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Use De Morgan's laws to find the negation of each of the following statements. a) Jan is rich and happy. b) Carlos will bicycle or run tomorrow. c) Mei walks or takes the bus to class. d) Ibrahim is smart and hard working.

Short Answer

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a) Jan is not rich or not happy. b) Carlos will not bicycle tomorrow and will not run tomorrow. c) Mei neither walks to class nor takes the bus. d) Ibrahim is not smart or not hard working.

Step by step solution

01

Understanding De Morgan's Laws

De Morgan's laws state that the negation of a conjunction is the disjunction of the negations, and the negation of a disjunction is the conjunction of the negations. In logical terms: egations of a) \( eg(A \land B) = eg A \lor eg B \)b) \( eg(A \lor B) = eg A \land eg B \)
02

Find negation for statement (a)

Given statement (a): 'Jan is rich and happy'. Let A be 'Jan is rich' and B be 'Jan is happy'. Using De Morgan's law: \( eg(A \land B) = eg A \lor eg B \) The negation is: 'Jan is not rich or not happy'.
03

Find negation for statement (b)

Given statement (b): 'Carlos will bicycle or run tomorrow'. Let A be 'Carlos will bicycle tomorrow' and B be 'Carlos will run tomorrow'. Using De Morgan's law: \( eg(A \lor B) = eg A \land eg B \) The negation is: 'Carlos will not bicycle tomorrow and will not run tomorrow'.
04

Find negation for statement (c)

Given statement (c): 'Mei walks or takes the bus to class'. Let A be 'Mei walks to class' and B be 'Mei takes the bus to class'. Using De Morgan's law: \( eg(A \lor B) = eg A \land eg B \) The negation is: 'Mei neither walks to class nor takes the bus'.
05

Find negation for statement (d)

Given statement (d): 'Ibrahim is smart and hard working'. Let A be 'Ibrahim is smart' and B be 'Ibrahim is hard working'. Using De Morgan's law: \( eg(A \land B) = eg A \lor eg B \) The negation is: 'Ibrahim is not smart or not hard working'.

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Key Concepts

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

Negation
Negation is a fundamental concept in logic that simply means the opposite of a given statement. For example, if a statement is true, its negation is false, and vice versa. We denote the negation of a statement A as \( eg A \).

For example, if A is 'The sky is blue', then \( eg A \) would be 'The sky is not blue'. In logical problems, understanding how to negate statements correctly is crucial. De Morgan's laws offer a standardized approach to this.

De Morgan's laws help us transform and simplify the negation of more complex logical statements that use 'and' (\( \land \)) or 'or' (\( \lor \)) operators.

Remember:
  • Negating a conjunction (\( A \land B \)) results in the disjunction of the negations (\( eg A \lor eg B \)).
  • Negating a disjunction (\( A \lor B \)) results in the conjunction of the negations (\( eg A \land eg B \)).
Let's apply this understanding to the provided exercise.
Logical Conjunction
Logical conjunction is a fancy term for the 'and' operation in logic. We denote it as \( A \land B \). This operation results in true only if both A and B are true. Otherwise, it is false.

Here's how it works:
  • If A is 'Jan is rich' and B is 'Jan is happy', then \( A \land B \) (Jan is rich and happy) is only true if both 'Jan is rich' and 'Jan is happy' are true.

Now, according to De Morgan's laws, negating a conjunction involves transforming it into a disjunction of their negations:

Let鈥檚 look at an example from the exercise:

Given the statement 'Jan is rich and happy', stated as \( A \land B \), to find its negation, we transform it using De Morgan's Law: \( eg (A \land B) = eg A \lor eg B \).

So, the negation 'Jan is not rich or not happy' means that at least one of these two conditions is true:
  • Jan is not rich
  • Jan is not happy

This brings more clarity as it simplifies how we understand and approach negating complex statements.
Logical Disjunction
Logical disjunction refers to the 'or' operation in logic, represented by \( A \lor B \). This operation is true if at least one of A or B is true. It鈥檚 only false if both A and B are false.

Here's an example from the exercise:

Given statement 'Carlos will bicycle or run tomorrow', represented as \( A \lor B \), where A is 'Carlos will bicycle tomorrow' and B is 'Carlos will run tomorrow'. To negate this statement, we use De Morgan's Law:

\( eg (A \lor B) = eg A \land eg B \).

The result 'Carlos will not bicycle tomorrow and will not run tomorrow' means:
  • Carlos will do neither activity tomorrow.
Let鈥檚 explore another example:

In the statement 'Mei walks or takes the bus to class', represented as \( A \lor B \), A is 'Mei walks to class' and B is 'Mei takes the bus to class'. Using De Morgan's Law, the negation is \( eg (A \lor B) = eg A \land eg B \), which simplifies to 'Mei neither walks to class nor takes the bus'.

This shows that learning and applying De Morgan's laws help in simplifying the process of negating logical statements accurately and efficiently.

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Most popular questions from this chapter

Let M(x, y) be 鈥渪 has sent y an e-mail message鈥 and T(x, y) be 鈥渪 has telephoned y,鈥 where the domain consists of all students in your class. Use quantifiers to express each of these statements. (Assume that all e-mail messages that were sent are received, which is not the way things often work.) a) Chou has never sent an e-mail message to Koko. b) Arlene has never sent an e-mail message to or tele- phoned Sarah. c) Jose has never received an e-mail message from Deborah. d) Every student in your class has sent an e-mail mes- sage to Ken. e) No one in your class has telephoned Nina. f ) Everyone in your class has either telephoned Avi or sent him an e-mail message. g) There is a student in your class who has sent every- one else in your class an e-mail message. h) There is someone in your class who has either sent an e-mail message or telephoned everyone else in your class. i) There are two different students in your class who have sent each other e-mail messages. j) There is a student who has sent himself or herself an e-mail message. k) There is a student in your class who has not received an e-mail message from anyone else in the class and who has not been called by any other student in the class. l) Every student in the class has either received an email message or received a telephone call from another student in the class. m) There are at least two students in your class such that one student has sent the other e-mail and the second student has telephoned the first student. n) There are two different students in your class who between them have sent an e-mail message to or telephoned everyone else in the class.

Use predicates, quantifiers, logical connectives, and mathematical operators to express the statement that every positive integer is the sum of the squares of four integers.

Exercises \(40-44\) deal with the translation between system specification and logical expressions involving quantifiers. Express each of these system specifications using predicates, quantifiers, and logical connectives. a) Every user has access to an electronic mailbox. b) The system mailbox can be accessed by everyone in the group if the file system is locked. c) The firewall is in a diagnostic state only if the proxy server is in a diagnostic state. d) At least one router is functioning normally if the throughput is between 100 kbps and 500 kbps and the proxy server is not in diagnostic mode.

Construct a truth table for each of these compound propositions. a) \(p \oplus p\) b) \(p \oplus \neg p\) c) \(p \oplus \neg q\) d) \(\neg p \oplus \neg q\) e) \((p \oplus q) \vee(p \oplus \neg q)\) f) \((p \oplus q) \wedge(p \oplus \neg q)\)

Construct a truth table for each of these compound propositions. a) \(p \rightarrow(\neg q \vee r)\) b) \(\neg p \rightarrow(q \rightarrow r)\) c) \((p \rightarrow q) \vee(\neg p \rightarrow r)\) d) \((p \rightarrow q) \wedge(\neg p \rightarrow r)\) e) \((p \rightarrow q) \vee(\neg q \rightarrow r)\) f) \((\neg p \leftrightarrow \neg q) \leftrightarrow(q \leftrightarrow r)\)

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