went to yesterday. • Q(x) : x was on vacation yesterday. Translate each of the following English statements into a logical expression. 1. Everyone who was well went to work yesterday. 2. Someone who was sick yesterday did not go to work yesterday. 3. Someone who missed work was neither sick nor on vacation. 4. Joshua was on vacation yesterday and he did not go to work. 5. Everyone was well and went to work yesterday.

went to yesterday. • Q(x) : x was on vacation yesterday. Translate each of the following English statements into a logical expression. 1. Everyone who was well went to work yesterday. 2. Someone who was sick yesterday did not go to work yesterday. 3. Someone who missed work was neither sick nor on vacation. 4. Joshua was on vacation yesterday and he did not go to work. 5. Everyone was well and went to work yesterday.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Let the domain be the set of all employees of a certain company and Joshua is an employee of that company. Define the following predicates: P(x) : x was sick yesterday. W(x) : x went to work yesterday. Q(x) : x was on vacation yesterday. Translate each of the following English statements into a logical expression. 1. Everyone who was well went to work yesterday. 2. Someone who was sick yesterday did not go to work yesterday. 3. Someone who missed work was neither sick nor on vacation. 4. Joshua was on vacation yesterday and he did not go to work. 5. Everyone was well and went to work yesterday. There are two ways to submit answers to this question: 1) Eneter in essay box directly. Note that logical expressions must be entered in Math mode, which begins with \(, and end with \). Below is a list of LaTex code for each logical operator. V \vee A \wedge - \neg → \to + \leftrightarrow Vx \forall x Ex \exists x
Computer Science The knowledge of an expert system is given as follow: male(john).male(sam).male(peter).male(david).female(mary).female(betty).female(jane).female(sarah). a). A new predicate, likes, with two arguments (i.e. likes(male,female)), is required to represent the likes relationship of all possible male and female. Using variable(s), write the rule(s) that can generate the required knowledge.b). Using variable(s), write the rule(s) that can also provide the knowledge, likes(female,male).c). Given that sam and mary are sibling, the likes relationship between sam and mary should be omitted from the knowledge. Modify the rule(s) defined above.
Using the predicate symbols shown and appropriate quantifiers, write each English language statement as a predicate wff. (The domain is the whole world.) D(x): x is a day S(x): x is sunny R(x): x is rainy M: Monday T: Tuesday 1.All animals are dogs. 3.Some animals are not cats. 4.Every animal that is a dog is not a cat. 5.Some animals are dogs and cats. 6.No animal is both a dog and a cat. 7.An animal is always a dog only if it is a cat. 8.No animals are dogs. 9.Meg is a cat; all animals are cats. 10.Meg and Alex are both cats. 11.If one animal is a cat, then every animal will be a cat.
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Using the predicate symbols shown and appropriate quantifiers, write each English language statement as a predicate wff. (The domain is the whole world.)B(x): x is a ballR(x): x is roundS(x): x is a soccer balla. All balls are round.b. Not all balls are soccer balls.c. All soccer balls are round.d. Some balls are not round.e. Some balls are round but soccer balls are not.f. Every round ball is a soccer ball.g. Only soccer balls are round balls.h. If soccer balls are round, then all balls are round.
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Using the predicate symbols shown and the appropriate quantifiers, write each English language statement as a predicate wff. (The domain is the whole world.) D(x): x is a day S(x): x is sunny R(x): x is rainy M: Monday T: Tuesday All days are sunny. which of these answers are right? a. (∀x)[D(x) ∧ S(x)] b. (∀x)[D(x) → S(x)] c. (∃x)[D(x) ∧ S(x)] d. (∃x)[D(x) → S(x)