Theorem un23 4147

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

Step	Hyp	Ref	Expression
1		unass 4145	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
2		un12 4146	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))
3		uncom 4131	. 2 ⊢ (𝐵 ∪ (𝐴 ∪ 𝐶)) = ((𝐴 ∪ 𝐶) ∪ 𝐵)
4	1, 2, 3	3eqtri 2761	1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐴 ∪ 𝐶) ∪ 𝐵)

Syntax hints:   = wceq 1539   ∪ cun 3922

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1542  df-ex 1779  df-sb 2064  df-clab 2713  df-cleq 2726  df-clel 2808  df-v 3459  df-un 3929

This theorem is referenced by:  ssunpr  4807  setscom  17184  cycpmco2rn  33054  poimirlem6  37571  poimirlem7  37572  poimirlem16  37581  poimirlem19  37584  iocunico  43160  dfrcl2  43623