/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 8\. Give the reasons for the ste... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

8\. Give the reasons for the steps verifying the following argument. \((\neg p \vee q) \rightarrow r\) \(r \rightarrow(s \vee t)\) \(\quad \neg s \wedge \neg u\) \(\frac{\neg u \rightarrow \neg t} \\\\{\therefore p} \end{array\) Steps Reasons 1) \(\neg s \wedge \neg u\) 2) \(\neg u\) 3) \(\neg u \rightarrow \neg t\) 4) \(\neg t\) 5) \(\neg s\) 6) \(\neg s \wedge \neg t\) 7) \(r \rightarrow(s \vee t)\) 8) \(\neg(s \vee t) \rightarrow \neg r\) 9) \((\neg s \wedge \neg t) \rightarrow \neg r\) 10) \(\neg r\) 11) \((\neg p \vee q) \rightarrow r\) 12) \(\neg r \rightarrow \neg(\neg p \vee q)\) 13) \(\neg r \rightarrow(p \wedge \neg q)\) 14) \(p \wedge \neg q\) 15) \(\therefore p\)

Short Answer

Expert verified
Therefore, based on the given premises, the conclusion \(p\) is logically true.

Step by step solution

01

Identify given premises and conclusion

The premises are \((\neg p \vee q) \rightarrow r\), \(r \rightarrow(s \vee t)\) and \(\neg s \wedge \neg u\), and the conclusion is \(p\).
02

Logical deduction from premise

From the premise, \(\neg s \wedge \neg u\) , it can be logically inferred that \(\neg u\) (logical conjunction elimination). Additionally, it can be inferred that \(\neg u \rightarrow \neg t\) (modus ponens). This results in \(\neg t\).
03

Further logical deduction

It also can be inferred from \(\neg s \wedge \neg u\) that \(\neg s\) (logical conjunction elimination). Together with \(\neg t\), we get \(\neg s \wedge \neg t\) (logical conjunction).
04

Application of Modus Tolens

Given the premise \(r \rightarrow(s \vee t)\), by contraposition or modus tollens, it becomes \(\neg(s \vee t) \rightarrow \neg r\). Thus, based on \(\neg s \wedge \neg t\), we can infer \(\neg r\).
05

Use of Double Negation

From the premise (\(\neg p \vee q) \rightarrow r\), it can be written in contraposition form as \(\neg r \rightarrow \neg(\neg p \vee q)\). Further using De Morgan's law of negation and double negation, we get \(\neg r \rightarrow(p \wedge \neg q)\). Based on \(\neg r\), it can infer that \(p \wedge \neg q\). Thus, the solution is \(p\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logical Deduction
Logical deduction is the process of deriving specific conclusions from general statements or premises. It forms the backbone of propositional logic, helping us validate the truth of an argument step by step.
In our exercise, we start with premises such as \( (eg p \vee q) \rightarrow r \) and \( r \rightarrow (s \vee t) \).
These premises set the stage for our deductions. We logically conclude each step by applying known principles.
For instance, from the conjunction \( eg s \wedge eg u \), we can independently infer both \( eg s \) and \( eg u \).
Logical deduction ensures every reason given for an argument follows a valid logical path, typically using rules like "if this happens, then that follows." It helps us rule out false assumptions and focus on proven facts.
In logical deduction, each derived statement must be substantiated by recognized logical rules, ensuring the argument is sound and valid.
Modus Ponens
Modus Ponens is a fundamental rule in propositional logic. It allows us to infer a conclusion from a conditional statement and its antecedent. You can think of it as "if \( P \) then \( Q \). \( P \) is true, so \( Q \) must also be true."
In the given exercise, we use Modus Ponens with the premise \( eg u \rightarrow eg t \).
Since we know \( eg u \) from our earlier deduction, we apply Modus Ponens to conclude that \( eg t \) is valid.
This step is crucial because it validates one part of our compound negations that we later use to derive further conclusions.
Modus Ponens is straightforward yet powerful, as it allows both simple reasoning and complex derivations by expanding from basic statements.
Modus Tollens
Modus Tollens is another crucial logical rule, used particularly when our propositions involve denying conclusions. Think of it like this: "if \( P \) then \( Q \). \( Q \) is not true, hence \( P \) is not true."
This rule is essential in situations where we have the contrapositive of a situation and need to infer a logical conclusion.
In our scenario, the premise \( r \rightarrow (s \vee t) \) allows us to apply Modus Tollens after deducing \( eg s \wedge eg t \).
Here, because neither \( s \) nor \( t \) is true, we infer that \( r \) cannot be true either.
Modus Tollens is invaluable in proving situations where deducing a false conclusion can affirm the falseness of its antecedent.
Conjunction Elimination
Conjunction Elimination, also known as simplification, is a logical rule that helps break down compound statements or conjunctions into their individual components.
It is the process where from a statement like \( eg s \wedge eg u \), we can conclude \( eg s \) and \( eg u \) separately.
This rule makes analysis more manageable by isolating individual parts of a complex proposition.

