Animated Solution for Mathematics - Sets and Relations: The Boolean expression $(\sim (p\wedge q))\vee q$ is equivalent to :

A Journey into Truth

Welcome, aspiring engineers! Today, we are going to peel back the layers of a fundamental Boolean algebra problem. Logic is the silent language of the digital world, the bedrock upon which every computer, smartphone, and algorithm is built.

When we look at an expression like $(\sim (p\wedge q))\vee q$, we aren't just looking at symbols; we are looking at a logical circuit waiting to be simplified. Let us embark on this journey to find its equivalent.

Deconstructing the Gate

Our journey begins with the expression $(\sim (p\wedge q))\vee q$. The first obstacle is the negation of the intersection, $\sim (p\wedge q)$.

Many students instinctively want to distribute the negation, but remember the golden rule: De Morgan's Law. It tells us that:

Imagine the negation as a key that unlocks the bracket, but in doing so, it flips the AND ($\wedge$) into an OR ($\vee$). Now, our expression transforms into $(\sim p\vee \sim q)\vee q$. We have successfully broken down the complex inner term into simpler, manageable components.

The Art of Rearrangement

Now that we have $(\sim p\vee \sim q)\vee q$, we look for patterns. In Boolean algebra, the Associative Law is our best friend. It allows us to regroup terms without changing the truth value of the expression.

We can rewrite our expression as:

Why do we do this? Because we see a beautiful symmetry between $\sim q$ and $q$. They are complements of each other. By grouping them, we are setting the stage for a powerful simplification.

The Tautology Revelation

Focus your attention on the bracketed term: $(\sim q\vee q)$. This is the Complement Law in action.

If you take a statement and combine it with its own negation using an OR gate, you are guaranteed to cover every possibility. If $q$ is false, $\sim q$ is true; if $q$ is true, $\sim q$ is false. In either case, the result is always true.

We call this a Tautology, denoted by $T$. Our expression now simplifies to $\sim p\vee T$. Here, the Domination Law takes over. Since $T$ is always true, the OR operation with any other variable $\sim p$ will always result in $T$. Thus, our entire original expression is a Tautology ($T$).

Testing the Options

We have established that our expression is equivalent to $T$. Now, we must find which of the given options also evaluates to $T$.

Let us examine Option (D): $p\to (p\vee q)$. To evaluate this, we use the Implication Rule:

Applying this to our option, we get $\sim p\vee (p\vee q)$. Using the Associative Law again, we group the terms:

Just as before, $(\sim p\vee p)$ is a Tautology ($T$). So, we are left with $T\vee q$. By the Domination Law, $T\vee q\equiv T$.

We have found our match! Option (D) is the correct equivalent expression. Through this process, we have not only solved the problem but also witnessed the beautiful, consistent structure of logic itself.