Problem 50
Question
For any two statements \(\mathrm{p}\) and \(\mathrm{q}\), the negation of the expression \(\mathrm{p} \vee(\sim \mathrm{p} \wedge \mathrm{q})\) is: (a) \(\sim \mathrm{p} \wedge \sim \mathrm{q}\) (b) \(\mathrm{p} \wedge q\) (c) \(\mathrm{p} \leftrightarrow q\) (d) \(\sim \mathrm{p} \vee \sim \mathrm{q}\)
Step-by-Step Solution
Verified Answer
The answer is (a) \( \sim \mathrm{p} \wedge \sim \mathrm{q} \).
1Step 1: Understand the Expression
The expression to negate is \( \mathrm{p} \vee (\sim \mathrm{p} \wedge \mathrm{q}) \). It consists of the disjunction (OR) of \( \mathrm{p} \) with \( \sim \mathrm{p} \wedge \mathrm{q} \).
2Step 2: Apply De Morgan's Laws
Applying De Morgan's laws to the expression \( \sim(\mathrm{p} \vee (\sim \mathrm{p} \wedge \mathrm{q})) \) gives the equivalent expression \( \sim \mathrm{p} \wedge \sim (\sim \mathrm{p} \wedge \mathrm{q}) \). This step involves negating the entire disjunction.
3Step 3: Simplify Inside the Conjunction
Next, we simplify the negation inside the conjunction: \( \sim (\sim \mathrm{p} \wedge \mathrm{q}) \) becomes \( \mathrm{p} \vee \sim \mathrm{q} \) using De Morgan's laws again.
4Step 4: Combine Parts
Combine the results from steps 2 and 3 into \( \sim \mathrm{p} \wedge (\mathrm{p} \vee \sim \mathrm{q}) \). Use distribution to expand this to \((\sim \mathrm{p} \wedge \mathrm{p}) \vee (\sim \mathrm{p} \wedge \sim \mathrm{q})\).
5Step 5: Simplify Using Contradictions
Since \( \sim \mathrm{p} \wedge \mathrm{p} \) is a contradiction, it simplifies to \( \text{False} \). Therefore, the expression reduces to \( \sim \mathrm{p} \wedge \sim \mathrm{q} \).
6Step 6: Identify the Correct Answer
The simplified expression \( \sim \mathrm{p} \wedge \sim \mathrm{q} \) matches option (a).
Key Concepts
De Morgan's lawsLogical ExpressionsDisjunction and Conjunction
De Morgan's laws
De Morgan's laws are fundamental rules in logic that are used to simplify and transform logical expressions. They connect two types of basic logical operations: negation and conjunction/disjunction. Understanding these laws helps you easily manipulate expressions involving AND (\(\wedge\)) and OR (\(\vee\)) operators, especially when negation is involved. Here’s a quick breakdown of the laws:
- De Morgan's first law states that the negation of a conjunction is equivalent to the disjunction of the negations. Formally, this can be written as \( \sim(p \wedge q) = \sim p \vee \sim q \).
- De Morgan's second law states that the negation of a disjunction is equivalent to the conjunction of the negations: \( \sim(p \vee q) = \sim p \wedge \sim q \).
Logical Expressions
A logical expression is composed of variables, constants, and logical operations that convey logical statements and propositions. These expressions are integral to fields like mathematics, computer science, and philosophy, as they form the basis of Boolean logic which underpins modern computing. Logical expressions commonly use logical connectives such as:
- AND (\(\wedge\))
- OR (\(\vee\))
- NOT (\(\sim\))
Disjunction and Conjunction
Disjunction and conjunction are two primary binary operations in logical expressions, often seen in expressions and conditions in mathematical logic.**Disjunction:** The disjunction (OR operation), symbolized by \(\vee\), combines two propositions to form a new proposition that is true if at least one of the propositions is true. For example, in \(\mathrm{p} \vee \mathrm{q}\), the expression is true if either \(\mathrm{p}\) or \(\mathrm{q}\) is true or both.**Conjunction:** The conjunction (AND operation), symbolized by \(\wedge\), combines two propositions such that the resulting proposition is only true if both propositions are true. For example, \(\mathrm{p} \wedge \mathrm{q}\) is only true if both \(\mathrm{p}\) and \(\mathrm{q}\) are true.When working with these operations:
- Remember that disjunction allows more flexibility in meeting the truth than conjunction.
- These operations are often coupled with negation, and simplification might require the use of De Morgan's laws to transform expressions, especially when expressions get nested or more complex.
Other exercises in this chapter
Problem 47
Negation of the statement: \(\sqrt{5}\) is an integer of 5 is irrational is: \(\quad\) (a) \(\sqrt{5}\) is not an integer or 5 is not irrational (b) \(\sqrt{5}\
View solution Problem 49
The negation of the Boolean expression \(\sim s \vee(\sim r \wedge s)\) is equivalent to: \(\quad\) (a) \(\sim s \wedge \sim r\) (b) \(r\) (c) \(s \vee r\) (d)
View solution Problem 51
The contrapositive of the statement "If you are born in India, then you are a citizen of India", is : (a) If you are not a citizen of India, then you are not bo
View solution Problem 52
Contrapositive of the statement "If two numbers are not equal, then their squares are not equal", is : (a) If the squares of two numbers are not equal, then the
View solution