Animated Solution for Mathematics - Sets and Relations: Let A, B, C and D be four non-empty sets. The contra-positive statement of " If $A\subset B$ and $B\subset D$, then $A\subset C$" is:

A Journey into Set Theory

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of logical structure that govern the universe of mathematics.

Set theory and formal logic are the silent languages of the cosmos. When we look at a statement like "If $A\subset B$ and $B\subset D$, then $A\subset C$", we aren't just looking at letters; we are looking at a hierarchy of containment.

Visualizing the Nested Reality

Imagine you are standing in a vast, open field. You see a massive boundary, a fence that encompasses everything we call set $D$.

Inside that fence, there is a smaller enclosure, set $B$. And tucked safely, snugly inside $B$, is the smallest enclosure, set $A$. This is the physical manifestation of the premise: $A\subset B$ and $B\subset D$.

Now, where is set $C$? The problem doesn't tell us, and that is the beauty of it. $C$ is a wild card; it could be anywhere. Our goal is to understand the logical "contrapositive" of the rule that binds these sets together.

The Logic Bridge

To handle this with the precision of a surgeon, we must translate the English into the symbolic language of logic. Let's define our atomic propositions:

- Let $P$ be the statement: $A\subset B$ - Let $Q$ be the statement: $B\subset D$ - Let $R$ be the statement: $A\subset C$

The original statement is a conditional implication: "If $P$ and $Q$, then $R$". In the crisp, elegant notation of logic, we write this as:

Here, the $\wedge$ symbol is our "AND", and the $\to$ is our "If... then..." implication. This is the bedrock of our problem.

The Power of the Contrapositive

Now, we invoke one of the most powerful tools in your arsenal: the Contrapositive Rule. For any implication $X\to Y$, the contrapositive is $\sim Y\to \sim X$.

Why is this so powerful? Because they are logically equivalent. They are the same truth, viewed from a different angle.

In our case, $X$ is the entire bracket $(P\wedge Q)$, and $Y$ is $R$. So, our contrapositive becomes:

We have successfully flipped the implication. We are now looking at the "Not" of our conclusion implying the "Not" of our premise.

The De Morgan's Twist

This is where many students stumble, but you won't. Look at the right side of our new implication: $\sim (P\wedge Q)$.

We cannot simply leave it as a negation of a bracket. We must distribute that negation. This is where De Morgan's Law comes to our rescue.

It tells us that the negation of an "AND" statement is the "OR" of the negations. Mathematically:

The "AND" ($\wedge$) flips into an "OR" ($\vee$). This is the moment of truth. If you forget to flip that operator, the entire logic collapses.

The Final Synthesis

Now, let's bring it all home. We substitute our expanded logic back into our contrapositive structure:

Now, we translate these symbols back into the language of sets:

- $\sim R$ becomes $A ot\subset C$ - $\sim P$ becomes $A ot\subset B$ - $\sim Q$ becomes $B ot\subset D$

Putting it all together, we get: "If $A ot\subset C$, then $A ot\subset B$ or $B ot\subset D$".

This is the mathematically rigorous, undeniable truth. If you encounter an exam option that uses "AND" instead of "OR", do not panic. You have derived the correct logical structure. Trust your derivation, identify the closest match, and move forward with the confidence of a master.