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Use rules of inference to show that if \(\forall x(P(x) \vee Q(x))\) and \(\forall x((\neg P(x) \wedge Q(x)) \rightarrow R(x))\) are true, then \(\forall x(\neg R(x) \rightarrow\) \(P(x)\) is also true, where the domains of all quantifiers are the same.

Short Answer

Expert verified
From the premises, using the rules of inference, \(\forall x(eg R(x) \rightarrow P(x))\) follows.

Step by step solution

01

Restate the Given Statements

Given the following premises:1. \(\forall x(P(x) \vee Q(x))\)2. \(\forall x((eg P(x) \wedge Q(x)) \rightarrow R(x))\)
02

Simplify the Premises

The first premise says that for all x, \(P(x) \vee Q(x)\) is true. The second premise says that for all x, \(eg P(x) \wedge Q(x) \rightarrow R(x)\) is true.
03

Use Modus Tollens on the Second Premise

Modus Tollens states that if \(A \rightarrow B\) and \eg B\ are true, then \eg A\ must be true. Applying this to the second premise:Given \(eg P(x) \wedge Q(x) \rightarrow R(x)\) and assuming \eg R(x)\, we get \[eg(eg P(x) \wedge Q(x)) \text{or equivalently} \egeg P(x) \vee eg Q(x)\]By double negation and distribution: \[P(x) \vee eg Q(x)\]
04

Combine with the First Premise

From the first premise, \(P(x) \vee Q(x)\), combine it with the result from Step 3:We already have \(P(x) \vee eg Q(x)\)Therefore, for \(eg Q(x)\) to hold true \(P(x)\), must be true.Thus, if \(eg R(x)\) is true, \(P(x)\) must be true for all \(x\).
05

Conclude the Result

Hence, we can conclude that \(\forall x(eg R(x) \rightarrow P(x))\) is true based on the premises given.

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Key Concepts

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

Quantifiers
Quantifiers are symbols used in logic to express statements about quantities. The two most common quantifiers are the universal quantifier \( \forall \), which means 'for all', and the existential quantifier \( \exists \), which means 'there exists'. When we use quantifiers, we want to make statements that apply to all members of a certain set or at least one member of a set.

For example, in the exercise, the statement \( \forall x (P(x) \vee Q(x)) \) means that for every element \(x \) in the domain, either \(P(x) \) is true or \(Q(x) \) is true. Quantifiers help in defining the scope of our logical assertions, making it clear whether we're talking about all elements or just some.

Remember, the domain of the quantifiers should be consistent throughout the problem. This consistency ensures the logical flow and correctness of our arguments.
Modus Tollens
Modus Tollens is a rule of inference that allows us to conclude the negation of a statement given if a conditional statement and the negation of its consequent are true. In simpler terms, if \(A \rightarrow B \) and \( eg B \) are true, then we can conclude \( eg A \) is true.

Here's a step-by-step application:
  • Start with the conditional statement: \( \forall x(eg P(x) \wedge Q(x) \rightarrow R(x)) \).
  • Assume that the consequent is false, \( eg R(x) \).
  • Using modus tollens, we can infer the negation of the antecedent, which gives us \( eg (eg P(x) \wedge Q(x)) \).
Modus Tollens is powerful because it helps us work backward from known falsity to deduce another falsity. This process ensures our logical pathway stays valid and consistent.
Logical Statements
Logical statements are sentences or expressions that can be either true or false. They form the building blocks of logical reasoning. In our exercise, we encounter several logical statements involving \(P(x) \), \(Q(x) \), and \(R(x) \).

Types of logical connectives used:
  • **Conjunction** (and): \(P(x) \wedge Q(x) \).
  • **Disjunction** (or): \( P(x) \vee Q(x) \).
  • **Negation** (not): \( eg P(x) \).
  • **Implication**: \( (eg P(x) \wedge Q(x)) \rightarrow R(x) \).
Logical statements are manipulated using rules of inference to derive new truths from known premises. Carefully considering the truth values and relationships between these statements allows us to logically deduce new conclusions, such as \( \forall x(eg R(x) \rightarrow P(x)) \).
Double Negation
Double negation is a logical law stating that the negation of a negation of a statement is equivalent to the original statement. Mathematically, this can be seen as \( eg (eg P(x)) \equiv P(x) \).

In our exercise, we used double negation to simplify \( eg (eg P(x) \wedge Q(x)) \) to \( P(x) \vee eg Q(x) \). The steps are:
  • Recognize that \( eg (eg P(x) \wedge Q(x)) \) can be written as \( eg eg P(x) \vee eg Q(x) \) using De Morgan's laws.
  • Apply double negation to get \( P(x) \vee eg Q(x) \).
Double negation is a helpful tool in proofs, allowing us to eliminate unnecessary complexity in expressions. Understanding and applying double negation ensures our logical statements are as straightforward as possible.

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Most popular questions from this chapter

For each of these arguments, explain which rules of inference are used for each step. a) 鈥淟inda, a student in this class, owns a red convertible. Everyone who owns a red convertible has gotten at least one speeding ticket. Therefore, someone in this class has gotten a speeding ticket.鈥 b) 鈥淓ach of five roommates, Melissa, Aaron, Ralph, Veneesha, and Keeshawn, has taken a course in discrete mathematics. Every student who has taken a course in discrete mathematics can take a course in algorithms. Therefore, all five roommates can take a course in algorithms next year.鈥 c) 鈥淎ll movies produced by John Sayles are wonder-ful. John Sayles produced a movie about coal miners. Therefore, there is a wonderful movie about coal miners.鈥 d) 鈥淭here is someone in this class who has been to France. Everyone who goes to France visits the Louvre. Therefore, someone in this class has visited the Louvre.鈥

The quadratic mean of two real numbers \(x\) and \(y\) equals \(\sqrt{\left(x^{2}+y^{2}\right) / 2}\) . By computing the arithmetic and quadratic means of different pairs of positive real numbers, formulate a conjecture about their relative sizes and prove your conjecture.

Let \(P(x)\) be the statement " \(x=x^{2} .\) "If the domain consists of the integers, what are these truth values? $$ \begin{array}{llll}{\text { a) } P(0)} & {\text { b) } P(1)} & {\text { c) } P(2)} \\ {\text { d) } P(-1)} & {\text { e) } \exists x P(x)} & {\text { f) } \forall x P(x)}\end{array} $$

Prove that there are infinitely many solutions in positive integers \(x, y,\) and \(z\) to the equation \(x^{2}+y^{2}=\) \(z^{2} .\left[\text { Hint: Let } x=m^{2}-n^{2}, y=2 m n, \text { and } z=m^{2}+n^{2}\right.\) where \(m\) and \(n\) are integers. \(]\)

Use resolution to show that the compound proposition \((p \vee q) \wedge(\neg p \vee q) \wedge(p \vee \neg q) \wedge(\neg p \vee \neg q)\) is not satisfiable.

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