Proof of An(BuC) = (AnB)u(AnC)

I need help understanding this proof:

I'm going to use 'e' to denote 'is an element of the set' (I can't find the proper symbol :redface:

)

Proof of An(BuC) = (AnB)u(AnC).

Suppose now that x e (AnB)u(AnC). If x e AnB, then x e A and x e B so that x e A and x e BuC which gives x e An(BuC) as required. On the other hand..... etc......

I don't get how the part in bold just becomes valid :confused:

Reply 1

john !!

if x is an element of B, then x is an element of the union of B and C... its simple if you read it aloud because the union of B and C contains all the elements of B and C, and x is an element of B so it goes in the union as well!