Theorem qdassr 4711

Description: Two ways to write an unordered quadruple. (Contributed by Mario Carneiro, 5-Jan-2016.)

Assertion

Ref	Expression
qdassr	⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})

Proof of Theorem qdassr

Step	Hyp	Ref	Expression
1		unass 4124	. 2 ⊢ (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2		df-pr 4583	. . 3 ⊢ {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
3	2	uneq1i 4116	. 2 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4		tpass 4709	. . 3 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
5	4	uneq2i 4117	. 2 ⊢ ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
6	1, 3, 5	3eqtr4i 2769	1 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})

Syntax hints:   = wceq 1541   ∪ cun 3899  {csn 4580  {cpr 4582  {ctp 4584

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3or 1087  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2715  df-cleq 2728  df-clel 2811  df-v 3442  df-un 3906  df-sn 4581  df-pr 4583  df-tp 4585

This theorem is referenced by:  en4  9182  ex-pw  30504