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Chapter 3: Problem 24

Construct a truth table for the given statement. \((p \leftrightarrow \sim q) \rightarrow(q \rightarrow \sim p)\)

Short Answer

Expert verified
The truth table is constructed successfully by evaluating negation, bi-conditional, and implication for each combination of truth values of p and q in systematic steps. The values for the final statement depend on the outcomes of these operations.

Step by step solution

01

Define the Initial Statements

Start with defining the statements: \(p\) and \(q\). Each of these statements can take the values T (true) or F (false). Thus, we have FOUR combinations namely: (T, T), (T, F), (F, T), (F, F). So, in the first and second column of our truth table, we put down these combinations.
02

Determine the Negations

First, we compute the negations: \(\sim p\) and \(\sim q\). To compute the negation, we invert the initial statement. If the initial statement is true (T), the negation is false (F), and vice versa. We record the values for \(\sim p\) and \(\sim q\) in the next columns of our truth table.
03

Evaluate the Bi-Conditional Expression

We evaluate the bi-conditional expression: \(p \leftrightarrow \sim q\). A bi-conditional is true if both its components have the same truth value, false otherwise. We keep track of the results in the following column.
04

Evaluate implication: \(q \rightarrow \sim p\)

The statement \(q \rightarrow \sim p\) is true unless \(q\) is true and \(\sim p\) is false. Write down the result in the next column of the truth table.
05

Evaluate the Full Statement

Now, evaluate the full given statement: \((p \leftrightarrow \sim q) \rightarrow (q \rightarrow \sim p)\). The implication \(A \rightarrow B\) is false only in the case where \(A\) is true and \(B\) is false, otherwise it's true. We apply this rule and determine the truth values for the full statement, and note these down for each row in the truth table.

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Most popular questions from this chapter

Use a truth table to determine whether the symbolic form of the argument is valid or invalid. $$ \begin{aligned} &q \rightarrow \sim p \\ &q \wedge r \\ &\therefore r \rightarrow p \end{aligned} $$

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