Theorem un4 4198

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 4196	. . 3 ⊢ (𝐵 ∪ (𝐶 ∪ 𝐷)) = (𝐶 ∪ (𝐵 ∪ 𝐷))
2	1	uneq2i 4188	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷))) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
3		unass 4195	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷)))
4		unass 4195	. 2 ⊢ ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷)) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
5	2, 3, 4	3eqtr4i 2778	1 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷))

Syntax hints:   = wceq 1537   ∪ cun 3974

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-v 3490  df-un 3981

This theorem is referenced by:  unundi  4199  unundir  4200  xpun  5773  resasplit  6791  addsdi  28199  mulsass  28210  ex-pw  30461  iunrelexp0  43664