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Chapter 1: Problem 19

Suppose that the domain of the propositional function \(P(x)\) consists of the integers \(1,2,3,4,\) and \(5 .\) Express these statements without using quantifiers, instead using only negations, disjunctions, and conjunctions. $$ \begin{array}{ll}{\text { a) } \quad \exists x P(x)} & {\text { b) } \forall x P(x)} \\ {\text { c) } \quad \neg \exists x P(x)} & {\text { d) } \neg \forall x P(x)}\end{array} $$ e) \(\quad \forall x((x \neq 3) \rightarrow P(x)) \vee \exists x \neg P(x)\)

Short Answer

Expert verified
Combine logical expressions about each element of the domain using conjunctions and disjunctions.

Step by step solution

01

Understand the Problem

Given a propositional function \(P(x)\) with a domain consisting of the integers 1, 2, 3, 4, and 5, the goal is to express the given logical statements without using quantifiers, instead using only negations, disjunctions, and conjunctions.
02

Express \( \forall x P(x) \)

The statement \( \forall x P(x) \) means that \(P(x)\) is true for all values in the domain. This can be written as: \[P(1) \text{ AND } P(2) \text{ AND } P(3) \text{ AND } P(4) \text{ AND } P(5).\]
03

Express \( eg \forall x P(x) \)

The negation of \( \forall x P(x) \) means that it is not true that \(P(x)\) is true for all values. This can be written as the negation of the conjunction: \[eg (P(1) \text{ AND } P(2) \text{ AND } P(3) \text{ AND } P(4) \text{ AND } P(5)).\]
04

Express \( eg \forall x P(x) \)

The statement \( eg \forall x P(x) \) means that there exists an x for which \(P(x)\) is not true. This can be written as: \[ eg P(1) \text{ OR } eg P(2) \text{ OR } eg P(3) \text{ OR } eg P(4) \text{ OR } eg P(5).\]
05

Express \( eg \forall x P(x) \)

The given proposition is \[ \forall x((x eq 3) \rightarrow P(x)) \text{ OR } eg P(x).\] Rephrasing the first part without quantifiers, for every x, if \(x\) is not equal to 3, then \(P(x)\) and combine it with the disjunction: \[(1 eq 3 \rightarrow P(1)) \text{ AND } (2 eq 3 \rightarrow P(2)) \text{ AND } (3 eq 3 \rightarrow P(3)) \text{ AND } (4 eq 3 \rightarrow P(4)) \text{ AND } (5 eq 3 \rightarrow P(5)) \text{ OR } P(1) \text{ OR } P(2) \text{ OR } P(3) \text{ OR } P(4) \text{ AND } (P(5)\]

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Key Concepts

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

Quantifiers
In propositional logic, quantifiers help to express the notion of 'there exists' or 'for all'. A common quantifier is the existential quantifier, denoted as \( \exists x \), which means 'there exists an x'. Another common quantifier is the universal quantifier, denoted as \( \forall x \), which means 'for all x'.

For example, \( \exists x P(x) \) means there is at least one x in the domain where the proposition \( P(x) \) is true. Conversely, \( \forall x P(x) \) means that the proposition \( P(x) \) is true for every x in the domain.

Quantifiers allow us to make precise and powerful statements about collections of objects in logic.
Negations
Negations in propositional logic are used to express the opposite of a given proposition. It is denoted by the symbol \( \eg \), which means 'not'.

If \( P(x) \) represents a proposition, then \( \eg P(x) \) means that the proposition is not true. For example, if \( P(1) \) says 'x is greater than 2', then \( \eg P(1) \) would say 'x is not greater than 2'.

Negations are essential for forming opposite or contradictory statements in logic. They help in constructing more complex propositions, making the logical relationships clearer and more comprehensive.
Disjunctions
Disjunctions are logical operations that combine two propositions with the word 'or'. It is expressed using the symbol \( \lor \). If we have two propositions \( P(x) \) and \( Q(x) \), their disjunction \( P(x) \lor Q(x) \) means that at least one of them is true.

For instance, \( \eg P(1) \lor \eg P(2) \lor \eg P(3) \lor \eg P(4) \lor \eg P(5) \) means that at least one of the propositions \( P(x) \) for x in the domain {1, 2, 3, 4, 5} is not true.

Disjunctions are beneficial for forming complex logical statements, where satisfying any one of multiple conditions is sufficient for the whole statement to be true.
Conjunctions
Conjunctions are logical operations that combine two propositions with the word 'and'. It is expressed using the symbol \( \land \). If we have two propositions \( P(x) \) and \( Q(x) \), their conjunction \( P(x) \land Q(x) \) means that both of them are true.

For instance, \( P(1) \land P(2) \land P(3) \land P(4) \land P(5) \) means that the propositions \( P(x) \) are true for all x in the domain {1, 2, 3, 4, 5}.

Conjunctions are significant in forming statements that require multiple conditions to be true simultaneously. They ensure that all combined propositions hold for the entire statement to be valid.

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Most popular questions from this chapter

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

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.

Use a proof by exhaustion to show that a tiling using dominoes of a \(4 \times 4\) checkerboard with opposite corners removed does not exist. [Hint: First show that you can assume that the squares in the upper left and lower right corners are removed. Number the squares of the original checkerboard from 1 to \(16,\) starting in the first row, moving right in this row, then starting in the leftmost square in the second row and moving right, and so on. Remove squares 1 and \(16 .\) To begin the proof, note that square 2 is covered either by a domino laid horizontally, which covers squares 2 and \(3,\) or vertically, which covers squares 2 and \(6 .\) Consider each of these cases separately, and work through all the subcases that arise. \(]\)

Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any two unequal bits you insert a 1 to produce nine new bits. Then you erase the nine original bits. Show that when you iterate this procedure, you can never get nine zeros. [Hint: Work backward, assuming that you did end up with nine zeros.]

Prove that there are no solutions in positive integers \(x\) and \(y\) to the equation \(x^{4}+y^{4}=625\)
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