Animated Solution for Mathematics - Basic Mathematics: Which one of the following statements is not a tautology ?

Unmasking the Tautology

Welcome, fellow truth-seekers! Today, we are diving into the elegant world of mathematical logic. We have a set of four logical statements, and our mission is to identify the one that is not a tautology.

A tautology is the ultimate truth—a statement that remains True regardless of the truth values of its individual components. It is a logical certainty.

Look closely at the options provided. Each one is an implication, structured as $p\to q$. In the realm of logic, the implication is a powerful, yet fragile, construct.

It is only False in one specific, catastrophic scenario: when the premise is True, but the conclusion is False. If you can find a way to force this $T\to F$ outcome, you have successfully broken the tautology.

We are not going to build massive, tedious truth tables. Instead, we are going to be detectives, hunting for that one specific counter-example that makes the statement collapse into Falsehood.

For Option 1, $(p\wedge q)\to p$, we want to make the left side True and the right side False. To make $(p\wedge q)$ True, both $p$ and $q$ must be True.

But if $p$ is True, the right side is True. We have a contradiction! We cannot force $T\to F$, so it is a tautology.

Moving to Option 2, $(p\wedge q)\to (\sim p)\vee q$, we again try to force the left side to be True, which forces $p=T$ and $q=T$.

Plugging these into the right side, we get $(\sim T)\vee T$, which simplifies to $F\vee T$, resulting in True. Again, we cannot force a False outcome; it is a tautology.

Option 3, $p\to (p\vee q)$, is even more resilient. If we set $p=T$, the right side becomes $(T\vee q)$, which is always True regardless of $q$. We are trapped in a cycle of truth!

Unmasking Option 4

Now, we arrive at Option 4: $(p\vee q)\to (p\vee \sim q)$. Let's apply our detective skills.

We want to force the right side, $(p\vee \sim q)$, to be False. For an 'or' statement to be False, both components must be False.

This forces $p=F$ and $\sim q=F$. If $\sim q=F$, then $q$ must be True.

We have our suspect: $p=F$ and $q=T$. Now, let's test the premise $(p\vee q)$ with these values.

We get $(F\vee T)$, which evaluates to True. We have done it! We have created the $T\to F$ scenario.

The premise is True, and the conclusion is False. Therefore, the implication is False. Option 4 is not a tautology.

Logic is not just about symbols; it is about understanding the structure of truth. By systematically testing for the one scenario that breaks the implication, we have navigated through the options and found our answer.

Keep this 'counter-example' mindset in your toolkit—it is the secret weapon for mastering logic problems in JEE Advanced. You have done excellent work today!