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Chapter 20: Problem 1910

\(\mathrm{p} \Rightarrow \mathrm{q} \mathrm{V} \mathrm{r}\) is false then the true values of \(\mathrm{p}, \mathrm{q}\) and \(\mathrm{r}\) are respectively. (a) \(\mathrm{F}, \mathrm{T}, \mathrm{T}\) (b) \(\mathrm{T}, \mathrm{T}, \mathrm{F}\) (c) \(\mathrm{T}, \mathrm{F}, \mathrm{F}\) (d) \(\mathrm{F}, \mathrm{F}, \mathrm{F}\)

Short Answer

Expert verified
The short answer is: \(\mathrm{(c) \ T, F, F}\).

Step by step solution

01

Analyzing the statement pattern p → q ∨ r

Here we have a conditional with the antecedent p and the consequent as a disjunction of q and r (q ∨ r). First, let's recall the truth table of a conditional p → q: p | q | p → q T | T | T T | F | F F | T | T F | F | T Now, let's recall the truth table of a disjunction q ∨ r: q | r | q ∨ r T | T | T T | F | T F | T | T F | F | F
02

Making the statement pattern p → q ∨ r false

For the given statement pattern to be false, the antecedent p must be true, and the whole statement q ∨ r must be false. Looking at our disjunction truth table, we see that q ∨ r is false when both q and r are false. This leads us to the following truth values for p, q, and r: p: True q: False r: False This matches the option (c) \(\mathrm{T}, \mathrm{F}, \mathrm{F}\).

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Most popular questions from this chapter

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

Which of the following a tautology ? (a) p \(\Lambda(\sim \mathrm{p})\) (b) \(\mathrm{p} \Lambda \mathrm{c}\) (c) \(p \mathrm{~V} \mathrm{t}\) (d) \(\mathrm{p} \Lambda \mathrm{p}\)

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

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) \((\mathrm{p} \Lambda \sim \mathrm{q}) \mathrm{V} \sim \mathrm{p}\)

The logically equivalent proposition of \(\sim q \Rightarrow p\) is (a) \(p \Rightarrow-q\) (b) \(\sim \mathrm{p} \Rightarrow \mathrm{q}\) (c) \(\sim \mathrm{q} \Rightarrow \sim \mathrm{p}\) (d) \(\sim \mathrm{p} \Rightarrow \sim \mathrm{q}\)
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