Theorem un12 3305

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 3291	. . 3 ⊢ (𝐴 ∪ 𝐵) = (𝐵 ∪ 𝐴)
2	1	uneq1i 3297	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐵 ∪ 𝐴) ∪ 𝐶)
3		unass 3304	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
4		unass 3304	. 2 ⊢ ((𝐵 ∪ 𝐴) ∪ 𝐶) = (𝐵 ∪ (𝐴 ∪ 𝐶))
5	2, 3, 4	3eqtr3i 2216	1 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))

Syntax hints:   = wceq 1363   ∪ cun 3139

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-10 1515  ax-11 1516  ax-i12 1517  ax-bndl 1519  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169

This theorem depends on definitions:  df-bi 117  df-tru 1366  df-nf 1471  df-sb 1773  df-clab 2174  df-cleq 2180  df-clel 2183  df-nfc 2318  df-v 2751  df-un 3145

This theorem is referenced by:  un23  3306  un4  3307