Theorem qdassr 3741

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 3338	. 2 ⊢ (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2		df-pr 3650	. . 3 ⊢ {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
3	2	uneq1i 3331	. 2 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4		tpass 3739	. . 3 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
5	4	uneq2i 3332	. 2 ⊢ ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
6	1, 3, 5	3eqtr4i 2238	1 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})

Syntax hints:   = wceq 1373   ∪ cun 3172  {csn 3643  {cpr 3644  {ctp 3645

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-3or 982  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2194  df-cleq 2200  df-clel 2203  df-nfc 2339  df-v 2778  df-un 3178  df-sn 3649  df-pr 3650  df-tp 3651

This theorem is referenced by: (None)