Animated Solution for Mathematics - Sets and Relations: Among the statements: (S1) $(p\Rightarrow q)\vee (p\wedge (\sim q))$ is a tautology, (S2) $((\sim p)\Rightarrow (\sim q))\wedge ((\sim p)\vee q)$ is a Contradiction. Then

Welcome, future engineers! Today, we are not just solving a problem; we are peeling back the layers of human reasoning. In the world of JEE Advanced, logic is the bedrock upon which all mathematics is built.

When you look at a complex logical expression, don't see a jumble of symbols. See a circuit—a logical machine that either glows with the light of truth or flickers into the darkness of contradiction. Let us embark on this journey to analyze two specific statements, $(S1)$ and $(S2)$, and determine their true nature.

We begin with the statement $(S1):(p\Rightarrow q)\vee (p\wedge \sim q)$. At first glance, this looks intimidating. But remember, every implication has a hidden identity.

We know that $p\Rightarrow q$ is logically equivalent to $\sim p\vee q$. This is our first key. By substituting this into our expression, we transform the statement into:

Now, look closely at the second term, $(p\wedge \sim q)$. If we apply De Morgan's Law, we can rewrite this as $\sim (\sim p\vee q)$. Suddenly, the structure of the entire expression becomes crystal clear.

If we let $X=(\sim p\vee q)$, our statement becomes $X\vee \sim X$. This is the moment of realization! According to the Law of Excluded Middle, a statement combined with its own negation via the OR operator is always true.

It is the logical equivalent of saying 'It is raining, or it is not raining.' There is no middle ground, and it is always true. Thus, $(S1)$ is a Tautology.

Now, let us turn our attention to $(S2):(\sim p\Rightarrow \sim q)\wedge (\sim p\vee q)$. We need to determine if this is a contradiction—a statement that is always false.

Let us simplify the first part, $(\sim p\Rightarrow \sim q)$. Using our implication rule, this becomes $\sim (\sim p)\vee \sim q$, which simplifies beautifully to $p\vee \sim q$. Substituting this back into $(S2)$, we get:

This expression is a classic in logic. If you look at it through the lens of conditional statements, it is equivalent to $(q\Rightarrow p)\wedge (p\Rightarrow q)$. This, my friends, is the definition of the biconditional $p\leftrightarrow q$.

Is $p\leftrightarrow q$ a contradiction? Let's test it. If $p$ is True and $q$ is True, then $p\leftrightarrow q$ is True.

Since we found a case where the statement is True, it cannot be a contradiction. A contradiction must be false for every single possible input. Therefore, $(S2)$ is not a contradiction; it is a contingency.

We have reached the end of our logical journey. We have verified that $(S1)$ is a Tautology and $(S2)$ is not a contradiction. Remember, logic is not about memorizing tables; it is about seeing the symmetry in the symbols.