# Answer on Discrete Mathematics Question for Maha

Question #17880

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

Expert's answer

(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.)

=(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.)

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## Comments

Assignment Expert09.11.2012 02:13You're welcome. We are glad to be helpful.

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Maha09.11.2012 00:46Thanks :)

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