OpenStudy (anonymous):

Prove: C-(AUB)=(C-A)N(C-B)

OpenStudy (anonymous):

By N I mean intersection

OpenStudy (amistre64):

truth tables help if you dont have the rules for the tautologies

OpenStudy (anonymous):

I do realize this is one of De Morgran's Laws. I need proof of it though.

OpenStudy (anonymous):

$C - (A \cup B ) = (C - A) n (C - B)$ to do this you want to prove the left is equal to the right vice versa let $x \in C - (A \cup B)$ then $x \in C$and $x \notin (A \cup B)$ so because$x \notin (A \cup B)$ then $x \notin A, x \notin B$ since $x \in C$ and $x \notin A$ then $x \in (C - A )$ and since $x \in C$and$x \notin B$ then $x \in (C - B )$ so since $x \in (C - A )$ and $x \in (C - B )$ then $x \in (C - A ) n (C - B )\$ now you prove the RHS to the LHS

OpenStudy (anonymous):

Thanks!

OpenStudy (anonymous):