# Prove logical equivalence....almost finished, but....!

I have almost finished, but I'm stuck at one place. Please guide.

Here is what I did:

Q. (¬p v q) ∧ (p ∧ (p ∧ q)) ⇔ p ∧ q

(¬p v q) ∧ ((p ∧ p) ∧ q) using Associative Law

(¬p v q) ∧ (p ∧ q) using idempotent Law

?

How to remove (¬p v q)?

Use some basic laws to rewrite $(\neg p\lor q)\land(p\land q)$ as
$$\big((\neg p\lor q)\land q\big)\land p$$
and show that $(\neg p\lor q)\land q$ is equivalent to $q$. How you do this will depend on exactly what laws you have available. It’s immediate if you have the absorption laws. Alternatively, you may have basic propositions $T$ and $F$ (or $1$ and $0$, or $\top$ and $\bot$) such that $q$ is equivalent to $F\lor q$. Then you can use a distributive law to change $(\neg p\lor q)\land(F\lor q)$ to $(\neg p\land F)\lor q$ and thence to $F\lor q$ and finally $q$.
• @GhostRider: You have something like $s\land(p\land q)$. Use commutativity inside the parentheses, and then use associativity, and you’ll have $(s\land q)\land p$, which is what you want. Commented Sep 11, 2016 at 1:05
• @GhostRider: You’re very welcome. It took me a while, but I finally realized that it was probably the fact that $\neg p\lor q$ was a compound proposition that was getting in your way. Commented Sep 11, 2016 at 1:46
• @GhostRider: Yes, any compound proposition within a larger compound proposition can be treated as a single entity. When you’re just beginning, it may be helpful actually to make a named substitution, as I did with $s$ for $\neg p\lor q$; with practice this becomes unnecessary. Commented Sep 11, 2016 at 1:54