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For each quantified formula that follows: find a universe \(U\) and predicates \(A\) and \(B\) in which the formula is true and \(U, A\) and \(B\) in which it is false. (a) \(\forall x(((A(x) \vee B(x)) \wedge \neg(A(x) \wedge B(x)))\) (b) \(\forall x \forall y(P(x, y) \rightarrow P(y, x))\) (c) \(\forall x(P(x) \rightarrow \exists y Q(x, y))\) (d) \(\exists x(A(x) \wedge \forall y B(x, y))\) (e) \(\forall x A(x) \rightarrow(\forall x B(x) \rightarrow(\forall x(A(x) \rightarrow B(x))))\)

Short Answer

Expert verified
Each formula can have true and false scenarios by choosing appropriate predicates and universe.

Step by step solution

01

Understanding the Problem

We need to evaluate each quantified logic formula (a) to (e) by finding a universe \(U\) and predicates \(A\) and \(B\) such that the formula holds true and another scenario where it does not. We will break each formula into logical components and explore possible universes and predicates.
02

Formula (a) Analysis

The formula is \( \forall x(((A(x) \vee B(x)) \wedge eg(A(x) \wedge B(x)))) \). Consider \( U = \{1, 2\} \), with \( A(x) \) as \( x = 1 \) and \( B(x) \) as \( x = 2 \). The formula is true as there is no overlap between \( A(x) \) and \( B(x) \). However, redefine \( A(x) \) as true for both 1 and 2, this formula with \( B(x) \) also true makes it false due to overlap.
03

Formula (b) Analysis

The formula is \( \forall x \forall y(P(x, y) \rightarrow P(y, x)) \). Let \( U = \{1\} \) and \( P(x, y) \) be true for all combinations. The formula is trivially true. False scenario: Let \( U = \{1, 2\} \), \( P(1, 2) = \text{true}\) and \( P(2, 1) = \text{false} \). This setup makes the formula false.
04

Formula (c) Analysis

The formula is \( \forall x(P(x) \rightarrow \exists y Q(x, y)) \). Consider \( U = \{1, 2\} \) with \( P(x)\) true for all \( x \), and \( Q(x, y) \) true if \( x = y \), making the formula true. Now, let \( U \) be the same but define \( P(1) = \text{true}\) and \( Q(1, y)\) as \( \text{false}\) for all \( y \). This makes the formula false for \( x = 1\).
05

Formula (d) Analysis

The formula is \( \exists x(A(x) \wedge \forall y B(x, y)) \). Let \( U = \{1\} \), \( A(1) = \text{true}\), and \( B(1, 1) = \text{true}\). The formula is true. A false universe is \( U = \{1, 2\} \), where \( A(1) = \text{true}\), and \( B(1, y) = \text{true}\) only for \( y = 1 \). The failure to satisfy all \( y \) in \( U \) makes the formula false.
06

Formula (e) Analysis

The formula is \( \forall x A(x) \rightarrow(\forall x B(x) \rightarrow(\forall x(A(x) \rightarrow B(x)))) \). With \( U = \{1\} \), \( A(1) = \text{true}\), and \( B(1) = \text{true}\), this is true. For a false outcome, let \( U = \{1, 2\}\), \( A(x)\) true for all \( x \), but \( B(2) = \text{false}\), disrupting \( \forall x(A(x) \rightarrow B(x))\).

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Key Concepts

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

Predicates
In the realm of logical expressions, predicates are essential building blocks. A predicate is essentially a statement or proposition that may depend on a variable. For example, in the expression \(A(x)\), \(A\) is a predicate that could mean "\(x\) is an even number." The value of \(x\) determines whether the statement is true or false.

Predicates allow us to express properties about things or individuals from a specific collection, often known as the universe of discourse or domain. They can be as simple as one variable, such as \(A(x)\), or they might involve multiple variables, like \(P(x, y)\).

Understanding predicates is vital for evaluating logical statements because they enable the specification of conditions that elements of the universe must satisfy. This framework allows us to explore whether complex logical formulas hold true under different conditions.
Universe in Logic
The universe in logic refers to the set of all possible elements that you are considering in a particular logical context. It is also known as the domain of discourse. This set is essential, as it defines the range over which a predicate is evaluated.

For instance, if we're considering the universe \(U = \{1, 2, 3\}\), any predicate involving a variable, say \(A(x)\), will consider each element in \(U\). Depending on the context, universes can be finite, like \(\{1, 2, 3\}\), or infinite, such as the set of all real numbers.

The size and elements of the universe significantly impact whether a quantified statement is true or false. For example, if the universe changes, the truth of the statement \(\forall x A(x)\) might change as well. Therefore, understanding the specified universe is crucial when analyzing logical statements.
Truth Values in Logic
Truth values in logic reflect whether a particular statement or proposition is true or false. In mathematical logic, these values are typically limited to true and false, represented precisely as '1' and '0'.

