Theorem un23 4168

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 4166	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
2		un12 4167	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))
3		uncom 4153	. 2 ⊢ (𝐵 ∪ (𝐴 ∪ 𝐶)) = ((𝐴 ∪ 𝐶) ∪ 𝐵)
4	1, 2, 3	3eqtri 2763	1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐴 ∪ 𝐶) ∪ 𝐵)

Syntax hints:   = wceq 1540   ∪ cun 3946

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-v 3475  df-un 3953

This theorem is referenced by:  ssunpr  4835  setscom  17120  cycpmco2rn  32569  poimirlem6  36810  poimirlem7  36811  poimirlem16  36820  poimirlem19  36823  iocunico  42275  dfrcl2  42740