The Boolean Expression (p ∧ ~ q) ∨ q ∨ (~ p ∧ q) is equivalent to: (a) p ∨ q (b) p ∨ ~ q (c) ~ p ∧ q

Q: Wait, why can we just group the last two terms together? Is that allowed?

Yes, absolutely! This is thanks to the Associative Law of Boolean Algebra. Just like in standard arithmetic where $(a + b) + c = a + (b + c)$, in Boolean algebra, the OR operator ($\vee$) is associative. This means $(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$ is equivalent to $(p \wedge \sim q) \vee [q \vee (\sim p \wedge q)]$. Grouping them allows us to isolate the part that can be simplified using the Absorption Law.

Q: What exactly is the Absorption Law and why does it work?

The Absorption Law states that $A \vee (B \wedge A) = A$. Think of it this way: if you have a set $A$, and you take the union of $A$ with the intersection of $A$ and $B$, you are just adding a subset of $A$ to $A$ itself. Since that subset is already inside $A$, it doesn't change the total area. In our problem, $q \vee (\sim p \wedge q)$ is just $q$, because the region $(\sim p \wedge q)$ is already entirely contained within the circle $q$.

Q: Can we use truth tables instead of these laws?

You certainly can! Truth tables are a foolproof way to verify any Boolean expression. However, for JEE Advanced, time is your most precious resource. Learning these algebraic laws allows you to simplify expressions in seconds, whereas a truth table requires you to map out all combinations of $p$ and $q$. Use the laws for speed, and use the truth table if you ever feel stuck or need to double-check your work.

Question

The Boolean Expression $(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$ is equivalent to:

Options:

(A) $p \vee q$

(B) $p \vee \sim q$

(C) $\sim p \wedge q$

(D) $p \wedge q$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to demystify a classic Boolean algebra problem. Our goal is to simplify the expression $(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$.

Think of this as a map of regions in a Venn diagram. Imagine two overlapping circles, $p$ and $q$. The expression asks us to combine three specific regions.

The first term, $(p \wedge \sim q)$, is the 'left moon'—the part of $p$ that does not overlap with $q$. The second term, $q$, is the entire circle $q$. The third term, $(\sim p \wedge q)$, is the 'right moon'—the part of $q$ that does not overlap with $p$.

### The Power of Grouping

When we look at the expression $(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$, we can use the Associative Law to our advantage. Let's group the last two terms:

$$(p \wedge \sim q) \vee [q \vee (\sim p \wedge q)]$$

Now, look closely at the bracketed part: $q \vee (\sim p \wedge q)$. We are taking the union of the entire circle $q$ and the 'right moon' (which is just a part of $q$).

Since the right moon is already inside $q$, adding it doesn't change anything. This is the essence of the **Absorption Law**: $A \vee (B \wedge A) = A$. In our case, the term $(\sim p \wedge q)$ is completely absorbed by $q$. Our expression has now shrunk to $(p \wedge \sim q) \vee q$.

### The Distributive Dance

We are making great progress, but we aren't done yet. We have $(p \wedge \sim q) \vee q$. To simplify this further, we need to use the **Distributive Law**.

The law states that $(A \wedge B) \vee C = (A \vee C) \wedge (B \vee C)$. Let's apply this to our expression, where $A = p$, $B = \sim q$, and $C = q$. This transforms our expression into:

$$(p \vee q) \wedge (\sim q \vee q)$$

### The Final Reveal

Now, look at the second part of that intersection: $(\sim q \vee q)$. This is a **Tautology**! It represents everything that is not $q$ combined with everything that is $q$, which is the entire universe of possibilities, or True ($T$).

So, our expression becomes $(p \vee q) \wedge T$. According to the **Identity Law**, anything ANDed with True is just itself.

Therefore, $(p \vee q) \wedge T = p \vee q$.

And there you have it! We started with a complex, three-part expression and distilled it down to the elegant **$p \vee q$**. Visually, this makes perfect sense: the left moon combined with the entire circle $q$ covers exactly the region defined by $p$ OR $q$.

Animated Solution for Mathematics - Sets and Relations: The Boolean Expression $(p \land \sim q) \lor q \lor (\sim p \land q)$ is equivalent to:

Select Answer:

Visualized Solution

Initial Expression Analysis

Visualizing the Terms

Grouping for Simplification

Applying the Absorption Law

Simplified Expression

Using the Distributive Law

Identity and Tautology

Conclusion and Final Answer

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Power of Grouping

The Distributive Dance

The Final Reveal

Similar Questions

The Boolean expression $(\sim (p \land q)) \lor q$ is equivalent to :

The Boolean expression $(p \land q) \Rightarrow ((r \land q) \land p)$ is equivalent to:

The negation of the Boolean expression $p \lor (\sim p \land q)$ is equivalent to :

Negation of the Boolean statement $(p \lor q) \Rightarrow ((\sim r) \lor p)$ is equivalent to:

The negation of the Boolean expression $((\sim q) \land p) \Rightarrow ((\sim p) \lor q)$ is logically equivalent to

The statement $(p \land (\sim q)) \lor ((\sim p) \land q) \lor ((\sim p) \land (\sim q))$ is equivalent to

The statement $\sim [p \lor (\sim (p \land q))]$ is equivalent to

The Boolean expression $((p \land q) \lor (p \land \sim q)) \lor (\sim p \land \sim q)$ is equivalent to:

Which of the following is equivalent to the Boolean expression $p \land \sim q$ ?

The negation of the boolean expression $\sim s \lor (\sim r \land s)$ is equivalent to :

Animated Solution for Mathematics - Sets and Relations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Boolean Expression (p∧∼q)∨q∨(∼p∧q) is equivalent to: