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I've been trying to solve this for about an hour now, but I keep getting stuck after a few steps. Here's what I have so far:

$(p \rightarrow \sim q)\wedge(\sim q \rightarrow p)$........................(Definition of $\leftrightarrow$)
$(\sim p \vee \sim q) \wedge (\sim\sim q \vee p)$....................(Definition of $\rightarrow$)
$(\sim p \vee \sim q) \wedge (q \vee p)$............................(Double Negation)
$\sim(p \wedge q) \wedge (q \vee p)$................................(Demorgan's Law)

It's about here that I get stuck, how do I get from $\sim(p \wedge q) \wedge (q \vee p)$ to $(\sim p \wedge q)\vee (p\wedge\sim q)$?

Mithlesh Upadhyay
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user214593
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2 Answers2

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In the last transition, use distribution of conjunction over disjunction

$$(\sim p \vee \sim q) \wedge (q \vee p)$$ $$((\sim p \vee \sim q) \wedge q) \vee ((\sim p \vee \sim q) \wedge p)$$ $$((\sim p \wedge q) \vee (q \wedge \sim q)) \vee ((\sim p \wedge p) \vee (\sim q \wedge p ))$$ $$(\sim p \wedge q) \vee (p \wedge \sim q)$$

benji
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You were so close. Things get a bit messy, but everything falls out fairly nice if you have the stamina and patience to do some tedious bookkeeping: \begin{align} \neg(p\land q)\land (q\lor p) &\equiv [\neg(p\land q)\land q]\lor [\neg(p\land q)\land p]\qquad\text{(distributivity)}\\[0.5em] &\equiv [(\neg p\lor \neg q)\land q]\lor [(\neg p\lor \neg q)\land p]\qquad\text{(DeMorgan)}\\[0.5em] &\equiv [(\neg p\land q)\lor (\neg q\land q)]\lor [(\neg p\land p)\lor (\neg q\land p)]\qquad\text{(distributivity)}\\[0.5em] &\equiv (\neg p\land q)\lor (p\land \neg q)\qquad\text{(elimination)} \end{align}

Daniel W. Farlow
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