Theorem un12 4168

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 4154	. . 3 ⊢ (𝐴 ∪ 𝐵) = (𝐵 ∪ 𝐴)
2	1	uneq1i 4160	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐵 ∪ 𝐴) ∪ 𝐶)
3		unass 4167	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
4		unass 4167	. 2 ⊢ ((𝐵 ∪ 𝐴) ∪ 𝐶) = (𝐵 ∪ (𝐴 ∪ 𝐶))
5	2, 3, 4	3eqtr3i 2769	1 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))

Syntax hints:   = wceq 1542   ∪ cun 3947

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 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3477  df-un 3954

This theorem is referenced by:  un23  4169  un4  4170  fresaun  6763  unfi  9172  reconnlem1  24342  poimirlem6  36494  poimirlem7  36495  asindmre  36571  frege133d  42516