Animated Solution for Mathematics - Sets and Relations: The negation of the statement $((A\wedge (B\vee C))\Rightarrow (A\vee B))\Rightarrow A$ is

Welcome, student. Today, we are going to tackle a problem that looks like a tangled mess of symbols, but I promise you, it is actually a beautiful, rhythmic dance of logic.

When you first look at the expression $((A\wedge (B\vee C))\Rightarrow (A\vee B))\Rightarrow A$, it is natural to feel a surge of anxiety. It looks like a wall of text. But remember: in mathematics, complexity is often just simplicity in disguise. Let us peel back the layers together.

The Strategy of Simplification

First, take a deep breath. We are not going to brute-force this with a truth table. That is the path of the amateur.

Instead, we are going to use the laws of logic to simplify the structure. Let us define the inner part of the expression as $P$. So, Now, our entire problem simplifies to the elegant form $P\Rightarrow A$.

Why do we do this? Because it allows us to focus on the structure rather than the noise. We know the fundamental identity of implication: $X\Rightarrow Y\equiv \sim X\vee Y$.

This is our master key. It allows us to turn those intimidating arrows into simple OR operations. Applying this to our main structure, we get $\sim P\vee A$. Now, we just need to find out what $\sim P$ is.

The Dance of De Morgan

Now, let us zoom into $P$. We have $(A\wedge (B\vee C))\Rightarrow (A\vee B)$. Applying our master key again, this becomes $\sim (A\wedge (B\vee C))\vee (A\vee B)$.

But wait, we need the negation of $P$ (which is $\sim P$). So, we take the negation of the entire expression:

This is where the magic happens. We invoke De Morgan's Laws. Remember, the negation of an OR is the AND of the negations.

So, the outer negation distributes, flipping the OR to an AND, and hitting both terms. The first term, which was already negated, gets a double negation, effectively cancelling it out. We are left with $(A\wedge (B\vee C))\wedge \sim (A\vee B)$.

The Beautiful Collapse

We are almost there. Look at the second part of our expression: $\sim (A\vee B)$. Applying De Morgan's Law again, this becomes $\sim A\wedge \sim B$.

Now, let us bring everything together. Our expression is now:

Look closely at the square bracket. Because all the operations are ANDs, we can rearrange them using the associative and commutative laws. Let us group $A$ and $\sim A$ together. We get $(A\wedge \sim A)\wedge (B\vee C)\wedge \sim B$.

Do you see it? $A\wedge \sim A$ is a contradiction. It is False. And in the world of logic, False AND anything is always False.

The entire bracket collapses into a single, silent False. We are left with $F\vee A$. By the identity law, this is simply $A$.

We have successfully reduced that massive, terrifying expression down to a single variable: $A$. But do not celebrate just yet!

The question asks for the negation of the statement. Since the statement is equivalent to $A$, its negation is $\sim A$.

See how we navigated that? We didn't panic. We broke it down, applied the laws, and watched as the complexity vanished. This is the power of logical thinking. You are not just solving a problem; you are mastering the language of truth itself.