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Chapter 2: Problem 6

Find the expression tree for the formula $$ ((\neg(p \wedge q)) \vee(\neg(q \wedge r))) \wedge((\neg(p \leftrightarrow(\neg(\neg s)))) \vee((r \wedge s) \vee(\neg q))) . $$ Evaluate the expression tree if proposition \(p\) is \(F\), proposition \(q\) is \(T\), proposition \(r\) is \(F\), and proposition \(s\) is \(T\).

Short Answer

Expert verified
The expression evaluates to True (T).

Step by step solution

01

Analyze the Formula Structure

The given formula is complex, and we need to organize it into a tree structure. The main connective is the outermost conjunction \( \wedge \). This divides the formula into two main parts: 1. \((eg(p \wedge q)) \vee(eg(q \wedge r))\)2. \((eg(p \leftrightarrow(eg(eg s)))) \vee((r \wedge s) \vee(eg q))\)
02

Break Down the Left Subformula

The left subformula is \((eg(p \wedge q)) \vee(eg(q \wedge r))\). Here, the main connective is \( \vee \), dividing it into two subparts:- \(eg(p \wedge q)\): a negation of the conjunction between \(p\) and \(q\).- \(eg(q \wedge r)\): a negation of the conjunction between \(q\) and \(r\).
03

Decompose the Right Subformula

The right subformula is \((eg(p \leftrightarrow(eg(eg s)))) \vee((r \wedge s) \vee(eg q))\). The main connective is \( \vee \), so we split it into:- \(eg(p \leftrightarrow(eg(eg s)))\): a negation of the biconditional between \(p\) and \(eg(eg s)\).- \((r \wedge s) \vee(eg q)\): where the main connective \( \vee \) separates two parts: \((r \wedge s)\) and \((eg q)\).
04

Construct the Expression Tree

Using the analysis, the tree structure from bottom to top is as follows: - Leaves containing propositions \(p, q, r, s\).- Unary operations \(eg\).- Binary operations within conjunctions \(\wedge\), disjunctions \(\vee\), and biconditionals \(\leftrightarrow\). - Root as the outermost \(\wedge\). This results in a layered structure reflecting the operations in their hierarchy.
05

Evaluate the Proposition Values

Given that \(p = F\), \(q = T\), \(r = F\), and \(s = T\), evaluate each part of the tree:- \(p \wedge q = F \wedge T = F\), so \(eg(p \wedge q) = T\).- \(q \wedge r = T \wedge F = F\), so \(eg(q \wedge r) = T\).- \(eg(eg s) = eg(eg T) = T\).- \(p \leftrightarrow T = F \leftrightarrow T = F\), so \(eg(p \leftrightarrow T) = T\).- \(r \wedge s = F \wedge T = F\).- \(T \vee eg q = T \vee F = T\).
06

Simplify and Conclude

Compute the results using the tree:1. For \((eg(p \wedge q)) \vee(eg(q \wedge r))\), we have \(T \vee T = T\).2. For \((eg(p \leftrightarrow(eg(eg s)))) \vee((r \wedge s) \vee(eg q))\), we find \(T \vee T = T\).Finally, the outer conjunction \(T \wedge T = T\). Thus, the evaluated expression is \(T\).

