Theorem qdass 4758

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 4182	. 2 ⊢ (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷}) = ({𝐴, 𝐵} ∪ ({𝐶} ∪ {𝐷}))
2		df-tp 4636	. . 3 ⊢ {𝐴, 𝐵, 𝐶} = ({𝐴, 𝐵} ∪ {𝐶})
3	2	uneq1i 4174	. 2 ⊢ ({𝐴, 𝐵, 𝐶} ∪ {𝐷}) = (({𝐴, 𝐵} ∪ {𝐶}) ∪ {𝐷})
4		df-pr 4634	. . 3 ⊢ {𝐶, 𝐷} = ({𝐶} ∪ {𝐷})
5	4	uneq2i 4175	. 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 1537   ∪ cun 3961  {csn 4631  {cpr 4633  {ctp 4635

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-v 3480  df-un 3968  df-pr 4634  df-tp 4636

This theorem is referenced by:  cnlmodlem1  25183  cnlmodlem2  25184  cnlmodlem3  25185  cnlmod4  25186  cnstrcvs  25188  ex-pw  30458