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Chapter 1: Problem 30

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) \(\neg \exists y \exists x P(x, y) \quad\) b) \(\neg \forall x \exists y P(x, y)\) c) \(\neg \exists y(Q(y) \wedge \forall x \neg R(x, y))\) d) \(\neg \exists y(\exists x R(x, y) \vee \forall x S(x, y))\) e) \(\neg \exists y(\forall x \exists z T(x, y, z) \vee \exists x \forall z U(x, y, z))\)

Short Answer

Expert verified
a) \(\forall y \forall x \eg P(x, y)\), b) \(\exists x \forall y \eg P(x, y)\), c) \(\forall y (\eg Q(y) \vee \exists x R(x, y))\), d) \(\forall y (\forall x \eg R(x, y) \wedge \exists x \eg S(x, y))\), e) \(\forall y (\forall x \exists z \eg T(x, y, z) \wedge \exists x \forall z \eg U(x, y, z))\)

Step by step solution

01

Negate Inside Statements (Part a)

For the first statement \(eg \exists y \exists x P(x, y)\), rewrite it using De Morgan's laws. The negation of an existential quantifier \((\exists y)\) becomes a universal quantifier \((\forall y)\), moving the negation inside: \(\forall y \forall x \eg P(x, y)\).
02

Negate Inside Statements (Part b)

For the second statement \(eg \forall x \exists y P(x, y)\), apply De Morgan's laws to negate the universal quantifier \((\forall x)\), which turns into an existential quantifier \((\exists x)\): \(\exists x \eg \exists y P(x, y)\). Simplify further by moving the negation inside: \(\exists x \forall y \eg P(x, y)\).
03

Negate Inside Statements (Part c)

For the third statement \(eg \exists y(Q(y) \wedge \forall x \eg R(x, y))\), apply De Morgan's laws to the negation of the conjunction \((Q(y) \wedge \forall x \eg R(x, y))\): \(\forall y \eg (Q(y) \wedge \forall x \eg R(x, y))\). Use De Morgan's laws again inside the negation: \(\forall y (\eg Q(y) \vee \exists x R(x, y))\).
04

Negate Inside Statements (Part d)

For the fourth statement \(eg \exists y(\exists x R(x, y) \vee \forall x S(x, y))\), apply De Morgan's laws to the negation of the disjunction \((\exists x R(x, y) \vee \forall x S(x, y))\): \(\forall y \eg (\exists x R(x, y) \vee \forall x S(x, y))\). Use De Morgan's laws inside the negation: \(\forall y (\forall x \eg R(x, y) \wedge \exists x \eg S(x, y))\).
05

Negate Inside Statements (Part e)

For the fifth statement \(eg \exists y(\forall x \exists z T(x, y, z) \vee \exists x \forall z U(x, y, z))\), apply De Morgan's laws to the negation of the disjunction \((\forall x \exists z T(x, y, z) \vee \exists x \forall z U(x, y, z))\): \(\forall y \eg (\forall x \exists z T(x, y, z) \vee \exists x \forall z U(x, y, z))\). Inside the negation, apply De Morgan's laws again: \(\forall y (\exists x \forall z \eg T(x, y, z) \wedge \forall x \exists z \eg U(x, y, z))\).

