Theorem qdass 4691

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

Step	Hyp	Ref	Expression
1		unass 4144	. 2 ⊢ (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
2		df-tp 4574	. . 3 ⊢ {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
3	2	uneq1i 4137	. 2 ⊢ ({𝐴, 𝐵, 𝐶} ∪ {𝐷}) = (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷})
4		df-pr 4572	. . 3 ⊢ {𝐶, 𝐷} = ({𝐶} ∪ {𝐷})
5	4	uneq2i 4138	. 2 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
6	1, 3, 5	3eqtr4ri 2857	1 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵, 𝐶} ∪ {𝐷})

Syntax hints:   = wceq 1537   ∪ cun 3936  {csn 4569  {cpr 4571  {ctp 4573

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-v 3498  df-un 3943  df-pr 4572  df-tp 4574

This theorem is referenced by:  cnlmodlem1  23742  cnlmodlem2  23743  cnlmodlem3  23744  cnlmod4  23745  cnstrcvs  23747  ex-pw  28210