Animated Solution for Mathematics - Sets and Relations: Which of the following Boolean expressions is not a tautology?

Welcome, future engineers! Today, we are diving into the beautiful, structured world of mathematical reasoning. Often, students look at Boolean expressions and see a chaotic mess of arrows and symbols.

But I want you to see something else: a puzzle waiting to be solved with the precision of a master architect. Our goal is to identify which of the given expressions is not a tautology.

A tautology is a statement that is universally true, a logical bedrock that never crumbles, regardless of the truth values of its components. While the brute force method of truth tables is a valid path, it is the slow, winding road. We are going to take the highway: Boolean Algebra.

The Implication Law

Before we touch the options, we must sharpen our primary tool: the Implication Law. This law is the bridge between the complex implication symbol ($\Rightarrow$) and the simpler, more manageable operators of negation ($\sim$) and disjunction ($\vee$).

The law states that:

Think of this as a translation. Whenever you see an arrow, you are essentially saying, 'If the first part is false, the whole thing is true, OR the second part must be true.' By applying this to every implication in our options, we strip away the complexity and reveal the underlying structure.

Analyzing Options 1, 2, and 3

Let us begin our investigation. For Option 1, $(p\Rightarrow q)\vee (\sim q\Rightarrow p)$, we apply our law.

The first bracket becomes $(\sim p\vee q)$, and the second becomes $(\sim (\sim q)\vee p)$, which simplifies to $(q\vee p)$. Combining these, we get $(\sim p\vee q)\vee (q\vee p)$.

By the Commutative and Associative laws, we can rearrange this to $(\sim p\vee p)\vee (q\vee q)$. Since $(\sim p\vee p)$ is always True ($T$), the entire expression collapses to $T\vee q$, which is always True. Option 1 is a tautology!

We repeat this process for Option 2, $(q\Rightarrow p)\vee (\sim q\Rightarrow p)$. This transforms into $(\sim q\vee p)\vee (q\vee p)$.

Grouping the $q$ terms gives us $(\sim q\vee q)\vee (p\vee p)$, which is $T\vee p$, again resulting in True. Option 2 is also a tautology.

Moving to Option 3, $(p\Rightarrow \sim q)\vee (\sim q\Rightarrow p)$, we get $(\sim p\vee \sim q)\vee (q\vee p)$. Rearranging gives us $(\sim p\vee p)\vee (\sim q\vee q)$, which is $T\vee T$, resulting in True. Options 1, 2, and 3 are all rock-solid tautologies.

Unmasking Option 4

Now, we arrive at Option 4: $(\sim p\Rightarrow q)\vee (\sim q\Rightarrow p)$. Let us apply the same rigor.

The first bracket, $(\sim p\Rightarrow q)$, becomes $(\sim (\sim p)\vee q)$, which simplifies to $(p\vee q)$. The second bracket, $(\sim q\Rightarrow p)$, becomes $(\sim (\sim q)\vee p)$, which simplifies to $(q\vee p)$.

When we combine these, we get $(p\vee q)\vee (q\vee p)$. Using the Idempotent Law, where $x\vee x\equiv x$, this entire expression simplifies down to just $(p\vee q)$.

Now, ask yourself: is $(p\vee q)$ always true? No! If we choose the case where $p$ is False and $q$ is False, the expression evaluates to False.

Because we have found a scenario where the statement is not true, Option 4 is not a tautology. We have successfully navigated the logic, avoided the trap of brute force, and arrived at the correct answer with mathematical elegance.