Theorem un12 3998

Description: A rearrangement of union. (Contributed by NM, 12-Aug-2004.)

Assertion

Ref	Expression
un12	⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))

Proof of Theorem un12

Step	Hyp	Ref	Expression
1		uncom 3984	. . 3 ⊢ (𝐴 ∪ 𝐵) = (𝐵 ∪ 𝐴)
2	1	uneq1i 3990	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐵 ∪ 𝐴) ∪ 𝐶)
3		unass 3997	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
4		unass 3997	. 2 ⊢ ((𝐵 ∪ 𝐴) ∪ 𝐶) = (𝐵 ∪ (𝐴 ∪ 𝐶))
5	2, 3, 4	3eqtr3i 2857	1 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))

Syntax hints:   = wceq 1656   ∪ cun 3796

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908  ax-5 2009  ax-6 2075  ax-7 2112  ax-9 2173  ax-10 2192  ax-11 2207  ax-12 2220  ax-ext 2803

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 879  df-tru 1660  df-ex 1879  df-nf 1883  df-sb 2068  df-clab 2812  df-cleq 2818  df-clel 2821  df-nfc 2958  df-v 3416  df-un 3803

This theorem is referenced by:  un23  3999  un4  4000  fresaun  6312  reconnlem1  22999  poimirlem6  33952  poimirlem7  33953  asindmre  34031  frege133d  38891