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Chapter 20: Problem 1918

\(p \Rightarrow(q \Rightarrow p) \Rightarrow r\) is (a) Contradiction (b) tautology (c) Neither contradiction Nor tautology (d) Both contradiction \& tautology

Short Answer

Expert verified
(c) Neither contradiction Nor tautology

Step by step solution

01

Setting up the truth table

To evaluate the truth of the given statement, we have to consider all possible values of \(p\), \(q\), and \(r\). We will create the following columns: 1. \(p\) 2. \(q\) 3. \(r\) 4. \(q \Rightarrow p\) 5. \((q \Rightarrow p) \Rightarrow r\) 6. \(p \Rightarrow ((q \Rightarrow p) \Rightarrow r)\) Now, we will fill up the table with all possible values of \(p\), \(q\), and \(r\).
02

Filling in all possible values

Fill in the table with all possible combinations of \(p\), \(q\), and \(r\). There are 2 possible values (True or False) for each variable, so there will be \(2^3 = 8\) rows in the table. $$ \begin{array}{ccc|c|c|c} p & q & r & q \Rightarrow p & (q \Rightarrow p) \Rightarrow r & p \Rightarrow ((q \Rightarrow p) \Rightarrow r)\\ \hline T & T & T & T & T & T \\ T & T & F & T & F & F \\ T & F & T & T & T & T \\ T & F & F & T & F & F \\ F & T & T & F & T & T \\ F & T & F & F & F & T \\ F & F & T & T & T & T \\ F & F & F & T & F & T \\ \end{array} $$
03

Analyzing the truth table

Now that we have the truth table completed, let's analyze the final column (the statement \(p \Rightarrow ((q \Rightarrow p) \Rightarrow r)\)). If every entry in the column is true, then the statement would be a tautology. If every entry is false, it's a contradiction. If there's a mix of true and false values, it's neither. In our case, we have a mix of true and false values in the final column. Therefore, the statement is neither a contradiction nor a tautology. #Answer#: (c) Neither contradiction Nor tautology

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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) \(\mathrm{p}\)

\(\sim(p V \sim q) V(\sim p \wedge q)=\ldots \ldots . .\) (a) \(\mathrm{p}\) (b) \(\sim \mathrm{p}\) (c) \(q\) (d) \(\sim \mathrm{q}\)

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

\(\mathrm{p} \Rightarrow(\mathrm{q} \Rightarrow \mathrm{p})\) is equivalent to (a) \(p \Rightarrow(p \Leftrightarrow q)\) (b) \(p \Rightarrow(p \Rightarrow q)\) (c) \(p \Rightarrow(p \vee q)\) (d) \(p \Rightarrow(p \wedge q)\)

The negation of compound proposition \(p V(\sim p V q)\) is \(\ldots \ldots \ldots .\) (a) \((p \Lambda \sim q) \Lambda \sim p\) (b) \((\mathrm{p} \Lambda \sim \mathrm{p}) \mathrm{V} \sim \mathrm{q}\) (c) \((p \Lambda \sim q) V(\sim p)\) (d) \((\mathrm{p} \Lambda \sim \mathrm{q}) \mathrm{V} \sim \mathrm{p}\)
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