1. ## set theory simplification

$\displaystyle \overline{\overline{((A\cup B)}\cap(C))} \cup \overline{(B)}$

I have the first part where you would use demorgans law to break up the a union b and then double negation to cancel the a union b double negation.

2. Originally Posted by aaronrj
$\displaystyle \overline{\overline{((A\cup B)}\cap(C))} \cup \overline{(B)}$

I have the first part where you would use demorgans law to break up the a union b and then double negation to cancel the a union b double negation.
It is more convenient to use A' instead of $\displaystyle \overline{A}$.

Hence ,$\displaystyle \overline{\overline{((A\cup B)}\cap(C))} \cup \overline{(B)}$ = $\displaystyle [(A\cup B)'\cap C]'\cup B'$ = $\displaystyle [(A'\cap B')\cap C]'\cup B'$ = $\displaystyle [A\cup B\cup C']\cup B'$ = $\displaystyle (A\cup C')\cup(B\cup B')$ = $\displaystyle ( A\cup C')\cup E$ = E

Where A,B,C are all subsets of E