Theorem un4 4176

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 4174	. . 3 ⊢ (𝐵 ∪ (𝐶 ∪ 𝐷)) = (𝐶 ∪ (𝐵 ∪ 𝐷))
2	1	uneq2i 4167	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷))) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
3		unass 4173	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷)))
4		unass 4173	. 2 ⊢ ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷)) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
5	2, 3, 4	3eqtr4i 2796	1 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷))

This proof depends on syntax axioms:   = wceq 1570   ∪ cun 4152

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-un 4154

This theorem is used by:  unundi  4177  unundir  4178  xpun  5730  resasplit  6777  addsdi  28574  mulsass  28585  ex-pw  31063  ttcun  37491  iunrelexp0  44905