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Chapter 3: Problem 36

Use a truth table to determine whether each statement is a tautology, a self- contradiction, or neither. \([(p \rightarrow q) \wedge \sim p] \rightarrow \sim q\)

Short Answer

Expert verified
Upon creating a truth table and seeing the mix of 'true' and 'false' values in the column of the full proposition, we can conclude that the given statement \( [(p \rightarrow q) \wedge \sim p] \rightarrow \sim q\) is neither a tautology nor a self-contradiction.

Step by step solution

01

Setup the truth table

List the possible combinations of truth values for the propositions \(p\) and \(q\) in a table. There are four possibilities to consider when we have two propositions: both true, both false, one true and the other false, and vice versa.
02

Calculate the intermediate values

Calculate the truth value of intermediate compound propositions \((p \rightarrow q)\), \(\sim p\), and \(\sim q\) based on the original truth values of \(p\) and \(q\). Remember that in logical notation, '\(\rightarrow\)' means 'implies', and '\(\sim\)' means 'not'.
03

Calculate the value of the full proposition

Calculate the truth value of the logical conjunction \((p \rightarrow q) \wedge \sim p\) and subsequently the final implication \((p \rightarrow q) \wedge \sim p) \rightarrow \sim q\). An implication is true except for the case where a true proposition implies a false one. A conjunction \(\wedge\) is true only when both propositions are true.
04

Determine whether the statement is a tautology, a self-contradiction, or neither

After filling all the truth values for the whole proposition according to the truth table, we can make a conclusion about the proposition. If all entries under the final proposition are 'true', then the statement is a tautology. If they're all 'false', it's a self-contradiction. If there's a mix of 'true' and 'false', the statement is considered neither.

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