In our exercise, we use it as a first step to derive separate propositions, allowing other logical rules, such as Modus Ponens and Modus Tollens, to be applied effectively.
Without conjunction elimination, we could not independently validate each conjunct of a combined statement which is crucial for further deductions.
This step is foundational in breaking down arguments into manageable parts that facilitate structured logical reasoning.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

3\. Let \(p, q\) be primitive statements for which the implication \(p \rightarrow q\) is false. Determine the truth values for each of the following. a) \(p \wedge q\) b) \(\neg p \vee q\) c) \(q \rightarrow p\) d) \(\neg q \rightarrow \neg p\)

16\. Prove that for every integer \(n, n^{2}\) is even if and only if \(n\) is even.

19\. For each of the following statements state the converse, inverse, and contrapositive. Also determine the truth value for each given statement, as well as the truth values for its converse, inverse, and contrapositive. (Here "divides" means "exactly divides.") a) [The universe comprises all positive integers.] If \(m>n\), then \(m^{2}>n^{2}\) b) [The universe comprises all integers.] If \(a>b\), then \(a^{2}>b^{2}\). c) [The universe comprises all integers.] If \(m\) divides \(n\) and \(n\) divides \(p\), then \(m\) divides \(p\). d) [The universe consists of all real numbers.] \(\forall x\left[(x>3) \rightarrow\left(x^{2}>9\right)\right]\) e) [The universe consists of all real numbers.] For all real numbers \(x\), if \(x^{2}+4 x-21>0\), then \(x>3\) or \(x<-7\)

12\. Write each of the following arguments in symbolic form. Then establish the validity of the argument or give a counterexample to show that it is invalid. a) If Rochelle gets the supervisor's position and works hard, then she'll get a raise. If she gets the raise, then she'll buy a new car. She has not purchased a new car. Therefore either Rochelle did not get the supervisor's position or she did not work hard. b) If Dominic goes to the racetrack, then Helen will be mad. If Ralph plays cards all night, then Carmela will be mad. If either Helen or Carmela gets mad, then Veronica (their attorney) will be notified. Veronica has not heard from either of these two clients. Consequently, Dominic didn't make it to the racetrack and Ralph didn't play cards all night. c) If there is a chance of rain or her red headband is missing, then Lois will not mow her lawn. Whenever the temperature is over \(80^{\circ} \mathrm{F}\), there is no chance for rain. Today the temperature is \(85^{\circ} \mathrm{F}\) and Lois is wearing her red headband. Therefore (sometime today) Lois will mow her lawn.

7\. a) If \(p, q\) are primitive statements, prove that $$ (\neg p \vee q) \wedge(p \wedge(p \wedge q)) \Leftrightarrow(p \wedge q) $$ b) Write the dual of the logical equivalence in part (a).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.