Theorem un12 4101

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 4087	. . 3 ⊢ (𝐴 ∪ 𝐵) = (𝐵 ∪ 𝐴)
2	1	uneq1i 4093	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐵 ∪ 𝐴) ∪ 𝐶)
3		unass 4100	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
4		unass 4100	. 2 ⊢ ((𝐵 ∪ 𝐴) ∪ 𝐶) = (𝐵 ∪ (𝐴 ∪ 𝐶))
5	2, 3, 4	3eqtr3i 2774	1 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))

Syntax hints:   = wceq 1539   ∪ cun 3885

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-un 3892

This theorem is referenced by:  un23  4102  un4  4103  fresaun  6645  unfi  8955  reconnlem1  23989  poimirlem6  35783  poimirlem7  35784  asindmre  35860  frege133d  41373