Theorem qdass 4686

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 4096	. 2 ⊢ (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
2		df-tp 4563	. . 3 ⊢ {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
3	2	uneq1i 4089	. 2 ⊢ ({𝐴, 𝐵, 𝐶} ∪ {𝐷}) = (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷})
4		df-pr 4561	. . 3 ⊢ {𝐶, 𝐷} = ({𝐶} ∪ {𝐷})
5	4	uneq2i 4090	. 2 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
6	1, 3, 5	3eqtr4ri 2777	1 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴, 𝐵, 𝐶} ∪ {𝐷})

Syntax hints:   = wceq 1539   ∪ cun 3881  {csn 4558  {cpr 4560  {ctp 4562

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-v 3424  df-un 3888  df-pr 4561  df-tp 4563

This theorem is referenced by:  cnlmodlem1  24205  cnlmodlem2  24206  cnlmodlem3  24207  cnlmod4  24208  cnstrcvs  24210  ex-pw  28694