The number of values of r in {p, q, sim p, si | vedclass

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(B) Let the given expression be $S = ((p \wedge q)$ $\Rightarrow (r \vee q)) \wedge ((p \wedge r)$ $\Rightarrow q)$.Using the identity $A \Rightarrow

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$2$

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(B) Let the given expression be $S = ((p \wedge q)$ $\Rightarrow (r \vee q)) \wedge ((p \wedge r)$ $\Rightarrow q)$.Using the identity $A \Rightarrow B \equiv \sim A \vee B$,we get:$S = (\sim(p \wedge q) \vee (r \vee q)) \wedge (\sim(p \wedge r) \vee q)$$S = (\sim p \vee \sim q \vee r \vee q) \wedge (\sim p \vee \sim r \vee q)$Since $\sim q \vee q = T$ (tautology),the first part becomes $\sim p \vee r \vee T = T$.Thus,$S = T \wedge (\sim p \vee \sim r \vee q) = \sim p \vee \sim r \vee q$.For $S$ to be a tautology,$\sim p \vee \sim r \vee q$ must be true for all truth values of $p$ and $q$.Case $1$: $r = p$. Then $\sim p \vee \sim p \vee q = \sim p \vee q$,which is not a tautology.Case $2$: $r = q$. Then $\sim p \vee \sim q \vee q = \sim p \vee T = T$. (Valid)Case $3$: $r = \sim p$. Then $\sim p \vee \sim(\sim p) \vee q = \sim p \vee p \vee q = T \vee q = T$. (Valid)Case $4$: $r = \sim q$. Then $\sim p \vee \sim(\sim q) \vee q = \sim p \vee q \vee q = \sim p \vee q$,which is not a tautology.Thus,there are $2$ values of $r$ for which the expression is a tautology.

The number of values of $r \in \{p, q, \sim p, \sim q\}$ for which $((p \wedge q)$ $\Rightarrow (r \vee q)) \wedge ((p \wedge r)$ $\Rightarrow q)$ is a tautology,is:

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