Theorem un4 3908

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 3906	. . 3 ⊢ (𝐵 ∪ (𝐶 ∪ 𝐷)) = (𝐶 ∪ (𝐵 ∪ 𝐷))
2	1	uneq2i 3899	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷))) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
3		unass 3905	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷)))
4		unass 3905	. 2 ⊢ ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷)) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
5	2, 3, 4	3eqtr4i 2784	1 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷))

Syntax hints:   = wceq 1624   ∪ cun 3705

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1863  ax-4 1878  ax-5 1980  ax-6 2046  ax-7 2082  ax-9 2140  ax-10 2160  ax-11 2175  ax-12 2188  ax-13 2383  ax-ext 2732

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1627  df-ex 1846  df-nf 1851  df-sb 2039  df-clab 2739  df-cleq 2745  df-clel 2748  df-nfc 2883  df-v 3334  df-un 3712

This theorem is referenced by:  unundi  3909  unundir  3910  xpun  5325  resasplit  6227  ex-pw  27589  iunrelexp0  38488