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Chapter 2: Problem 15

Let \(\phi=(\neg(p \wedge q)) \leftrightarrow(\neg r \vee \neg s)\). For each of the following interpretations of \(p, q, r,\) and \(s,\) compute \(I(\phi)\) using the truth tables for \(\neg, \vee, \wedge, \rightarrow,\) and \(\leftrightarrow\) (a) \(I(p)=T, I(q)=T, I(r)=F,\) and \(I(s)=T\) (b) \(I(p)=T, I(q)=F, I(r)=F,\) and \(I(s)=F\) (c) \(I(p)=F, I(q)=T, I(r)=F,\) and \(I(s)=T\) (d) \(I(p)=F, I(q)=F, I(r)=F,\) and \(I(s)=T\)

Short Answer

Expert verified
(a) F, (b) T, (c) T, (d) T.

Step by step solution

01

Understanding the Expression

The given logical expression is \( \phi = (eg(p \wedge q)) \leftrightarrow(eg r \vee eg s) \). To solve this, we need to evaluate the logical operators \( eg \), \( \vee \), \( \wedge \), and \( \leftrightarrow \) using truth tables for each interpretation of variables.
02

Set Truth Table for Basic Operators

Recall the truth table definitions: - \( eg A \) is true if \( A \) is false and false if \( A \) is true.- \( A \vee B \) is true if at least one of \( A \) or \( B \) is true.- \( A \wedge B \) is true only if both \( A \) and \( B \) are true.- \( A \leftrightarrow B \) is true if \( A \) and \( B \) have the same truth value.
03

Evaluate for Interpretation (a)

Given \( I(p)=T, I(q)=T, I(r)=F, \) and \( I(s)=T \):- Calculate \( p \wedge q = T \wedge T = T \)- Hence, \( eg(p \wedge q) = eg T = F \)- Calculate \( eg r = eg F = T \) and \( eg s = eg T = F \)- Hence, \( eg r \vee eg s = T \vee F = T \)- Finally, \( eg(p \wedge q) \leftrightarrow (eg r \vee eg s) = F \leftrightarrow T = F \)
04

Evaluate for Interpretation (b)

Given \( I(p)=T, I(q)=F, I(r)=F, \) and \( I(s)=F \):- Calculate \( p \wedge q = T \wedge F = F \)- Hence, \( eg(p \wedge q) = eg F = T \)- Calculate \( eg r = eg F = T \) and \( eg s = eg F = T \)- Hence, \( eg r \vee eg s = T \vee T = T \)- Finally, \( eg(p \wedge q) \leftrightarrow (eg r \vee eg s) = T \leftrightarrow T = T \)
05

Evaluate for Interpretation (c)

Given \( I(p)=F, I(q)=T, I(r)=F, \) and \( I(s)=T \):- Calculate \( p \wedge q = F \wedge T = F \)- Hence, \( eg(p \wedge q) = eg F = T \)- Calculate \( eg r = eg F = T \) and \( eg s = eg T = F \)- Hence, \( eg r \vee eg s = T \vee F = T \)- Finally, \( eg(p \wedge q) \leftrightarrow (eg r \vee eg s) = T \leftrightarrow T = T \)
06

Evaluate for Interpretation (d)

Given \( I(p)=F, I(q)=F, I(r)=F, \) and \( I(s)=T \):- Calculate \( p \wedge q = F \wedge F = F \)- Hence, \( eg(p \wedge q) = eg F = T \)- Calculate \( eg r = eg F = T \) and \( eg s = eg T = F \)- Hence, \( eg r \vee eg s = T \vee F = T \)- Finally, \( eg(p \wedge q) \leftrightarrow (eg r \vee eg s) = T \leftrightarrow T = T \)

