Animated Solution for Mathematics - Sets and Relations: Which of the following is a tautology?

Welcome, future engineers! Today, we are going to embark on a journey into the heart of logic. Often, students view Boolean algebra as a dry, mechanical process of flipping bits and symbols.

But I want you to see it differently. Logic is the architecture of thought itself. When we solve a problem like this, we are not just manipulating symbols; we are uncovering the fundamental truths that govern the universe of information.

Let us begin by demystifying the term 'Tautology.' Imagine a statement that is so robust, so fundamentally sound, that no matter what values you plug into it—whether the components are true or false—the result is always, unequivocally, true.

It is the logical equivalent of a mathematical identity like It is a bedrock of truth. Our mission is to test the given options and find the one that stands firm as a tautology.

Unlocking the Implication

Let us focus on the first option: $(P\wedge (P\to Q))\to Q$. At first glance, it looks like a dense thicket of brackets and operators. But do not be intimidated; we will solve this from the inside out.

The first tool in our arsenal is the Implication Rule: $A \to B \equiv eg A \lor B$. This rule is our skeleton key. It allows us to transform the implication into an OR statement.

Applying this to our inner term, $(P\to Q)$, the expression transforms into: See how the structure is already becoming clearer?

Now, we encounter the term $P \land ( eg P \lor Q)$. This is a classic setup for the Distributive Law: $A\wedge (B\vee C)\equiv (A\wedge B)\vee (A\wedge C)$.

Just as you would distribute a number in standard algebra, we distribute $P$ across the bracket. This gives us: We are peeling back the layers of the onion.

Here is where the magic happens. Look at the term $(P \land eg P)$. Can a statement be true and false at the same time? Of course not!

This is the Law of Contradiction. It tells us that $(P \land eg P)$ is always False ($F$). Substituting this into our expression, we get: Now, apply the Identity Law: $F\vee A\equiv A$. Since adding 'False' to an OR operation does nothing, our expression simplifies beautifully to $(P\wedge Q)\to Q$.

We are almost at the finish line. We have one last implication to resolve. Applying the Implication Rule again, where $A=(P\wedge Q)$ and $B=Q$, we get $ eg(P \land Q) \lor Q$.

Now, we invoke De Morgan's Law: $ eg(A \land B) \equiv eg A \lor eg B$. This flips our AND to an OR, giving us $( eg P \lor eg Q) \lor Q$.

Because all our operators are now OR, the Associative Law allows us to regroup the terms as $ eg P \lor ( eg Q \lor Q)$.

Finally, we arrive at the Law of Excluded Middle: $ eg Q \lor Q \equiv T$. A statement must be either true or false; there is no middle ground.

Substituting $T$ into our expression, we are left with $ eg P \lor T$. By the Domination Law, anything OR True is always True.

Our complex expression has collapsed into a single, elegant $T$. You have just proven that the first option is a tautology. Take a moment to appreciate the beauty of this reduction. Logic is not just about rules; it is about finding the simple, undeniable truth hidden within the complexity.