Theorem un23 3340

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 3338	. 2 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = (𝐴 ∪ (𝐵 ∪ 𝐶))
2		un12 3339	. 2 ⊢ (𝐴 ∪ (𝐵 ∪ 𝐶)) = (𝐵 ∪ (𝐴 ∪ 𝐶))
3		uncom 3325	. 2 ⊢ (𝐵 ∪ (𝐴 ∪ 𝐶)) = ((𝐴 ∪ 𝐶) ∪ 𝐵)
4	1, 2, 3	3eqtri 2232	1 ⊢ ((𝐴 ∪ 𝐵) ∪ 𝐶) = ((𝐴 ∪ 𝐶) ∪ 𝐵)

Syntax hints:   = wceq 1373   ∪ cun 3172

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2194  df-cleq 2200  df-clel 2203  df-nfc 2339  df-v 2778  df-un 3178

This theorem is referenced by:  setscom  12987