/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Determine whether each of the fo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 12

Determine whether each of the following pairs of statements is logically equivalent. [Here \(p, q, r\) are primitive statements and \((p \downarrow q) \Leftrightarrow \neg(p \vee q)\) while \((p \uparrow q) \Leftrightarrow \neg(p \wedge q) .]\) a) \(p \downarrow(q \downarrow r)\), b) \(p \uparrow(q \downarrow r), \quad(p \uparrow q) \downarrow(p \uparrow r)\) c) \(p \downarrow(q \uparrow r), \quad(p \downarrow q) \uparrow(p \downarrow r)\)

Short Answer

Expert verified
The pairs of statements in 'b' and 'c' are logically equivalent, but 'a' has no pair to compare so it cannot be determined.

Step by step solution

01

Analyzing the Statements

The first step is to identify the operation involved in each primitive statement. From there, rewrite them using 'and' (\(\wedge\)) and 'or' (\(\vee\)) notation, and then applying De Morgan's Laws.
02

Equivalence of Statement a

For the first pair, \(p \downarrow(q \downarrow r)\), we rewrite it using 'and' and 'or'. This yields \(\neg p \vee (\neg q \wedge \neg r)\). There is no other statement to compare to, so no equivalence can be determined for 'a'.
03

Equivalence of Statement b

For the second pair, \(p \uparrow(q \downarrow r)\) and \((p \uparrow q) \downarrow(p \uparrow r)\), we rewrite them as \((\neg p \wedge (q \vee r))\) and \((\neg p \wedge q) \vee (\neg p \wedge r)\). Using distributive law, we can rewrite the second statement as \(\neg p \wedge (q \vee r)\), which is the same as the first statement in the pair. Therefore, 'b' are equivalent.
04

Equivalence of Statement c

For the third pair, \(p \downarrow(q \uparrow r)\) and \((p \downarrow q) \uparrow(p \downarrow r)\), we rewrite them as \(\neg p \vee (q \wedge r)\) and \((\neg p \vee q) \wedge (\neg p \vee r)\). Using distributive law on the second statement, we get \(\neg p \vee (q \wedge r)\), which is the same as the first statement in the pair. Therefore, 'c' are equivalent.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Logical Operators
Logical operators are foundational tools in mathematical logic and computer science. They are used to form complex expressions from simpler ones by combining statements. The most common logical operators include:
  • AND Operator ( \(\wedge\) ): Combines two statements and returns true if both statements are true. Otherwise, it returns false.
  • OR Operator ( \(\vee\) ): Returns true if at least one of the statements is true. It only returns false if both statements are false.
  • NOT Operator ( \(eg\) ): Negates a statement. If a statement is true, the NOT operator makes it false, and vice versa.
These operators are used to create complex expressions. By applying these operators, you can evaluate whether statements are logically equivalent, meaning they always result in the same truth value under all circumstances.
De Morgan's Laws
De Morgan's Laws offer a useful way to simplify complex logical statements, especially when dealing with negations. They are two transformation rules that help in breaking down logical expressions involving NOT, AND, and OR operators.
  • The first law states: The negation of a conjunction is equivalent to the disjunction of the negations. \[eg (A \wedge B) \equiv (eg A \vee eg B)\]
  • The second law states: The negation of a disjunction is equivalent to the conjunction of the negations. \[eg (A \vee B) \equiv (eg A \wedge eg B)\]
Applying these laws can significantly simplify logical expressions, making it easier to determine logical equivalences. For instance, in the given exercise, we can transform expressions like \((p \,\uparrow\, q)\), which is \(eg(p \,\wedge\, q)\), using De Morgan's Laws to check for logical equivalences.
Distributive Law
The distributive law in logic is akin to the distributive law in arithmetic, allowing the distribution of one operation over another within a logical expression. It is particularly useful when dealing with expressions that have both OR and AND operators.The distributive laws in logic are:
  • AND distributes over OR: \[A \wedge (B \vee C) \equiv (A \wedge B) \vee (A \wedge C)\]
  • OR distributes over AND: \[A \vee (B \wedge C) \equiv (A \vee B) \wedge (A \vee C)\]
Understanding this concept allows us to rearrange and simplify expressions efficiently. For example, in the solution provided for the original exercise, the use of the distributive law helped rewrite complex expressions into simpler ones that reveal logical equivalence. This transformation is crucial for both simplifying expressions and for comparing whether two logical expressions are equivalent.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For primitive statements \(p, q\), a) verify that \(p \rightarrow[q \rightarrow(p \wedge q)]\) is a tautology. b) verify that \((p \vee q) \rightarrow[q \rightarrow q]\) is a tautology by using the result from part (a) along with the substitution rules and the laws of logic. c) is \((p \vee q) \rightarrow[q \rightarrow(p \wedge q)]\) a tautology?

\begin{aligned} &\text { Rewrite each of the following statements (with prescribed universes) as an implication in } \\ &\text { the if-then form. Then write the converse, inverse, and contrapositive of your implica- } \\ &\text { tion. For each result in parts (a) and (d) give the truth value for the implication and the } \\ &\text { truth values for its converse, inverse, and contrapositive. (In part (a) "divisibility" } \\ &\text { requires a remainder of } 0 .) \\ &\text { a) [The universe comprises all positive integers.] } \\ &\text { Divisibility by } 21 \text { is a sufficient condition for divisibility by } 7 . \\ &\text { b) [The universe consists of all the present residents of the United States.] } \\ &\text { Having a sizable stock-market portfolio is a necessary condition for a person to be } \\ &\text { happy. } \\ &\text { c) [The universe comprises all snakes presently slithering about the jungles of Asia.] } \\ &\text { Being a cobra is a sufficient condition for a snake to be dangerous. } \\ &\text { d) [The universe consists of all complex numbers.] } \\ &\text { For every complex number } z, z \text { being real is necessary for } z^{2} \text { to be real. } \end{aligned}

Use truth tables to verify that each of the following is a logical implication. a) \([(p \rightarrow q) \wedge(q \rightarrow r)] \rightarrow(p \rightarrow r)\) b) \([(p \rightarrow q) \wedge \neg q] \rightarrow \neg p\) c) \([(p \vee q) \wedge \neg p] \rightarrow q\) d) \([(p \rightarrow r) \wedge(q \rightarrow r)] \rightarrow[(p \vee q) \rightarrow r]\)

Negate and simplify each of the following. a) \(\exists x[p(x) \vee q(x)]\) b) \(\forall x[p(x) \wedge \neg q(x)]\) c) \(\forall x[p(x) \rightarrow q(x)]\) d) \(3 x[(p(x) \vee q(x)) \rightarrow r(x)]\)

Identify the bound variables and the free variables in each of the following expressions (or statements). In part (a) the universe comprises all real numbers except for... \(-5 \pi / 2,-3 \pi / 2,-\pi / 2, \pi / 2,3 \pi / 2,5 \pi 2, \ldots .\) In all other cases the universe comprises all real numbers. a) \(\forall x \forall y\left[\sec ^{2} x-\sec ^{2} y=\tan ^{2} x-\tan ^{2} y\right]\) b) \(\forall y \exists z[\cos (x+y)=\sin (z-x)]\) c) \(\exists x \exists y\left[x^{2}-y^{2}=z\right]\) d) \(\exists x[x y=y]\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.