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Chapter 20: Problem 1905

\((p \wedge \sim q) \Lambda(\sim p \wedge q)\) is (a) a contradiction (b) a tautology (c) neither a tautology nor a contradiction (d) both tautology and contradiction

Short Answer

Expert verified
The expression \((p \wedge \sim q) \Lambda(\sim p \wedge q)\) is (a) a contradiction.

Step by step solution

01

Construct a truth table for the expression

First, let's create a truth table with columns for p, q, \(\sim q\), \(p \wedge \sim q\), \(\sim p\), \( \sim p \wedge q\), and \((p \wedge \sim q) \Lambda(\sim p \wedge q)\). We will have four rows representing all possible combinations of truth values for p and q.
02

Fill in the truth values for p and q

We will start by filling in the truth values for p and q, systematically, for each row: 1. p = T, q = T 2. p = T, q = F 3. p = F, q = T 4. p = F, q = F
03

Calculate the truth values for negations and conjunctions

Next, let's fill in the truth values for the negations and conjunctions in the expression: 1. For the first row, where p = T and q = T: \(\sim q = F, p \wedge \sim q = F, \sim p = F, \sim p \wedge q = F\) 2. For the second row, where p = T and q = F: \(\sim q = T, p \wedge \sim q = T, \sim p = F, \sim p \wedge q = F\) 3. For the third row, where p = F and q = T: \(\sim q = F, p \wedge \sim q = F, \sim p = T, \sim p \wedge q = T\) 4. For the fourth row, where p = F and q = F: \(\sim q = T, p \wedge \sim q = F, \sim p = T, \sim p \wedge q = F\)
04

Calculate the truth values for the entire expression

Finally, we will fill in the truth values for the entire expression \((p \wedge \sim q) \Lambda(\sim p \wedge q)\), using the calculated values for each conjunction: 1. For the first row, where \(p \wedge \sim q = F\) and \(\sim p \wedge q = F\): \((p \wedge \sim q) \Lambda(\sim p \wedge q) = F\) 2. For the second row, where \(p \wedge \sim q = T\) and \(\sim p \wedge q = F\): \((p \wedge \sim q) \Lambda(\sim p \wedge q) = F\) 3. For the third row, where \(p \wedge \sim q = F\) and \(\sim p \wedge q = T\): \((p \wedge \sim q) \Lambda(\sim p \wedge q) = F\) 4. For the fourth row, where \(p \wedge \sim q = F\) and \(\sim p \wedge q = F\): \((p \wedge \sim q) \Lambda(\sim p \wedge q) = F\)
05

Analyze the truth table to determine the classification of the expression

Looking at the truth values for the entire expression in the truth table, we see that the expression is always False, regardless of the truth values of p and q. Therefore, the expression \((p \wedge \sim q) \Lambda(\sim p \wedge q)\) is a contradiction. The correct answer is (a) a contradiction.

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Most popular questions from this chapter

Negation of for all \(\mathrm{x}, \mathrm{p}^{\prime}\) is \(\ldots \ldots \ldots\) (a) there exists \(\mathrm{x}, \sim \mathrm{p}\) (b) for all \(\mathrm{x} ; \sim \mathrm{p}\) (c) \(\sim \mathrm{p}\) (d) \(\mathrm{p}\)

Which one of the following is false (a) p \(\Lambda(\sim \mathrm{p})\) is a contradiction (b) \((\mathrm{p} \Rightarrow \mathrm{q}) \Leftrightarrow(\sim \mathrm{q} \Rightarrow \sim \mathrm{p})\) is a contradiction (c) \(\sim(\sim p) \Leftrightarrow p\) is a tautology (d) \(p \mathrm{~V}(\sim \mathrm{p})\) is a tautology

If both \(\mathrm{p}\) and \(\mathrm{q}\) are true (a) \(\mathrm{p} \Lambda \mathrm{q}\) is true (b) \(\mathrm{p} \mathrm{V} \mathrm{q}\) is false (c) \(\mathrm{p} \Rightarrow \mathrm{q}\) is false (d) None of them

If both \(\mathrm{p}\) and \(\mathrm{q}\) are false then (a) \(\mathrm{p} \Lambda \mathrm{q}\) is true (b) \(\mathrm{p} \mathrm{V} \mathrm{q}\) is false (c) \(p \Rightarrow q\) is false (d) \((\sim p) V q\) is false

If \(\mathrm{p}=\mathrm{He}\) is intelligent, \(\mathrm{q}=\mathrm{He}\) is strong. Then symbolic form of the statement "It is wrong that he is intelligent or strong" is (a) \(\sim \mathrm{p} \mathrm{V} \sim \mathrm{q}\) (b) \(\sim(\mathrm{p} \Lambda \mathrm{q})\) (c) \(\sim(p \mathrm{~V} q)\) (d) \(\mathrm{p} \Lambda \sim \mathrm{q}\)
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