# Simplify $p\lor q\lor(\neg p\land\neg q\land r)$ to $p\lor q\lor r$

I am trying to simplify the logic expression: $$p\lor q\lor(\neg p\land\neg q\land r)$$ with laws of logic to get: $$p\lor q\lor r$$ I have no idea how to get to the result by using the laws of logic. Help!

• Use Distributivity. Jan 30, 2018 at 8:51
• Can you show me? Jan 30, 2018 at 8:52

$$p\lor q\lor(\neg p\land\neg q\land r)$$ Use the distributive law: $$=(p\lor q\lor\neg p)\land(p\lor q\lor\neg q)\land(p\lor q\lor r)$$ Complementarity ($p\lor\neg p=1$) reduces the first two terms to 1: $$=1\land1\land(p\lor q\lor r)$$ Identity ($p\land1=p$) eliminates those 1s: $$=p\lor q\lor r$$
• Yea sorry but I saw that there is $p \lor \lnot p$ and $q \lor \lnot q$ In the first two terms. Should I use the inverse law and then then the domination law or just the absorption law? Jan 30, 2018 at 9:04
• Yes I do, I just did it differently. Using the commutative law on the first term then the inverse law on both terms to make it $p \lor T_0$ and $q \lor T_0$. I then used the domination law and then identify law to get where you are. It’s how my book wanted it Jan 30, 2018 at 9:28
• Absorption. $a\vee(\neg a\wedge b)=a\vee b$ Jan 30, 2018 at 9:30