Theorem un23 4154

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 4152	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
2		un12 4153	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))
3		uncom 4138	. 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 3929

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810  df-v 3466  df-un 3936

This theorem is referenced by:  ssunpr  4815  setscom  17204  cycpmco2rn  33141  poimirlem6  37655  poimirlem7  37656  poimirlem16  37665  poimirlem19  37668  iocunico  43202  dfrcl2  43665