Animated Solution for Mathematics - Sets and Relations: Let $F_1(A,B,C)=(A\wedge \sim B)\vee [\sim C\wedge (A\vee B)]\vee \sim A$ and $F_2(A,B)=(A\vee B)\vee (B\to \sim A)$ be two logical expressions. Then :

Mastering Logical Expressions

Welcome, future engineers! Today, we are going to peel back the curtain on one of the most elegant topics in the JEE syllabus: Boolean Algebra.

Often, students look at logical expressions like $F_1=(A\wedge \sim B)\vee [\sim C\wedge (A\vee B)]\vee \sim A$ and feel a sense of dread. But I want you to shift your perspective. Do not see these as dry, abstract symbols; see them as a circuit board.

A tautology is simply a circuit that is always 'ON', no matter what the input is. Let us embark on this journey to decode these expressions.

The Art of Disproof

When we are asked to check if an expression is a tautology, we are essentially being asked: "Is this statement always True?" To prove it is NOT a tautology, we do not need to check every single possibility. We only need to find one single "crack" in the armor—one set of inputs that makes the expression False.

Look at $F_1$. It is a massive OR ($\vee$) chain. In logic, an OR statement is only False if every single component is False. This is our golden opportunity.

We have three blocks: $(A\wedge \sim B)$, $[\sim C\wedge (A\vee B)]$, and $\sim A$. To make $F_1$ False, we must force all three to be False simultaneously.

Let us start with the simplest block: $\sim A$. For this to be False, $A$ must be True. Now, with $A=T$, look at the first block: $(A\wedge \sim B)$. Since $A$ is True, the only way to make this block False is to make $\sim B$ False, which means $B$ must be True.

Finally, look at the middle block: $[\sim C\wedge (A\vee B)]$. Since we have already established $A=T$ and $B=T$, the term $(A\vee B)$ is definitely True. To make the entire middle block False, $\sim C$ must be False, which forces $C$ to be True.

By setting $A=T,B=T,C=T$, we have successfully forced $F_1$ to be False. The expression is not a tautology. We have dismantled it with pure logic.

The Implication Trap

Now, let us turn our attention to $F_2=(A\vee B)\vee (B\to \sim A)$. This looks intimidating because of the implication arrow ($\to$).

Many students freeze here. But remember, in the JEE, every operator has a translation. The implication $P\to Q$ is mathematically equivalent to $\sim P\vee Q$. This is your secret weapon.

Let us apply this to the term $(B\to \sim A)$. It transforms into $(\sim B\vee \sim A)$. Now, substitute this back into $F_2$:

Notice something beautiful? The entire expression is now just a series of OR operators. Because of the Associative and Commutative laws, we can rearrange these terms however we please. Let us group the $A$ terms and the $B$ terms together:

The Elegance of Symmetry

Look at what we have uncovered. We have $(A\vee \sim A)$ and $(B\vee \sim B)$. This is the Law of Excluded Middle.

A variable OR-ed with its own negation is always True. It is the fundamental bedrock of logic. So, $F_2$ simplifies to $T\vee T$, which is simply True.

No matter what values $A$ and $B$ take, $F_2$ will always output True. It is a tautology, rock-solid and unbreakable.

We have analyzed, we have manipulated, and we have conquered. Remember, logic is not about memorizing tables; it is about seeing the underlying structure. Keep practicing, keep questioning, and you will master this.