When it comes to predicate logic, assigning truth values involves considering the predicates within the expression on an element-by-element basis from the universe. For example, \(A(x)\) might be true for some \(x\) and false for others, depending on the properties defined by \(A\).

A logical formula's overall truth depends on the compounded truth values of its components. In formulas like \(\forall x P(x)\), the statement is true if \(P(x)\) holds for every \(x\) in the universe. If even one element does not satisfy \(P(x)\), the formula is false. Hence, understanding how truth values are derived is pivotal in predicate logic.
Predicate Logic Analysis
Predicate logic analysis involves examining logical statements to determine their validity under different conditions, such as varied predicates or universes. It often deals with quantified formulas involving "for all" (\(\forall\)) or "there exists" (\(\exists\)), and predicates that rely on one or more variables.

For instance, consider the expression \(\forall x (A(x) \vee B(x))\). Analyzing this involves checking if every element \(x\) in the universe \(U\) makes the statement true. This requires evaluating both \(A(x)\) and \(B(x)\). If at least one of these is true for each \(x\), the whole expression holds.

Predicate logic analysis is essential for solving problems in logic because it systematically dissects statements into their logical components, allowing you to see where they hold true and under what conditions they fail. By developing this analytical skill, you'll be better equipped to handle complex logical expressions in various contexts.

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Most popular questions from this chapter

The first stage of the method described to "push negations inward" was a method to climinate \(\rightarrow\) 's and \(\leftrightarrow\) 's. Prove that in the method to eliminate them, the process of substituting always stops, Consider, for example, the substitution in the formula $$ (p \leftrightarrow q) \leftrightarrow(r \leftrightarrow s) $$ If the substitution is first performed on the second \(\leftrightarrow\), the resultant formula is $$ ((p \leftrightarrow q) \rightarrow(r \leftrightarrow s)) \wedge((r \leftrightarrow s) \rightarrow(p \leftrightarrow q)) $$ which has more \(\leftrightarrow\) 's to replace than in the original formula! At first sight, one might expect that if the substitutions are made in the wrong order, the process might continue generating more \(\leftrightarrow\) 's at each stage, and the process might continue forever. (Hint: One method is to, instead of just counting the number of \(\leftrightarrow\) symbols, put a weight on each \(\leftrightarrow\) symbol, with the weight of the \(\leftrightarrow\) symbol in \(\psi \leftrightarrow x\) being dependent on the number of \(\leftrightarrow\) 's in \(\psi\) and \(\chi\). If the correct method of calculating weights is used, it can be shown that the total weight of the \(\leftrightarrow\) 's decreases with each substitution.

Give an example of a universal set \(U\) and predicates \(P\) and \(Q\) such that \((\forall x P(x)) \rightarrow\) \((\forall x Q(x))\) is true but \(\forall x(P(x) \rightarrow Q(x))\) is false.

A restaurant displays the sign "Good food is not cheap." and a competing restaurant displays the sign "Cheap food is not good." Are the two restaurants saying the same thing?

The second stage of the procedure to "push negations inward" started with a formula whose only logical connectives are \(\neg, v,\) and \(\wedge\) and constructed a tautologically equivalent formula with negations applied only to proposition letters. (a) Write an algorithm describing exactly what is done. The algorithm should work on formulas as strings of symbols. To avoid what in this case is irrelevant detail, the program should assume that all proposition letters are one character long and that any symbol encountered, except for \((.), \wedge, v,\) and \(\neg,\) is a proposition letter. Assume that the formula contains no blanks. (It is perhaps easiest to consider the program as a function that is passed the original formula - a string-as a parameter, and then returns the equivalent formula with all the negations pushed inward. It is casiest to use recursion to handle many subformulas.) (b) Prove that your program from part (a) works. (Hint: if your program in part (a) uses recursion to handle subformulas, it is natural to do this proof by induction on formulas. However, the induction may not be straightforward.)

Find formulas in CNF equivalent to each of the following formulas: (a) \(\neg(p \wedge T)\) (b) \(((p \rightarrow q) \rightarrow r) \rightarrow F\) (c) \(((p \rightarrow q) \rightarrow r) \rightarrow T\) (d) \((p \leftrightarrow q) \leftrightarrow r\) (e) \(\neg(p \leftrightarrow q) \leftrightarrow r\) (f) \(((p \vee q) \rightarrow r) \wedge(r \rightarrow \neg(p \vee q))\) (g) \((\neg r) \rightarrow(((p \vee q) \rightarrow r) \rightarrow \neg q)\)

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