The statement ~[p ∨ (~(p ∧ q))] is equivalent to (1) ~(p ∧ q) (2) (p ∧ q) ∧ (~p) (3) (~(p ∧ q)) ∧ q

Q: Wait, why does the OR symbol change to AND when we apply De Morgan's Law?

Great question! Think of De Morgan's Law as a logical filter. When you negate a disjunction (an OR statement), you are saying that 'neither A nor B is true.' For that to be true, both A must be false AND B must be false. Hence, the OR flips to an AND. It is a fundamental rule of logical consistency.

Q: Can I just distribute the negation inside the bracket without using De Morgan's Law?

You cannot simply 'distribute' a negation like a negative sign in algebra. Logic has its own specific distributive rules. If you try to distribute without flipping the operator, you will end up with an incorrect logical statement. Always rely on the formal laws like De Morgan's to ensure your steps are valid.

Q: Why do we need the Double Negation Law here? Can't we just ignore the two negations?

In logic, two negations do not just 'disappear'—they cancel each other out to return to the original truth value. It is exactly like saying 'I am not not going'—it means you are going. Mathematically, $\sim(\sim X) \equiv X$. It is a crucial step to clean up the expression and reach the final simplified form.

Question

The statement $\sim[p \vee (\sim(p \wedge q))]$ is equivalent to

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with simplifying the logical expression $\sim[p \vee (\sim(p \wedge q))]$. Our goal is to find its equivalent form by systematically breaking down the logical structure from the outside in.

Think of the expression as a nested puzzle where the outer negation acts as a shell that must be addressed first.

### The Power of De Morgan's Law

When we face a negation outside a bracket containing an OR ($\vee$) operation, we utilize **De Morgan's Law**. This law states that:

$$\sim(A \vee B) \equiv \sim A \wedge \sim B$$

Notice that the negation distributes to both terms, and the OR operator flips into an AND operator ($\wedge$). Applying this to our expression, where $A = p$ and $B = \sim(p \wedge q)$, we get:

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

We have successfully dismantled the outer shell of the expression.

### The Beauty of Double Negation

Now, consider the second part of our new expression: $\sim[\sim(p \wedge q)]$. We encounter two negations acting upon the same term.

According to the **Double Negation Law**, which states that $\sim(\sim X) \equiv X$, these two negations cancel each other out. This simplifies the term to $(p \wedge q)$.

Substituting this back into our expression, we obtain:

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

### The Final Rearrangement

We are now at the final stage of simplification. We have the expression $\sim p \wedge (p \wedge q)$.

By applying the **Associative Property** of the AND operator, we can regroup the terms:

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

Since the term $(\sim p \wedge p)$ is a contradiction (it is always false), the entire expression simplifies to:

$$F \wedge q \equiv F$$

Thus, the expression is logically equivalent to a **Contradiction (False)**. Through the application of rigorous logical laws, we have revealed the underlying truth of the expression.

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

Select Answer:

Visualized Solution

Analyze the Expression

Recall De Morgan's Law

Identify Terms for Substitution

Apply De Morgan's Law

Double Negation Law

Simplify the Expression

Commutative Property

Final Conclusion

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Power of De Morgan's Law

The Beauty of Double Negation

The Final Rearrangement

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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 statement ∼[p∨(∼(p∧q))] is equivalent to