Animated Solution for Mathematics - Sets and Relations: 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:

I know, looking at this expression feels like staring into a storm of symbols: $((p\wedge q)\Rightarrow (r\vee q))\wedge ((p\wedge r)\Rightarrow q)$. It is easy to feel overwhelmed, but in the world of JEE logic, complexity is often just a mask for simplicity.

The secret is to never fight the whole monster at once. Instead, we break it down into manageable pieces.

Taming the Implication

Our first step is to recognize the structure. We have two major components joined by an 'AND' ($\wedge$) operator. Let's call them $P_1$ and $P_2$.

Our goal is to find the values of $r$ that make the entire expression a tautology—a statement that is true under every possible combination of truth values for $p,q,$ and $r$.

Before we touch the algebra, we must wield our most powerful tool: the implication rule. The statement $A\Rightarrow B$ is logically equivalent to $\sim A\vee B$.

This is the bridge that turns complex arrows into simple 'OR' ($\vee$) statements. Let's apply this to $P_1=(p\wedge q)\Rightarrow (r\vee q)$.

Using the rule, $P_1$ becomes $\sim (p\wedge q)\vee (r\vee q)$. Now, we invoke De Morgan's Law. The negation of the conjunction $(p\wedge q)$ becomes the disjunction of the negations: $\sim p\vee \sim q$.

So, $P_1$ simplifies to $\sim p\vee \sim q\vee r\vee q$. Look closely at the terms $\sim q$ and $q$.

Since everything is connected by 'OR', we can group them: $\sim p\vee r\vee (\sim q\vee q)$. Because $\sim q\vee q$ is always True, the entire expression $P_1$ becomes a tautology. It is always True, regardless of $p,q,$ or $r$!

The Search for Truth

Since $P_1$ is always True, our original expression $P_1\wedge P_2$ simplifies beautifully to $T\wedge P_2$, which is just $P_2$. Now, the entire problem rests on $P_2=(p\wedge r)\Rightarrow q$.

We apply the same logic:

Again, using De Morgan's Law, we get: This is our battlefield. We need to test our four candidates for $r$: $p,q,\sim p,$ and $\sim q$.

1. Case $r=p$: Substituting this into $P_2$, we get $\sim p\vee \sim p\vee q$, which simplifies to $\sim p\vee q$. This is not a tautology because it depends on the values of $p$ and $q$.

2. Case $r=q$: Substituting this, we get $\sim p\vee \sim q\vee q$. Since $\sim q\vee q$ is True, the whole expression becomes $\sim p\vee T$, which is True. Success! This is a valid value.

3. Case $r=\sim p$: Substituting this, we get $\sim p\vee \sim (\sim p)\vee q$. The double negation $\sim (\sim p)$ is just $p$. So we have $\sim p\vee p\vee q$. Since $\sim p\vee p$ is True, the expression is True. Another success!

4. Case $r=\sim q$: Substituting this, we get $\sim p\vee \sim (\sim q)\vee q$, which simplifies to $\sim p\vee q\vee q$, or $\sim p\vee q$. Like the first case, this is not a tautology.

By systematically testing each candidate, we found that only $r=q$ and $r=\sim p$ make the expression a tautology. That gives us exactly 2 valid values.

Logic is not about memorizing truth tables; it is about seeing the underlying structure and simplifying it until the truth reveals itself. Keep practicing this, and you will find that even the most intimidating problems have a simple, elegant core.