# Simplify (p v (r v q)) ∧ ~(~q ∧ ~r)

I understand that ~(~q ∧ ~r) simplifies down to (q v r), but I don't understand how the answer to this question is q v r.

For (p v (r v q)), I can simplify it to be (p v r v q). I thought maybe I could use the Universal Bound Laws or Identity Laws, but neither of the two equations leftover are tautologies or contradictions. Is there a specific property that I'm missing to handle (p v r v q) ∧ (q v r).

Thanks!

You're missing out something really simple!

I don't know exactly how you're treating stuff formally, but note that q v r implies p v q v r and hence (p v r v q) ∧ (q v r) can be rewritten as (q v r).

In the book "How to Prove It" by Daniel Velleman this law is called an Absorption Law.

$$a$$ implies $$a\vee b$$, since if $$a$$ is true $$a\vee b$$ is also true.

$$a\Rightarrow b$$ implies $$a\wedge b = a$$, since if $$a$$ is true, $$b$$ is also true and thus both are true. If $$a$$ is false, $$a\wedge b$$ could only be false.

Now substitute $$a$$ with $$q\vee r$$ and $$b$$ with $$p$$ in the first rule and you get $$(q\vee r)\Rightarrow (p\vee q\vee r)$$.

Then you can apply the second rule, you ($$q\vee r)\wedge (p\vee q\vee r) = q\vee r$$

Thanks to the implication $$q\vee r\Rightarrow p\vee q\vee r$$, the part $$p\vee q\vee r$$ is actually redundant.

$$(p \lor (r \lor q)) \land \lnot(\lnot q \land \lnot r)$$

$$\iff (p \lor (r \lor q)) \land (\lnot \lnot q \lor \lnot \lnot r) \hspace{1in} \text{De Morgan}$$

$$\iff (p \lor (r \lor q)) \land (q \lor r) \hspace{1in} \text{Double negation}$$

$$\iff (p \land (q\lor r)) \lor ((r\lor q) \land (q\lor r)) \hspace{1in} \text{distributive law}$$

$$\iff (p \land (q\lor r))\lor (q\lor r)\hspace{1in}\text{Idempotent law}$$

$$\iff q\lor r \hspace{1in} \text{Absorption}$$

$$(A\lor B)\land B$$ is true exactly when $$B$$ is true , whatever the value of $$A$$ may be. So it is simply $$B$$.

Others have pointed out that this is an instance of the Absorption Law. But if you don't have Absorption in your allowed set of rules, you can do this:

$$(p \lor q \lor r) \land (q \lor r) \overset{Identity}{\Leftrightarrow}$$

$$(p \lor q \lor r) \land (\bot \lor q \lor r) \overset{Distribution}{\Leftrightarrow}$$

$$(p \land \bot) \lor (q \lor r) \overset{Annihilation}{\Leftrightarrow}$$

$$\bot \lor (q \lor r) \overset{Identity}{\Leftrightarrow}$$

$$q \lor r$$