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):

your welcome