		tpass 3491	. . 3 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
5	4	uneq2i 3119	. 2 ⊢ ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
6	1, 3, 5	3eqtr4i 2084	1 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})

Syntax hints:   = wceq 1257   ∪ cun 2940  {csn 3400  {cpr 3401  {ctp 3402

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 638  ax-5 1350  ax-7 1351  ax-gen 1352  ax-ie1 1396  ax-ie2 1397  ax-8 1409  ax-10 1410  ax-11 1411  ax-i12 1412  ax-bndl 1413  ax-4 1414  ax-17 1433  ax-i9 1437  ax-ial 1441  ax-i5r 1442  ax-ext 2036

This theorem depends on definitions:  df-bi 114  df-3or 895  df-tru 1260  df-nf 1364  df-sb 1660  df-clab 2041  df-cleq 2047  df-clel 2050  df-nfc 2181  df-v 2574  df-un 2947  df-sn 3406  df-pr 3407  df-tp 3408

This theorem is referenced by: (None)