Animated Solution for Mathematics - Sets and Relations: The number of ordered triplets of the truth values of $p,q$ and $r$ such that the truth value of the statement $(p\vee q)\wedge (p\vee r)\Rightarrow (q\vee r)$ is True, is equal to

Welcome, future engineer! Today, we are going to explore the elegant world of mathematical logic.

Often, when we see a problem involving truth values, our first instinct is to draw a massive truth table. While that is a valid strategy, it is the brute-force approach. As JEE aspirants, we want to cultivate the 'mathematician's eye'—the ability to see the underlying structure and simplify the problem before we even touch our pens.

We are given the logical expression: $(p\vee q)\wedge (p\vee r)\Rightarrow (q\vee r)$.

Our goal is to find the number of ordered triplets $(p,q,r)$ that make this statement True. Since each variable $p$, $q$, and $r$ can be either True ($T$) or False ($F$), we are dealing with a total of $2^3=8$ possible combinations.

Instead of checking all eight, let's use the power of logic to find our answer.

Look at the left side of the implication: $(p\vee q)\wedge (p\vee r)$. It looks a bit bulky, doesn't it?

This is where the Distributive Law of logic comes to our rescue. Just as we factor out common terms in algebra, we can factor out the $(p\vee )$ part.

This transforms our expression into:

Now, our entire implication looks much cleaner:

Here is the secret weapon: an implication $A\Rightarrow B$ is only False in one specific scenario—when the premise $A$ is True and the conclusion $B$ is False. This is the only way to 'break' the implication.

So, instead of counting the seven cases where the statement is True, we will hunt for the one case where it is False. Let $A=p\vee (q\wedge r)$ and $B=q\vee r$. We want to find when $A\Rightarrow B$ is False.

For $B=q\vee r$ to be False, both $q$ and $r$ must be False. There is no other way for an 'OR' statement to be False.

So, we have our first two clues: $q=F$ and $r=F$. Now, we substitute these into our premise $A$:

Since $(F\wedge F)$ is simply False, our premise simplifies to $A=p\vee F$, which is just $p$. For the implication to be False, we need $A$ to be True, which means $p$ must be True.

We have found our culprit! The only triplet that makes the statement False is $(T,F,F)$.

We started with eight possible worlds. We found exactly one world where the statement is False.

Therefore, in the remaining $8-1=7$ worlds, the statement must be True. It is that simple!

By using the laws of logic, we turned a potentially tedious task into a quick, satisfying victory. The final answer is 7. Keep practicing this mindset, and you will find that even the most complex problems have a hidden, beautiful simplicity waiting to be discovered.