Answer to Question #17880 in Discrete Mathematics for Maha

Question

let A,B and C be sets. show that (A-B)-C=(A-C)-(B-C)

Answer 1

(A-C)-(B-C) = (A n C') n (B n C')'
=(A n C') n (B' u C'')
=(A n C') n (B' u C)
= A n (C' n (B' u C))
= A n ((C' n B') u (C' n C))
= A n ((C' n B') u emptyset)
= A n (C' n B')
= A n (B' n C')
= (A n B') n C'
= (A-B)-C

All we are using here is the definition of ', identities like X'' = X, the associativity and commutativity of n, "DeMorgan's law" (X n Y)' = X' u Y', a few 'distributive laws' like X n (Y u Z) = (X n Y) u (X n Z), and a property of the empty set.

If you read this "backwards" you get a proof that starts from the left hand side; if you got stuck going that way, it is probably because the "step" writing C' n B' as (C' n B') union the empty set and then writing the empty set in a new way is not at all obvious. (If you had drawn a venn diagram or something representing the situation, it might make sense from that, but from algebra it is not at all clear.)