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Chapter 1: Problem 31

Suppose that the domain of \(Q(x, y, z)\) consists of triples \(x, y, z,\) where \(x=0,1,\) or \(2, y=0\) or \(1,\) and \(z=0\) or \(1 .\) Write out these propositions using disjunctions and conjunctions. $$ \begin{array}{ll}{\text { a) } \forall y Q(0, y, 0)} & {\text { b) } \exists x Q(x, 1,1)} \\ {\text { c) } \exists z \neg Q(0,0, z)} & {\text { d) } \exists x \neg Q(x, 0,1)}\end{array} $$

Short Answer

Expert verified
a) \[ Q(0, 0, 0) \land Q(0, 1, 0) \] b) \[ Q(0, 1, 1) \lor Q(1, 1, 1) \lor Q(2, 1, 1) \] c) \[ eg Q(0, 0, 0) \lor eg Q(0, 0, 1) \] d) \[ eg Q(0, 0, 1) \lor eg Q(1, 0, 1) \lor eg Q(2, 0, 1) \]

Step by step solution

01

Identify the domain

The domain of the predicate consists of triples (x, y, z) where \[x \in \{0, 1, 2\}, \ y \in \{0, 1\}, \ z \in \{0, 1\} \].
02

Statement a: Expressing \( \forall y Q(0, y, 0) \)

The statement \( \forall y Q(0, y, 0) \) means that for all values of y within the domain, \( Q(0, y, 0) \) holds true.There are two possible values for y: 0 and 1. Hence, the statement can be written as:\[ Q(0, 0, 0) \land Q(0, 1, 0) \]
03

Statement b: Expressing \( \exists x Q(x, 1, 1) \)

The statement \( \exists x Q(x, 1, 1) \) means that there exists at least one x such that \( Q(x, 1, 1) \) holds true.There are three possible values for x: 0, 1, and 2. Hence, the statement can be written as:\[ Q(0, 1, 1) \lor Q(1, 1, 1) \lor Q(2, 1, 1) \]
04

Statement c: Expressing \( \exists z \eg Q(0, 0, z) \)

The statement \( \exists z \eg Q(0, 0, z) \) means that there exists at least one z such that \( \eg Q(0, 0, z) \) holds true.There are two possible values for z: 0 and 1. Hence, the statement can be written as:\[ \eg Q(0, 0, 0) \lor \eg Q(0, 0, 1) \]
05

Statement d: Expressing \( \exists x \eg Q(x, 0, 1) \)

The statement \( \exists x \eg Q(x, 0, 1) \) means that there exists at least one x such that \( \eg Q(x, 0, 1) \) holds true.There are three possible values for x: 0, 1, and 2. Hence, the statement can be written as:\[ \eg Q(0, 0, 1) \lor \eg Q(1, 0, 1) \lor \eg Q(2, 0, 1) \]

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Key Concepts

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

Universal Quantifiers
Universal quantifiers are represented by the symbol \( \forall \), which means 'for all' or 'for every.' They are used in propositions to state that a certain condition applies to all elements within a specific domain. For instance, consider the expression \( \forall y Q(0, y, 0) \). This means that the proposition \( Q(0, y, 0) \) holds true for every value of \( y \) in the domain.
In the given exercise, since the domain of \( y \) is \{0, 1\}, the statement \( \forall y Q(0, y, 0) \) can be expanded to \( Q(0, 0, 0) \land Q(0, 1, 0) \). This indicates that both \( Q(0, 0, 0) \) and \( Q(0, 1, 0) \) are true.
Existential Quantifiers
Existential quantifiers use the symbol \( \exists \), which translates to 'there exists' or 'for at least one.' They are used to express that there is at least one element in the domain for which the predicate holds true. Take the example \( \exists x Q(x, 1, 1) \). This means that there is at least one value of \( x \) for which \( Q(x, 1, 1) \) is true.
Given the domain of \( x \) as \{0, 1, 2\}, the statement \( \exists x Q(x, 1, 1) \) can be expanded to \( Q(0, 1, 1) \lor Q(1, 1, 1) \lor Q(2, 1, 1) \). Essentially, this tells us that at least one of \( Q(0, 1, 1) \), \( Q(1, 1, 1) \), or \( Q(2, 1, 1) \) holds true.
Logical Conjunction
Logical conjunction is denoted by the symbol \( \land \), also known as 'and.' It is used to combine two or more propositions that must all be true simultaneously. For example, in this problem, the expression \( Q(0, 0, 0) \land Q(0, 1, 0) \) means that both \( Q(0, 0, 0) \) and \( Q(0, 1, 0) \) are true.
In general, if you have propositions \( A \) and \( B \), \( A \land B \) is true only if both \( A \) and \( B \) are true. If either \( A \) or \( B \) is false (or both), then \( A \land B \) is false.
Logical Disjunction
Logical disjunction is represented by the symbol \( \lor \), which means 'or.' It is used to combine propositions such that at least one of them must be true. For instance, the statement \( Q(0, 1, 1) \lor Q(1, 1, 1) \lor Q(2, 1, 1) \) means that at least one of these propositions is true.
In general, if you have propositions \( A \) and \( B \), \( A \lor B \) is true if at least one of \( A \) or \( B \) is true. It is false only if both \( A \) and \( B \) are false.
Negation in Logic
Negation in logic involves reversing the truth value of a given proposition and is symbolized by \( \eg \). If a proposition \( Q \) is true, then \( \eg Q \) is false, and if \( Q \) is false, then \( \eg Q \) is true. Consider the statement \( \exists z \eg Q(0, 0, z) \), which means there exists at least one value of \( z \) such that \( Q(0, 0, z) \) is not true.
In the exercise, each value of \( z \) can either be 0 or 1. Consequently, the expression \( \exists z \eg Q(0, 0, z) \) can be expanded to \( \eg Q(0, 0, 0) \lor \eg Q(0, 0, 1) \). This tells us that either \( Q(0, 0, 0) \) is false, or \( Q(0, 0, 1) \) is false, or both.

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Most popular questions from this chapter

Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expression involving logical connectives). a) \(\neg \forall x \forall y P(x, y) \quad\) b) \(\neg \forall y \exists x P(x, y)\) c) \(\neg \forall y \forall x(P(x, y) \vee Q(x, y))\) d) \(\neg(\exists x \exists y \neg P(x, y) \wedge \forall x \forall y Q(x, y))\) e) \(\quad \neg \forall x(\exists y \forall z P(x, y, z) \wedge \exists z \forall y P(x, y, z))\)

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. \(]\)

Prove that \(\sqrt[3]{2}\) is irrational.

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