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
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How to remove (¬p v q)?
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1 Answer
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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 $q$.
answered Sep 11, 2016 at 0:09
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$\begingroup$ Thanks! one question that I think, you probably wrote (¬p ∨ q) ∧ (p ∧ q) as ( (¬p ∨ q) ∧ q ) ∧ p using the Distributive Rule, but I think we can only use the distributive rule when it is like this for example in this case (¬p ∨ q) ∧ (p ∧ q) should have been (¬p ∧ q) ∧ (p ∧ q) to use it. But you used it for (¬p ∨ q) ∧ (p ∧ q) as well. I might be wrong, but please clarify. $\endgroup$ Commented Sep 11, 2016 at 0:34
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1$\begingroup$ @GhostRider: The only place that I explicitly used a distributive law was where I mentioned it near the end. Try using commutative and associative laws to do the rewriting mentioned at the beginning. $\endgroup$ Commented Sep 11, 2016 at 0:45
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1$\begingroup$ @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. $\endgroup$ Commented Sep 11, 2016 at 1:05
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1$\begingroup$ @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. $\endgroup$ Commented Sep 11, 2016 at 1:46
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1$\begingroup$ @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. $\endgroup$ Commented Sep 11, 2016 at 1:54