Theorem un4 4165

Description: A rearrangement of the union of 4 classes. (Contributed by NM, 12-Aug-2004.)

Assertion

Ref	Expression
un4	⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷))

Proof of Theorem un4

Step	Hyp	Ref	Expression
1		un12 4163	. . 3 ⊢ (𝐵 ∪ (𝐶 ∪ 𝐷)) = (𝐶 ∪ (𝐵 ∪ 𝐷))
2	1	uneq2i 4156	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷))) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
3		unass 4162	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷)))
4		unass 4162	. 2 ⊢ ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷)) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
5	2, 3, 4	3eqtr4i 2765	1 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷))

Syntax hints:   = wceq 1534   ∪ cun 3942

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2698

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-ex 1775  df-sb 2061  df-clab 2705  df-cleq 2719  df-clel 2805  df-v 3471  df-un 3949

This theorem is referenced by:  unundi  4166  unundir  4167  xpun  5745  resasplit  6761  addsdi  28048  mulsass  28059  ex-pw  30232  iunrelexp0  43104