Theorem qdass 4288

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 3770	. 2 ⊢ (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
2		df-tp 4182	. . 3 ⊢ {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
3	2	uneq1i 3763	. 2 ⊢ ({𝐴, 𝐵, 𝐶} ∪ {𝐷}) = (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷})
4		df-pr 4180	. . 3 ⊢ {𝐶, 𝐷} = ({𝐶} ∪ {𝐷})
5	4	uneq2i 3764	. 2 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
6	1, 3, 5	3eqtr4ri 2655	1 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵, 𝐶} ∪ {𝐷})

Syntax hints:   = wceq 1483   ∪ cun 3572  {csn 4177  {cpr 4179  {ctp 4181

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-un 3579  df-pr 4180  df-tp 4182

This theorem is referenced by:  cnlmodlem1  22936  cnlmodlem2  22937  cnlmodlem3  22938  cnlmod4  22939  cnstrcvs  22941  ex-pw  27286