Theorem un4 3287

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 3285	. . 3 ⊢ (𝐵 ∪ (𝐶 ∪ 𝐷)) = (𝐶 ∪ (𝐵 ∪ 𝐷))
2	1	uneq2i 3278	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷))) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
3		unass 3284	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = (𝐴 ∪ (𝐵 ∪ (𝐶 ∪ 𝐷)))
4		unass 3284	. 2 ⊢ ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷)) = (𝐴 ∪ (𝐶 ∪ (𝐵 ∪ 𝐷)))
5	2, 3, 4	3eqtr4i 2201	1 ⊢ ((𝐴 ∪ 𝐵) ∪ (𝐶 ∪ 𝐷)) = ((𝐴 ∪ 𝐶) ∪ (𝐵 ∪ 𝐷))

Syntax hints:   = wceq 1348   ∪ cun 3119

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-un 3125

This theorem is referenced by:  unundi  3288  unundir  3289  xpun  4672  resasplitss  5377