Using logical laws, I would like to simplify the following expression:
$\neg a \lor \neg b \lor (a \wedge b \wedge \neg c)$
1) Distribution law:
$(\neg a \lor a) \land (\neg a \lor b) \land (\neg a \lor \neg c) \land (\neg b \lor a) \land (\neg b \lor b) \land (\neg b \lor \neg c)$
2) Inverse law:
$(T) \land (\neg a \lor b) \land (\neg a \lor \neg c) \land (\lnot b \lor a) \land (T) \land (\neg b \lor \neg c)$
3) Identity law:
$(\neg a \lor b) \land (\neg a \lor \neg c) \land (\lnot b \lor a) \land (\neg b \lor \neg c)$
from here I'm not sure what to do next.
Best Answer
Try distributing like this:
\begin{align*} &\neg a \vee \neg b \vee (a\wedge b\wedge \neg c) & \\ &\equiv \left[\neg a \vee \neg b\right] \vee (a\wedge b\wedge \neg c) &\text{(Associativity)}\\ &\equiv (\left[\neg a \vee \neg b\right]\vee a) \wedge (\left[\neg a \vee \neg b\right]\vee b) \wedge (\left[\neg a \vee \neg b\right]\vee \neg c) &\text{(Distribution law)}\\ &\equiv ([\neg a \vee a] \vee \neg b)\wedge ([\neg b \vee b] \vee \neg a) \wedge (\neg a \vee \neg b\vee \neg c) &\text{(Assoc. and commut.)}\\ &\equiv (T\vee \neg b)\wedge (T\vee \neg a)\wedge (\neg a \vee \neg b\vee \neg c)&\text{(Identity laws)}\\ &\equiv (T)\wedge (T)\wedge (\neg a \vee \neg b\vee \neg c) &\text{(Identity laws)}\\ &\equiv (\neg a \vee \neg b\vee \neg c) &\text{(Identity laws)} \end{align*}