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Chapter 1: Problem 21

Each of Exercises \(20-32\) asks you to show that two compound propositions are logically equivalent. To do this, either show that both sides are true, or that both sides are false, for exactly the same combinations of truth values of the propositional variables in these expressions (whichever is easier). Show that \(\neg(p \leftrightarrow q)\) and \(p \leftrightarrow \neg q\) are logically equivalent.

Short Answer

Expert verified
They are logically equivalent because \(eg(p \leftrightarrow q)\) and \(p \leftrightarrow \eg q\) have identical truth values for all combinations of \(p\) and \(q\).

Step by step solution

01

- Understand the problem

We need to show that \(eg(p \leftrightarrow q)\) and \(p \leftrightarrow \eg q\) are logically equivalent by demonstrating that they produce identical truth values for all combinations of truth values of \(p\) and \(q\).
02

- Construct the truth table

List all possible combinations of truth values for \(p\) and \(q\).
03

- Calculate \(p \leftrightarrow q\)

Determine the truth value of \(p \leftrightarrow q\) (i.e., true when both \(p\) and \(q\) are the same, false otherwise) for each combination from the truth table.
04

- Calculate \(eg(p \leftrightarrow q)\)

Determine the truth value of \(p \leftrightarrow q\) from Step 3 and negate it to find \(eg(p \leftrightarrow q)\).
05

- Calculate \(p \leftrightarrow \eg q\)

For \(p \leftrightarrow \eg q\), first calculate \(eg q\) (negation of \(q\)), and then determine when \(p\) and \(eg q\) have the same truth value and when they don't.
06

- Compare columns

Compare the columns of \(eg(p \leftrightarrow q)\) and \(p \leftrightarrow \eg q\) to verify that their truth values match for all combinations of \(p\) and \(q\).

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Key Concepts

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

Truth Tables
To understand logical equivalence, we need to use truth tables.
A truth table is a tool used in propositional logic to determine the truth value of a compound proposition for all possible truth values of its component propositions.
Each row in a truth table shows a possible combination of truth values for the component propositions and the resulting truth value of the compound proposition.
Truth tables are especially helpful when comparing two different compound propositions to see if they are logically equivalent.
In our exercise, we'll use a truth table to compare the truth values of \(eg(p \leftrightarrow q)\) and \(p \leftrightarrow eg q\) to show they are logically equivalent.
Compound Propositions
A compound proposition is made up of two or more simple propositions connected by logical connectives like AND (\(\land\)), OR (\(\lor\)), and NOT (\(eg\)).
In our exercise, \(eg(p \leftrightarrow q)\) and \(p \leftrightarrow eg q\) are compound propositions.
The components \(p\) and \(q\) are simple propositions that can either be true (T) or false (F).
The connectives used to form our compound propositions include bi-conditional (\(\leftrightarrow\)) and negation (\(eg\)).
Compound propositions become helpful in creating complex logical statements and in problems like ours where we need to reason about the equivalence of expressions.
Propositional Logic
Propositional logic is a branch of logic that deals with propositions and their relationships to each other.
In propositional logic, we use symbols to represent propositions and logical connectives to form compound propositions.
Propositional logic helps us understand the structure of logical arguments and provides tools like truth tables to analyze these arguments.
Our exercise uses propositional logic to determine if two compound propositions, \(eg(p \leftrightarrow q)\) and \(p \leftrightarrow eg q\), are logically equivalent.
By creating a truth table for these expressions, we apply propositional logic to compare their truth values for all possible truth value combinations of \(p\) and \(q\).
Negation
Negation (\(eg\)) is a logical connective that flips the truth value of a proposition.
If a proposition \(p\) is true, then \(eg p\) (not \(p\)) is false. Conversely, if \(p\) is false, then \(eg p\) is true.
In our exercise, we use negation in both compound propositions: \(eg(p \leftrightarrow q)\) negates the result of \(p \leftrightarrow q\), and \(p \leftrightarrow eg q\) uses the negated form of \(q\) in the bi-conditional statement.
Understanding how negation works helps us in constructing and interpreting the truth table for our problem to verify logical equivalence.
Biconditional Statements
A biconditional statement (\(\leftrightarrow\)) is true when both component propositions have the same truth value.
For example, \(p \leftrightarrow q\) is true when both \(p\) and \(q\) are true or both are false.
In our problem, we analyze two biconditional statements: \(p \leftrightarrow q\) and \(p \leftrightarrow eg q\).
By constructing a truth table, we can see that \(eg(p \leftrightarrow q)\) and \(p \leftrightarrow eg q\) yield the same truth values for all possible combinations of \(p\) and \(q\), thus proving their logical equivalence.

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Most popular questions from this chapter

Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expression involving logical connectives). a) \(\neg \forall x \forall y P(x, y) \quad\) b) \(\neg \forall y \exists x P(x, y)\) c) \(\neg \forall y \forall x(P(x, y) \vee Q(x, y))\) d) \(\neg(\exists x \exists y \neg P(x, y) \wedge \forall x \forall y Q(x, y))\) e) \(\quad \neg \forall x(\exists y \forall z P(x, y, z) \wedge \exists z \forall y P(x, y, z))\)

What is wrong with this argument? Let \(S(x, y)\) be "\(x\) is shorter than \(y\) ." Given the premise \(\exists s S(s, \text { Max })\) , it follows that \(S(\text { Max, Max })\) . Then by existential generalization it follows that \(\exists x S(x, x),\) so that someone is shorter than himself.

Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives. a) Something is not in the correct place. b) All tools are in the correct place and are in excellent condition. c) Everything is in the correct place and in excellent condition. d) Nothing is in the correct place and is in excellent condition. e) One of your tools is not in the correct place, but it is in excellent condition.

For each of these arguments, explain which rules of inference are used for each step. a) 鈥淒oug, a student in this class, knows how to write programs in JAVA. Everyone who knows how to write programs in JAVA can get a high-paying job. Therefore, someone in this class can get a high-paying job.鈥� b) 鈥淪omebody in this class enjoys whale watching. Every person who enjoys whale watching cares about ocean pollution. Therefore, there is a person in this class who cares about ocean pollution.鈥� c) 鈥淓ach of the 93 students in this class owns a personal computer. Everyone who owns a personal computer can use a word processing program. Therefore, Zeke, a student in this class, can use a word processing pro- gram.鈥� d) 鈥淓veryone in New Jersey lives within 50 miles of the ocean. Someone in New Jersey has never seen the ocean. Therefore, someone who lives within 50 miles of the ocean has never seen the ocean.鈥�

A statement is in prenex normal form (PNF) if and only if it is of the form $$ Q_{1} x_{1} Q_{2} x_{2} \cdots Q_{k} x_{k} P\left(x_{1}, x_{2}, \ldots, x_{k}\right) $$ where each \(Q_{i}, i=1,2, \ldots, k,\) is either the existential quantifier or the universal quantifier, and \(P\left(x_{1}, \ldots, x_{k}\right)\) is a predicate involving no quantifiers. For example, \(\exists x \forall y(P(x, y) \wedge Q(y))\) is in prenex normal form, whereas \(\exists x P(x) \vee \forall x Q(x)\) is not (because the quantifiers do not all occur first). Every statement formed from propositional variables, predicates, \(\mathbf{T},\) and \(\mathbf{F}\) using logical connectives and quantifiers is equivalent to a statement in prenex normal form. Exercise 51 asks for a proof of this fact. Show how to transform an arbitrary statement to a statement in prenex normal form that is equivalent to the given statement. (Note: A formal solution of this exercise requires use of structural induction, covered in Section \(5.3 . )\)
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