Theorem qdass 4692

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 4108	. 2 ⊢ (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
2		df-tp 4567	. . 3 ⊢ {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
3	2	uneq1i 4101	. 2 ⊢ ({𝐴, 𝐵, 𝐶} ∪ {𝐷}) = (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷})
4		df-pr 4565	. . 3 ⊢ {𝐶, 𝐷} = ({𝐶} ∪ {𝐷})
5	4	uneq2i 4102	. 2 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
6	1, 3, 5	3eqtr4ri 2774	1 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵, 𝐶} ∪ {𝐷})

Syntax hints:   = wceq 1547   ∪ cun 3888  {csn 4562  {cpr 4564  {ctp 4566

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-v 3434  df-un 3895  df-pr 4565  df-tp 4567

This theorem is referenced by:  cnlmodlem1  25128  cnlmodlem2  25129  cnlmodlem3  25130  cnlmod4  25131  cnstrcvs  25133  ex-pw  30524