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Key Concepts

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

Logical Operators
Logical operators are foundational elements in propositional logic, providing the means to combine simpler statements into more complex expressions.
These operators include the well-known symbols such as \( eg \) (NOT), \( \vee \) (OR), \( \wedge \) (AND), \( \rightarrow \) (IMPLIES), and \( \leftrightarrow \) (BICONDITIONAL). Each operator has its own rules for how it manipulates truth values.
Let's break them down:
  • Negation (\( eg \)): This operator takes a single proposition and flips its truth value. If a proposition \( p \) is true, \( eg p \) will be false, and vice versa.
  • Conjunction (\( \wedge \)): This combines two propositions, \( p \) and \( q \). The result is true only when both propositions are true.
  • Disjunction (\( \vee \)): This operator is true if at least one of the propositions \( p \) or \( q \) is true. It only becomes false when both propositions are false.
  • Implication (\( \rightarrow \)): A proposition \( p \rightarrow q \) states that if \( p \) is true, then \( q \) must be true. If \( p \) is false, \( q \) can be either true or false for the implication to hold.
  • Biconditional (\( \leftrightarrow \)): This operator states that two propositions \( p \) and \( q \) are logically equivalent, meaning they share the same truth value.
Understanding these operators through truth tables helps in evaluating logical expressions effectively.
Logical Equivalence
Logical equivalence is a key concept in propositional logic. It refers to two statements or expressions that always have the same truth value in every possible scenario.
When two expressions, say \( A \) and \( B \), are logically equivalent, it is represented as \( A \equiv B \). This means no matter how you assign truth values to the variables within the expressions, \( A \) and \( B \) will always evaluate to the same truth value.
To determine if two propositions are logically equivalent, truth tables come in handy. By listing all possible truth values for the component variables, you can see if the columns under both expressions match for each possible combination.
  • Examples of Logical Equivalence:
  • Double Negation: \( eg(eg A) \equiv A \), meaning taking the negation of a negation returns you to the original truth value.
  • De Morgan's Laws: \( eg (A \wedge B) \equiv (eg A \vee eg B) \) and \( eg (A \vee B) \equiv (eg A \wedge eg B) \).
  • Implication as Biconditional: \( A \rightarrow B \equiv eg A \vee B \).
Logical equivalence helps in simplifying complex logical expressions and proofs.
Propositional Logic
Propositional Logic, also known as sentential logic or statement logic, is a branch of logic that deals with propositions, which are statements that can either be true or false but not both.
Each proposition is represented by a variable, typically letters like \( p, q, \, \text{or} \, r \), and can be connected using logical operators to form complex statements.
Propositional logic does not concern itself with the content of the propositions but focuses solely on the logical form. This abstraction allows us to perform logical reasoning independent of the context.
  • Advantages of Propositional Logic:
  • Clarity and Simplicity: By representing statements with variables, complex arguments can be broken down and simplified.
  • Framework for Reasoning: Provides a formal foundation for constructing valid arguments and proving logical statements.
  • Application in Computer Science: Forms the basis for designing circuits, programming, and developing algorithms.
Understanding propositional logic is essential for delving deeper into more advanced areas of mathematics and computer science.

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Most popular questions from this chapter

For the following formulns find equivalent formulas in CNF and DNF form. Draw combinatorial networks corresponding to the original formulas and their equivalent CNF forms. (a) \((p \wedge q) \leftrightarrow(p \wedge r)\) (b) \(((p \rightarrow q) \rightarrow r) \rightarrow p\)

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) \((\neg r) \rightarrow(((p \vee q) \rightarrow r) \rightarrow \neg q)\)

Simplify the following boolean expressions: (a) \((x \wedge y) \vee(x \wedge \neg y) \vee(\neg x \wedge y) \vee(\neg x \wedge \neg y)\) (b) \((x \wedge y \wedge z) \vee(x \wedge \neg y \wedge z) \vee(\neg x \wedge y \wedge \neg z) \vee(\neg x \wedge \neg y \wedge z)\) (c) \((x \wedge y \wedge \neg z) \vee(x \wedge \neg y \wedge z) \vee(x \wedge \neg y \wedge \neg z)\)

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.

Given an array Values with \(n\) elements Values[0], Values[1],.... Values \([n-1 \mid\) each containing a real number, the following algorithm finds the sum of all the positive values in Values. Write an invariant for the loop. rollingSum \(=0\) for \(i=\phi, 2, \ldots, n-1\) if Values \([t]>0\) rolling Sum \(=\) rollingSum \(+\) Values \([i]\) Output rollingSum.
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