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Theorem un23 4172
Description: A rearrangement of union. (Contributed by NM, 12-Aug-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
Assertion
Ref Expression
un23 ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐴 ∪ 𝐶) ∪ 𝐵)

Proof of Theorem un23
StepHypRef Expression
1 unass 4170 . 2 ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
2 un12 4171 . 2 (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))
3 uncom 4157 . 2 (𝐵 ∪ (𝐴 ∪ 𝐶)) = ((𝐴 ∪ 𝐶) ∪ 𝐵)
41, 2, 33eqtri 2787 1 ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐴 ∪ 𝐶) ∪ 𝐵)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   = wceq 1569   ∪ cun 4149
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-un 4151
This theorem is used by:  ssunpr  4788  setscom  17364  cycpmco2rn  33733  poimirlem6  38638  poimirlem7  38639  poimirlem16  38648  poimirlem19  38651  iocunico  44307  dfrcl2  44768
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