Theorem qdassr 4603

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 4069	. 2 ⊢ (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2		df-pr 4481	. . 3 ⊢ {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
3	2	uneq1i 4062	. 2 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4		tpass 4601	. . 3 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
5	4	uneq2i 4063	. 2 ⊢ ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
6	1, 3, 5	3eqtr4i 2831	1 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})

Syntax hints:   = wceq 1525   ∪ cun 3863  {csn 4478  {cpr 4480  {ctp 4482

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1781  ax-4 1795  ax-5 1892  ax-6 1951  ax-7 1996  ax-8 2085  ax-9 2093  ax-10 2114  ax-11 2128  ax-12 2143  ax-ext 2771

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-3or 1081  df-tru 1528  df-ex 1766  df-nf 1770  df-sb 2045  df-clab 2778  df-cleq 2790  df-clel 2865  df-nfc 2937  df-v 3442  df-un 3870  df-sn 4479  df-pr 4481  df-tp 4483

This theorem is referenced by:  en4  8609  ex-pw  27896