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Key Concepts

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

Understanding De Morgan's Laws
De Morgan's laws are fundamental rules in logic. They help us transform logical expressions with NOT operators. These laws show the relationship between conjunctions (AND, \( \wedge \)) and disjunctions (OR, \( \vee \)) when negation is involved. One of De Morgan's laws states: \( \eg (A \wedge B) = (\eg A \vee \eg B) \). This means the negation of conjunction is equivalent to the disjunction of negations. The other law is: \( \eg (A \vee B) = (\eg A \wedge \eg B) \), meaning the negation of a disjunction equals the conjunction of the negations. When applying these laws to quantifiers like \( \exists (there exists) \) and \( \forall (for all) \), they help us push the negation inside. For example, \( \eg \exists x P(x) \) becomes \( \forall x \eg P(x) \), and \( \eg \forall x P(x) \) changes to: \( \exists x \eg P(x) \).
Grasping Quantifiers
Quantifiers in logical expressions define the scope of the variables. There are two main types of quantifiers: existential (\( \exists \)) and universal (\( \forall \)). An existential quantifier expresses that there is at least one element in a set for which the predicate holds true. For instance, \( \exists x P(x) \) means there is some x for which P(x) is true. A universal quantifier dictates that the predicate holds true for every element in the set. For example, \( \forall x P(x) \) means P(x) is true for all x. When combined with negations, their meanings change. The negation \( \eg \exists x P(x) \) transforms to \( \forall x \eg P(x) \), emphasizing that P(x) is false for every x. Conversely, \( \eg \forall x P(x) \) alters to: \( \exists x \eg P(x) \), highlighting that there is some x where P(x) is false.
Logical Connectives
Logical connectives link together logical statements. The primary connectives are AND (\( \wedge \)), OR (\( \vee \)), and NOT (\( \eg \)). AND is true only if both statements it connects are true. OR is true if at least one of the connected statements is true. NOT inverts the truth value of a statement. Using AND and OR together with NOT forms compound statements. An example with AND: \( P(x) \wedge Q(x) \) is only true if both P(x) and Q(x) are true. For OR: \( P(x) \vee Q(x) \) is true if either P(x) or Q(x) (or both) are true. NOT reverses a statement, making \( \eg A \) true when A is false and vice versa. Mastering logical connectives and their interactions helps in understanding complex logical structures, especially when simplifying expressions using rules like De Morgan's laws.

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Most popular questions from this chapter

Find a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all integers. $$ \begin{array}{l}{\text { a) } \forall x\left(x^{2} \geq x\right)} \\ {\text { b) } \forall x(x>0 \vee x<0)} \\ {\text { c) } \forall x(x=1)}\end{array} $$

Show that the equivalence \(p \wedge \neg p \equiv \mathbf{F}\) can be derived using resolution together with the fact that a conditional statement with a false hypothesis is true. [Hint: Let \(q=\) \(r=\mathbf{F}\) in resolution.

Suppose the domain of the propositional function \(P(x, y)\) consists of pairs \(x\) and \(y,\) where \(x\) is \(1,2,\) or 3 and \(y\) is \(1,2,\) or \(3 .\) Write out these propositions using disjunctions and conjunctions. $$ \begin{array}{ll}{\text { a) } \exists x P(x, 3)} & {\text { b) } \forall y P(1, y)} \\ {\text { c) } \exists y \neg P(2, y)} & {\text { d) } \forall x \neg P(x, 2)}\end{array} $$

For each of these arguments determine whether the argument is correct or incorrect and explain why. a) All students in this class understand logic. Xavier is a student in this class. Therefore, Xavier understands logic. b) Every computer science major takes discrete math- ematics. Natasha is taking discrete mathematics. Therefore, Natasha is a computer science major. c) All parrots like fruit. My pet bird is not a parrot. Therefore, my pet bird does not like fruit. d) Everyone who eats granola every day is healthy. Linda is not healthy. Therefore, Linda does not eagranola every day.

The Logic Problem, taken from \(W F F^{\prime} N\) PROOF, The Game of Logic, has these two assumptions: 1\. "Logic is difficult or not many students like logic." 2\. "If mathematics is easy, then logic is not difficult." By translating these assumptions into statements involving propositional variables and logical connectives, de- termine whether each of the following are valid conclusions of these assumptions: a) That mathematics is not easy, if many students like logic. b) That not many students like logic, if mathematics is not easy. c) That mathematics is not easy or logic is difficult. d) That logic is not difficult or mathematics is not easy. e) That if not many students like logic, then either mathematics is not easy or logic is not difficult.
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