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Chapter 2: Problem 6

6\. Negate each of the following and simplify the resulting statement. a) \(p \wedge(q \vee r) \wedge(\neg p \vee \neg q \vee r)\) b) \((p \wedge q) \rightarrow r\) c) \(p \rightarrow(\neg q \wedge r)\) d) \(p \vee q \vee(\neg p \wedge \neg q \wedge r)\)

Short Answer

Expert verified
a) \((\neg p \lor (\neg q \land \neg r)) \lor (p \land q \land \neg r)\), b) \((p \land q) \land \neg r\), c) \((p \land q) \land r\), d) \((\neg p \land \neg q \land (\neg p \lor \neg q \lor r))\)

Step by step solution

01

Applying De Morgan's Laws to simplify each statement

We apply De Morgan's Laws to simplify the statements. 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. We also remember that the negation of an implication can be represented as a conjunction. Therefore, the negations of the given statements would be: \na) \(\neg((p \land (q \lor r)) \land (\neg p \lor \neg q \lor r))\)\nb) \(\neg((p \land q) \rightarrow r)\)\nc) \(\neg(p \rightarrow (\neg q \land r))\)\nd) \(\neg(p \lor q \lor (\neg p \land \neg q \land r))\)
02

Simplifying the negations

Next step is to simplify the negations. \na) For the first expression, we just distribute the negation across, which gives: \((\neg p \lor \neg(q \lor r)) \lor (\neg(\neg p \lor \neg q \lor r))\)\nb) For the second expression, negating an implication gives a conjunction with the antecedent and the negation of the consequent: \((p \land \neg r)\) \nc) For the third expression, again, negating an implication gives: \((p \land (\neg q \land r))\)\nd) For the last expression, we again distribute the negation across to get: \((\neg p \land \neg q \land (\neg p \lor \neg q \lor r))\)
03

Further Simplifying the expressions

For the expressions that still contain negations of disjunctions or conjunctions, we apply De Morgan鈥檚 Laws again:\na) \((\neg p \lor (\neg q \land \neg r)) \lor (p \land q \land \neg r)\)\nd) \((\neg p \land \neg q \land (\neg p \lor \neg q \lor r))\)\nThe expressions b) and c) are already simplified and do not need further adjustments.

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Key Concepts

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

De Morgan's Laws
When dealing with logical negations, De Morgan's Laws come in handy. These laws provide a way to transform the negation of complex logical statements. Specifically, De Morgan's Laws include two fundamental rules in logic:

  • The negation of a conjunction is the disjunction of the negations: \[ eg(p \land q) = (eg p \lor eg q) \]
  • The negation of a disjunction is the conjunction of the negations: \[ eg(p \lor q) = (eg p \land eg q) \]
These transformations are particularly useful when simplifying logical expressions. In the original exercise, we used De Morgan's Laws to simplify the negation of complex expressions. For example, given a statement like \(p \land(q \lor r)\), its negation using De Morgan's Laws would be \(eg p \lor (eg q \land eg r)\). Learning to apply these laws can significantly ease your work with logical expressions and is an essential skill in mathematical logic.
Implication in Logic
Implication in logic, often denoted with a right arrow \(\rightarrow\), is a critical concept to understand. An implication \(p \rightarrow q\) is read as "if \(p\), then \(q\)." It asserts that \(q\) must be true whenever \(p\) is true. In logical terms, the implication is only false if the antecedent \(p\) is true and the consequent \(q\) is false.

To negate an implication, we employ a useful transformation: \[eg(p \rightarrow q) = p \land eg q\]This equation tells us that the negation \(eg(p \rightarrow q)\) is equivalent to asserting that \(p\) is true while \(q\) is false. This method was used in the original exercise to simplify expressions like \(eg((p \land q) \rightarrow r) = (p \land eg r)\). Understanding how to handle negations of implications will help you manipulate logical statements effectively.
Disjunction and Conjunction
In logic, disjunction and conjunction are key operations.

