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Theorem un23 3267
Description: A rearrangement of union. (Contributed by NM, 12-Aug-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Assertion
Ref Expression
un23 ((𝐴𝐵) ∪ 𝐶) = ((𝐴𝐶) ∪ 𝐵)

Proof of Theorem un23
StepHypRef Expression
1 unass 3265 . 2 ((𝐴𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵𝐶))
2 un12 3266 . 2 (𝐴 ∪ (𝐵𝐶)) = (𝐵 ∪ (𝐴𝐶))
3 uncom 3252 . 2 (𝐵 ∪ (𝐴𝐶)) = ((𝐴𝐶) ∪ 𝐵)
41, 2, 33eqtri 2182 1 ((𝐴𝐵) ∪ 𝐶) = ((𝐴𝐶) ∪ 𝐵)
Colors of variables: wff set class
Syntax hints:   = wceq 1335  cun 3100
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139
This theorem depends on definitions:  df-bi 116  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-v 2714  df-un 3106
This theorem is referenced by:  setscom  12326
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