Theorem un23 3182

Description: A rearrangement of union. (Contributed by NM, 12-Aug-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
un23	⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐴 ∪ 𝐶) ∪ 𝐵)

Proof of Theorem un23

Step	Hyp	Ref	Expression
1		unass 3180	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
2		un12 3181	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))
3		uncom 3167	. 2 ⊢ (𝐵 ∪ (𝐴 ∪ 𝐶)) = ((𝐴 ∪ 𝐶) ∪ 𝐵)
4	1, 2, 3	3eqtri 2124	1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐴 ∪ 𝐶) ∪ 𝐵)

Syntax hints:   = wceq 1299   ∪ cun 3019

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 671  ax-5 1391  ax-7 1392  ax-gen 1393  ax-ie1 1437  ax-ie2 1438  ax-8 1450  ax-10 1451  ax-11 1452  ax-i12 1453  ax-bndl 1454  ax-4 1455  ax-17 1474  ax-i9 1478  ax-ial 1482  ax-i5r 1483  ax-ext 2082

This theorem depends on definitions:  df-bi 116  df-tru 1302  df-nf 1405  df-sb 1704  df-clab 2087  df-cleq 2093  df-clel 2096  df-nfc 2229  df-v 2643  df-un 3025

This theorem is referenced by:  setscom  11781