  • **Disjunction (\(\lor\))**: Represents a logical 'or'. - An expression like \(p \lor q\) is true if \(p\) is true, \(q\) is true, or both are true.
  • **Conjunction (\(\land\))**: Represents a logical 'and'. - An expression like \(p \land q\) is true only if both \(p\) and \(q\) are true.
In simplifying logical expressions, De Morgan鈥檚 Laws directly affect these operations, transforming conjunctions into disjunctions and vice versa under negation. For example, when simplifying the negation of a disjunction like \(p \lor q \lor(r)\), we convert it to a conjunction of negations: \(eg p \land eg q \land eg r\). Similarly, the negation of a conjunction \(p \land q \) would turn into a disjunction: \(eg p \lor eg q\). Grasping these operations allows you to construct or deconstruct logical statements with ease.

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Most popular questions from this chapter

10\. For the following program segment, \(m\) and \(n\) are integer variables. The variable \(A\) is a two-dimensional array \(A[1,1]\), \(A[1,2], \ldots, A[1,20], \ldots, A[10,1], \ldots, A[10,20]\), with 10 rows (indexed from 1 to 10 ) and 20 columns (indexed from 1 to 20 ). for \(m:=1\) to 10 do \(\quad\) for \(n:=1\) to 20 do \(\quad A[m, n]:=m+3 * n\) Write the following statements in symbolic form. (The universe for the variable \(m\) contains only the integers from 1 to 10 inclusive; for \(n\) the universe consists of the integers from 1 to 20 inclusive.) a) All entries of \(A\) are positive. b) All entries of \(A\) are positive and less than or equal to 70 . c) Some of the entries of \(A\) exceed 60 . d) The entries in each row of \(A\) are sorted into (strictly) ascending order. e) The entries in each column of \(A\) are sorted into (strictly) ascending order. f) The entries in the first three rows of \(A\) are distinct.

18\. Let \(m, n\) be two positive integers. Prove that if \(m, n\) are perfect squares, then the product \(m n\) is also a perfect square.

15\. Provide a proof by contradiction for the following: For every integer \(n\), if \(n^{2}\) is odd, then \(n\) is odd.

3\. Verify that each of the following is a logical implication by showing that it is impossible for the conclusion to have the truth yalue 0 while the hypothesis has the truth value 1 . a) \((p \wedge q) \rightarrow p\) b) \(p \rightarrow(p \vee q)\) c) \([(p \vee q) \wedge \neg p] \rightarrow q\) d) \([(p \rightarrow q) \wedge(r \rightarrow s) \wedge(p \vee r)] \rightarrow(q \vee s)\) e) \([(p \rightarrow q) \wedge(r \rightarrow s) \wedge(\neg q \vee \neg s)] \rightarrow(\neg p \vee \neg r)\)

15\. For cach of the following pairs of statements determine whether the proposed negation is correct. If correct, determine which is true: the original statement or the proposed negation. If the proposed negation is wrong, write a correct version of the negation and then determine whether the original statement or your corrected version of the negation is true. a) Statement: For all real numbers \(x, y\), if \(x^{2}>y^{2}\), then \(x>y\) Proposed negation: There exist real numbers \(x, y\) such that \(x^{2}>y^{2}\) but \(x \leq y\) b) Statement: There exist real numbers \(x, y\) such that \(x\) and \(y\) are rational but \(x+y\) is irrational. Proposed negation: For all real numbers \(x, y\), if \(x+y\) is rational, then each of \(x, y\) is rational. c) Statement: For all real numbers \(x\), if \(x\) is not 0 , then \(x\) has a multiplicative inverse. Proposed negation: There exists a nonzero real number that does not have a multiplicative inverse. d) Statement: There exist odd integers whose product is odd. Proposed negation: The product of any two odd integers is odd. 16\. Write the negation of each of the following statements as an English sentence-without symbolic notation. (Here the universe consists of all the students at the university where Professor Lenhart teaches.) a) Every student in Professor Lenhart's C++ class is majoring in computer science or mathematics. b) At least one student in Professor Lenhart's C++ class is a history major.
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