Theorem qdass 3820

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 3421	. 2 ⊢ (({A, B} ∪ {C}) ∪ {D}) = ({A, B} ∪ ({C} ∪ {D}))
2		df-tp 3744	. . 3 ⊢ {A, B, C} = ({A, B} ∪ {C})
3	2	uneq1i 3415	. 2 ⊢ ({A, B, C} ∪ {D}) = (({A, B} ∪ {C}) ∪ {D})
4		df-pr 3743	. . 3 ⊢ {C, D} = ({C} ∪ {D})
5	4	uneq2i 3416	. 2 ⊢ ({A, B} ∪ {C, D}) = ({A, B} ∪ ({C} ∪ {D}))
6	1, 3, 5	3eqtr4ri 2384	1 ⊢ ({A, B} ∪ {C, D}) = ({A, B, C} ∪ {D})

Syntax hints:   = wceq 1642   ∪ cun 3208  {csn 3738  {cpr 3739  {ctp 3740

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-un 3215  df-pr 3743  df-tp 3744

This theorem is referenced by: (None)