Question

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

Answer 1

By N I mean intersection

Answer 2

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

Answer 3

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

Answer 4

\[C - (A \cup B ) = (C - A) n (C - B)\]

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

Answer 5

Thanks!

Answer 6

your welcome