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6\. Let \(p, q, r\) denote primitive statements. Write the converse, nverse, and contrapositive of a) \(p \rightarrow(q \wedge r)\) b) \((p \vee q) \rightarrow r\)

Short Answer

Expert verified
The converses are \((q \wedge r) \rightarrow p\) and \(r \rightarrow (p \vee q)\), inverses are \(\neg p \rightarrow \neg (q \wedge r)\) and \(\neg (p \vee q) \rightarrow \neg r\), and contrapositives are \(\neg(q \wedge r) \rightarrow \neg p\) and \(\neg r \rightarrow \neg (p \vee q)\).

Step by step solution

01

Understanding the Definitions

Converse of a conditional statement \(p \rightarrow q\) is \(q \rightarrow p\). Inverse of a conditional statement \(p \rightarrow q\) is \(\neg p \rightarrow \neg q\). Contrapositive of a conditional statement \(p \rightarrow q\) is \(\neg q \rightarrow \neg p\).
02

Converse, Inverse, and Contrapositive of \(p \rightarrow(q \wedge r)\)

The converse is \((q \wedge r) \rightarrow p\), the inverse is \(\neg p \rightarrow \neg (q \wedge r)\), and the contrapositive is \(\neg(q \wedge r) \rightarrow \neg p\).
03

Converse, Inverse, and Contrapositive of \((p \vee q) \rightarrow r\)

The converse is \(r \rightarrow (p \vee q)\), the inverse is \(\neg (p \vee q) \rightarrow \neg r\), and the contrapositive is \(\neg r \rightarrow \neg (p \vee q)\).

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Key Concepts

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

Converse Statement
When learning about propositional logic, we come across the term "Converse Statement". In the world of logical conditionals, a statement often begins with an implication such as \( p \rightarrow q \). To find the converse of this statement, we simply reverse the roles of \( p \) and \( q \), resulting in \( q \rightarrow p \). This is essentially swapping the hypothesis and conclusion.For example, let's take the compound conditional proposition \( p \rightarrow (q \wedge r) \). Its converse will be \( (q \wedge r) \rightarrow p \). If dealing with another proposition like \( (p \vee q) \rightarrow r \), its converse becomes \( r \rightarrow (p \vee q) \).Understanding the converse is important because it tests a different aspect of the logical relationship. While a true statement does not guarantee a true converse, exploring them unveils different facets of logical reasoning.
Inverse Statement
The inverse statement is another transformation of the original conditional statement. To form an inverse, we take our initial conditional \( p \rightarrow q \) and apply a negation to both \( p \) and \( q \). The inverse thus becomes \( eg p \rightarrow eg q \).As a practical example, consider the statement \( p \rightarrow (q \wedge r) \). Its inverse would be \( eg p \rightarrow eg (q \wedge r) \), meaning if \( p \) is not true, then \( q \) and \( r \) are not both true. Similarly, for \( (p \vee q) \rightarrow r \), the inverse is \( eg (p \vee q) \rightarrow eg r \), indicating if neither \( p \) nor \( q \) is true, then \( r \) is not true.An inverse can help in testing the boundaries of logical implications, although like the converse, the truth value of the inverse doesn't necessarily follow from the truth of the original statement.
Contrapositive Statement
In propositional logic, the contrapositive is a very powerful transformation due to its equivalence to the original statement. For any given \( p \rightarrow q \), its contrapositive would be \( eg q \rightarrow eg p \). Finding the contrapositive involves reversing the order and also negating each component of the proposition.For instance, from our example \( p \rightarrow (q \wedge r) \), the contrapositive would turn into \( eg (q \wedge r) \rightarrow eg p \). Similarly, for \( (p \vee q) \rightarrow r \), its contrapositive is \( eg r \rightarrow eg (p \vee q) \).The beauty of contrapositive lies in its logical equivalence to the original statement. If a statement is true, its contrapositive is also true, making it a critical tool in proofs and deductions.

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Most popular questions from this chapter

9\. Let \(p(x), q(x)\), and \(r(x)\) be the following open statements. $$ \begin{array}{ll} p(x): & x^{2}-7 x+10=0 \\ q(x): & x^{2}-2 x-3=0 \\ r(x): & x<0 \end{array} $$ a) Determine the truth or falsity of the following statements, where the universe is all integers. If a statement is false, provide a counterexample or explanation. i) \(\forall x[p(x) \rightarrow \neg r(x)]\) ii) \(\forall x[q(x) \rightarrow r(x)]\) iii) \(\exists x[q(x) \rightarrow r(x)]\) iv) \(\exists x[p(x) \rightarrow r(x)]\) b) Find the answers to part (a) when the universe consists of all positive integers. c) Find the answers to part (a) when the universe contains only the integers 2 and 5 .

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.

20\. Rewrite each of the following statements in the if-then form. Then write the converse, inverse, and contrapositive of your implication. For each result in parts (a) and (c) give the truth value for the implication and the truth values for its converse, inverse, and contrapositive. [In part (a) "divisibility" requires a remainder of \(0 .]\) a) [The universe comprises all positive integers.] Divisibility by 21 is a sufficient condition for divisibility by \(7 .\) b) [The universe comprises all snakes presently slithering about the jungles of Asia.] Being a cobra is a sufficient condition for a snake to be dangerous. c) [The universe consists of all complex numbers.] For every complex number \(z, z\) being real is necessary for \(z^{2}\) to be real.

12\. Give a direct proof (as in Theorem \(2.3\) ) for each of the following. a) For all integers \(k\) and \(l\), if \(k, l\) are both even, then \(k+l\). is even. b) For all integers \(k\) and \(l\), if \(k, l\) are both even, then \(k l\) is even.

19\. Prove or disprove: If \(m, n\) are positive integers and \(m, n\) are perfect squares, then \(m+n\) is a perfect square.

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