Animated Solution for Mathematics - Sets and Relations: Among the statements: (S1) $((p\vee q)\Rightarrow r)\iff (p\Rightarrow r)$, (S2) $((p\vee q)\Rightarrow r)\iff ((p\Rightarrow r)\vee (q\Rightarrow r))$

A Journey into Tautologies

Welcome, future engineers! Today, we are going to peel back the curtain on one of the most elegant and foundational topics in mathematics: Mathematical Reasoning.

Specifically, we are going to master the concept of a tautology. Imagine a statement that is so robust, so perfectly constructed, that it remains true regardless of the chaos of the variables within it. That, my friends, is a tautology.

In the world of logic, a tautology is a compound statement that is always true, no matter what truth values you assign to its individual components ($p$, $q$, and $r$). Think of it as a mathematical identity, like:

It does not matter if $\theta$ is $30^{\circ }$ or $300^{\circ }$; the identity holds. Similarly, a tautology is a logical identity.

Our mission today is to evaluate two statements:

Are they tautologies? Let's find out.

Counter-Examples

Many students immediately reach for their pens to draw a massive eight-row truth table. While that is a valid approach, it is the 'brute force' method. In the JEE, we prefer the 'sniper' approach.

To prove a statement is a tautology, you must show it is true for all 8 cases. But to prove it is not a tautology? You only need one counter-example where the Left Hand Side (LHS) does not equal the Right Hand Side (RHS).

If we find a single case where the bi-conditional $\iff$ fails, the statement is not a tautology. Let's start hunting.

Let's look at $S_1:((p\vee q)\Rightarrow r)\iff (p\Rightarrow r)$. We want to find a case where the LHS and RHS differ.

The easiest way to break an implication is to make the premise True and the conclusion False. Let's try to make the LHS False. For $(p\vee q)\Rightarrow r$ to be False, we need $(p\vee q)$ to be True and $r$ to be False.

Let's choose $p=\text{F}$, $q=\text{T}$, and $r=\text{F}$.

1. Evaluate LHS: $(p\vee q)\Rightarrow r$. Since $p=\text{F}$ and $q=\text{T}$, $(p\vee q)$ is $\text{T}$. Then, $\text{T}\Rightarrow \text{F}$ is False.

2. Evaluate RHS: $p\Rightarrow r$. Since $p=\text{F}$ and $r=\text{F}$, $\text{F}\Rightarrow \text{F}$ is True.

Since $\text{LHS}=\text{False}$ and $\text{RHS}=\text{True}$, the bi-conditional fails. $S_1$ is not a tautology!

Now for the more complex $S_2:((p\vee q)\Rightarrow r)\iff ((p\Rightarrow r)\vee (q\Rightarrow r))$. Again, we hunt for a counter-example.

Let's try to make the LHS False again. We need $(p\vee q)$ to be True and $r$ to be False. Let's try $p=\text{T}$, $q=\text{F}$, and $r=\text{F}$.

1. Evaluate LHS: $(p\vee q)\Rightarrow r$. With $p=\text{T}$ and $q=\text{F}$, $(p\vee q)$ is $\text{T}$. Then, $\text{T}\Rightarrow \text{F}$ is False.

2. Evaluate RHS: $(p\Rightarrow r)\vee (q\Rightarrow r)$. - $(p\Rightarrow r)$ is $\text{T}\Rightarrow \text{F}$, which is False. - $(q\Rightarrow r)$ is $\text{F}\Rightarrow \text{F}$, which is True. - The OR of these is $\text{False}\vee \text{True}$, which is True.

Again, $\text{LHS}=\text{False}$ and $\text{RHS}=\text{True}$. The bi-conditional fails. $S_2$ is not a tautology either!

We have successfully navigated the logic maze. By using the counter-example strategy, we avoided the tedious work of drawing full truth tables and arrived at the conclusion that neither $S_1$ nor $S_2$ is a tautology.

Remember, logic is not just about memorizing tables; it is about understanding the flow of truth. Keep practicing, stay curious, and you will master these concepts in no time!