Metamath Proof Explorer


Theorem un23

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

Ref Expression
Assertion un23 ( ( 𝐴𝐵 ) ∪ 𝐶 ) = ( ( 𝐴𝐶 ) ∪ 𝐵 )

Proof

Step Hyp Ref Expression
1 unass ( ( 𝐴𝐵 ) ∪ 𝐶 ) = ( 𝐴 ∪ ( 𝐵𝐶 ) )
2 un12 ( 𝐴 ∪ ( 𝐵𝐶 ) ) = ( 𝐵 ∪ ( 𝐴𝐶 ) )
3 uncom ( 𝐵 ∪ ( 𝐴𝐶 ) ) = ( ( 𝐴𝐶 ) ∪ 𝐵 )
4 1 2 3 3eqtri ( ( 𝐴𝐵 ) ∪ 𝐶 ) = ( ( 𝐴𝐶 ) ∪ 𝐵 )