Animated Solution for Mathematics - Sets and Relations: Consider the two statements: (S1) : $(p\to q)\vee (\sim q\to p)$ is a tautology. (S2) : $(p\wedge \sim q)\wedge (\sim p\vee q)$ is a fallacy. Then :

Mathematical reasoning is the bedrock of all science. Today, we will dissect two logical statements, $S_1$ and $S_2$, to determine their true nature: whether they are tautologies (always true) or fallacies (always false).

The Tautology

Our first statement is $S_1:(p\to q)\vee (\sim q\to p)$. To simplify this, we utilize the Implication Rule, which states that $a\to b\equiv \sim a\vee b$.

Applying this rule to both components: 1. The first part, $p\to q$, transforms into $\sim p\vee q$. 2. The second part, $\sim q\to p$, becomes $\sim (\sim q)\vee p$, which simplifies to $q\vee p$.

Now, our expression $S_1$ becomes:

Because all operators are logical ORs ($\vee$), we apply the Associative and Commutative laws to rearrange the terms:

By the Complement Law, $\sim p\vee p$ is always True ($T$). By the Idempotent Law, $q\vee q$ is simply $q$. Thus, $S_1$ simplifies to:

According to the Domination Law, True OR anything is always True. Therefore, $S_1$ is a tautology.

The Fallacy

Now, we examine $S_2:(p\wedge \sim q)\wedge (\sim p\vee q)$. Observe the second bracket, $\sim p\vee q$.

Applying De Morgan's Law in reverse, we recognize that $\sim p\vee q$ is equivalent to $\sim (p\wedge \sim q)$. Substituting this back into $S_2$, we obtain:

Let us define $X=p\wedge \sim q$. The expression then simplifies to:

This is the classic Contradiction Law. A statement AND its negation can never be true simultaneously; it is always False ($F$). Therefore, $S_2$ is a fallacy.

We have navigated the maze of logic to reach our conclusion. We have determined that $S_1$ is a tautology and $S_2$ is a fallacy.

The beauty of mathematical reasoning lies in this process: when you break down complexity, the truth reveals itself with elegant simplicity.