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Key Concepts

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

Logical Connectives
Logical connectives play a crucial role in understanding expression trees and propositional logic. These are symbols that connect simple propositions to form more complex ones. In the context of the exercise, we encounter several basic logical connectives:
  • Negation (\(eg\)): This connective flips the truth value of a proposition. If a proposition is true, negation makes it false, and vice versa.
  • Conjunction (\(\wedge\)): Also known as "and", a conjunction is true if and only if both propositions it connects are true.
  • Disjunction (\(\vee\)): This is the "or" connective, which is true if at least one of the propositions is true.
  • Biconditional (\(\leftrightarrow\)): True if both propositions have the same truth value鈥攅ither both true or both false.
Recognizing and understanding these connectives is vital to correctly structuring expression trees and evaluating propositions logically.
Propositional Logic
Propositional logic is a branch of logic dealing with propositions and their connectives. It is concerned with the combinations of simple statements and their truth values. This type of logic is foundational for constructing expression trees, which graphically represent logical formulas. In our exercise, we work with a formula containing several propositions:
  • Propositions are basic, declarative statements, such as ``p is false.鈥�
  • The formula combines these propositions using logical connectives to capture more complex logical relationships.
The task is to build an expression tree that visually represents how these connectives combine the propositions. By doing so, you can clearly see the hierarchical structure of the formula. Building such trees makes it easier to understand how different logical parts interact and help in evaluating the compound proposition effectively.
Boolean Evaluation
Boolean evaluation involves determining the truth value of an expression based on the truth values of its individual propositions. In the context of propositional logic, this evaluation does not rely on the content of the propositions but rather on their assigned truth values.
  • For example, a proposition might be assigned true (T) or false (F).
  • The structure of the expression tree helps to systematically plug in these truth values from the leaves (propositions) up through the tree.
  • The result at the root of the tree is the overall truth value of the entire logical formula.
In our exercise, once we constructed the tree, we substituted the given truth values:
  • $p = F$, $q = T$, $r = F$, and $s = T$.
Ultimately, after processing through the connectives, the formula evaluated to true (T). This example highlights how Boolean evaluation can simplify complicated logical expressions into clear true or false outcomes.

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Most popular questions from this chapter

For the following predicates with universal set \(\mathbb{R}\), state the meaning of the predicate in a sentence. If it is false, give an example to show why. (Example: \(\forall x(\exists y(xz \wedge z>y))))\) (h) \(\forall x(\forall y(\forall z)(x>y \wedge y>z) \rightarrow x>z))\) )

(a) Show that the following formula in CNF is unsatisfiable: $$ (p \vee q) \wedge(p \vee \neg q) \wedge(\neg p \vee q) \wedge(\neg p \vee \neg q) $$ (b) Show that the following formula in CNF is unsatisfiable: $$ \begin{array}{c} (p \vee q \vee r) \wedge(p \vee \neg q \vee r) \wedge(\neg p \vee q \vee r) \wedge(\neg p \vee \neg q \vee r) \\ \wedge(p \vee q \vee \neg r) \wedge(p \vee \neg q \vee \neg r) \wedge(\neg p \vee q \vee \neg r) \wedge(\neg p \vee \neg q \vee \neg r) \end{array} $$ Can you find an easier argument than just writing the entire truth table? (c) Generalize the above to some class of CNF formulas on an arbitrary number \(n \geq 1\) of proposition letters, and prove it by induction on \(n\).

(a) The conjunction of \(n\) formulas \(p_{1}, p_{2}, \ldots, p_{n}\) is defined to be the formula \(\left(\ldots\left(\left(p_{1} \wedge p_{2}\right) \wedge p_{3}\right) \wedge \ldots\right) \wedge p_{n} .\) For \(n=0,\) there is a special case: The conjunction of zero formulas is defined to be \(T\). For \(n=1\), that conjunction simplifies to \(p_{1}\). Let \(\phi\) be the conjunction of \(p_{1}, p_{2}, \ldots, p_{n} .\) Prove that for any interpretation \(I, I(\phi)=T\) if and only if \(I\left(p_{i}\right)=T\) for each \(i\) such that \(1 \leq i \leq n .\) (Hint: Use induction.) (b) Let \(\phi\) be the formula $$ \left(\ldots\left(\left(p_{1} \leftrightarrow p_{2}\right) \leftrightarrow p_{3}\right) \leftrightarrow \ldots\right) \leftrightarrow p_{n} $$ for \(n \geq 1 .\) For what interpretations \(I\) is \(I(\phi)=T ?\) (Hint: The answer involves counting how many of the \(p_{i}\) 's are true in \(I\). Prove the result by induction on \(n\).)

Find formulas in DNF 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) \((\neg r) \rightarrow(((p \vee q) \rightarrow r) \rightarrow \neg q)\)

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.
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