Theorem un12 3280

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

Assertion

Ref	Expression
un12	⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))

Proof of Theorem un12

Step	Hyp	Ref	Expression
1		uncom 3266	. . 3 ⊢ (𝐴 ∪ 𝐵) = (𝐵 ∪ 𝐴)
2	1	uneq1i 3272	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐵 ∪ 𝐴) ∪ 𝐶)
3		unass 3279	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
4		unass 3279	. 2 ⊢ ((𝐵 ∪ 𝐴) ∪ 𝐶) = (𝐵 ∪ (𝐴 ∪ 𝐶))
5	2, 3, 4	3eqtr3i 2194	1 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))

Syntax hints:   = wceq 1343   ∪ cun 3114

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 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-v 2728  df-un 3120

This theorem is referenced by:  un23  3281  un4  3282