Theorem qdassr 3720

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 3320	. 2 ⊢ (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2		df-pr 3629	. . 3 ⊢ {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
3	2	uneq1i 3313	. 2 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4		tpass 3718	. . 3 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
5	4	uneq2i 3314	. 2 ⊢ ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
6	1, 3, 5	3eqtr4i 2227	1 ⊢ ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})

Syntax hints:   = wceq 1364   ∪ cun 3155  {csn 3622  {cpr 3623  {ctp 3624

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-3or 981  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-v 2765  df-un 3161  df-sn 3628  df-pr 3629  df-tp 3630

This theorem is referenced by: (None)