Animated Solution for Mathematics - Sets and Relations: Negation of $p\wedge (q\wedge \sim (p\wedge q))$ is

Imagine you are standing on the edge of a logical landscape, looking at the expression $p\wedge (q\wedge \sim (p\wedge q))$. It looks intimidating, but in the world of JEE Advanced, we dismantle complexity by stripping away layers to reveal the core truth.

Let's embark on this journey to simplify the expression step-by-step.

The Inside-Out Strategy

When you face a nested logical expression, always start from the innermost bracket. We have $\sim (p\wedge q)$, which is the negation of the intersection of $p$ and $q$.

By applying De Morgan's Law, we can rewrite this as:

This is our first step toward clarity, representing everything outside the overlapping region of $p$ and $q$.

The Algebraic Dance

Now, let's incorporate $q$ into the expression: $q\wedge (\sim p\vee \sim q)$. We apply the Distributive Law to distribute $q$ across the bracket:

Look closely at the second term, $q\wedge \sim q$. This is a classic contradiction, as a statement cannot be both $q$ and not $q$ simultaneously.

Thus, $q\wedge \sim q\equiv F$. Our expression simplifies beautifully to $q\wedge \sim p$, which represents the region inside $q$ but strictly outside $p$.

The Collapse

Now, we take this result and intersect it with $p$. The full expression becomes $p\wedge (q\wedge \sim p)$. By the associative property, we rearrange this as:

Again, we see the familiar $p\wedge \sim p$, which is a contradiction ($F$). Consequently, the entire expression simplifies to $F\wedge q$, which is simply $F$.

The entire structure has collapsed into a single, elegant False.

The Final Twist

The problem asks for the negation of this expression. Since we found the expression is $F$, the negation of $F$ is $T$ (a Tautology).

We are looking for an option that is always True.

Verifying the Options

Let's test Option 1: $(\sim (p\wedge q))\vee p$. Using De Morgan's Law, this becomes:

We can rearrange this using the commutative and associative properties as:

Since $\sim p\vee p$ is always True ($T$), we have $T\vee \sim q$. Any statement OR-ed with True is always True.

Thus, Option 1 is a Tautology. We have found our answer; it is a perfect match. Keep practicing this algebraic rhythm, and you will find that even the most complex logical problems become second nature.