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Chapter 2: Problem 7

Find the expression tree for the formula $$ \neg(p \wedge q) \leftrightarrow(\neg p \vee \neg q) $$ Evaluate the expression tree for all possible pairs of truth values for \(p\) and \(q\). Use these evaluations to prove this formula is a tautology.

Short Answer

Expert verified
The formula is a tautology as it evaluates to True for all truth values of \(p\) and \(q\).

Step by step solution

01

Understand the Formula

The given formula is \( eg(p \wedge q) \leftrightarrow (eg p \vee eg q) \). It involves logical operators such as NOT (\( eg \)), AND (\( \wedge \)), OR (\( \vee \)) and BICONDITIONAL (\( \leftrightarrow \)). Our task is to build its expression tree and evaluate it for different truth values of \(p\) and \(q\).
02

Construct the Expression Tree

The expression tree is a hierarchical representation of the formula. The root is \( \leftrightarrow \), the left child is \( eg \) with a subtree of \( \wedge \) whose children are \( p \) and \( q \). The right child of the root is \( \vee \), with two children \( eg p \) and \( eg q \).
03

Define Truth Values

List all possible truth values for \(p\) and \(q\): (1) True, True (2) True, False (3) False, True (4) False, False. These combinations will be used to evaluate the expression tree.
04

Evaluate for (True, True)

For \(p = \text{True}\) and \(q = \text{True}\), compute:1. \(p \wedge q = \text{True}\) 2. \(eg(p \wedge q) = \text{False}\) 3. \(eg p = \text{False}, eg q = \text{False}\) 4. \(eg p \vee eg q = \text{False}\) 5. \(eg(p \wedge q) \leftrightarrow (eg p \vee eg q) = \text{True}\)
05

Evaluate for (True, False)

For \(p = \text{True}\) and \(q = \text{False}\), compute:1. \(p \wedge q = \text{False}\)2. \(eg(p \wedge q) = \text{True}\) 3. \(eg p = \text{False}, eg q = \text{True}\) 4. \(eg p \vee eg q = \text{True}\) 5. \(eg(p \wedge q) \leftrightarrow (eg p \vee eg q) = \text{True}\)
06

Evaluate for (False, True)

For \(p = \text{False}\) and \(q = \text{True}\), compute:1. \(p \wedge q = \text{False}\) 2. \(eg(p \wedge q) = \text{True}\) 3. \(eg p = \text{True}, eg q = \text{False}\) 4. \(eg p \vee eg q = \text{True}\) 5. \(eg(p \wedge q) \leftrightarrow (eg p \vee eg q) = \text{True}\)
07

Evaluate for (False, False)

For \(p = \text{False}\) and \(q = \text{False}\), compute:1. \(p \wedge q = \text{False}\)2. \(eg(p \wedge q) = \text{True}\) 3. \(eg p = \text{True}, eg q = \text{True}\) 4. \(eg p \vee eg q = \text{True}\) 5. \(eg(p \wedge q) \leftrightarrow (eg p \vee eg q) = \text{True}\)
08

Analyze the Results

For all combinations of \(p\) and \(q\), the expression \(eg(p \wedge q) \leftrightarrow (eg p \vee eg q)\) evaluates to True. Therefore, the formula is a tautology.

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Key Concepts

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

Expression Tree
An expression tree is a diagram that represents a logical formula in a hierarchical manner. It is beneficial for visualizing how a formula is structured and is often used to evaluate parts of a formula systematically. In our case, the root of the tree is the biconditional operator (\( \leftrightarrow \)). It compares the two sides of the formula: the left being \( eg(p \wedge q) \) and the right \( (eg p \vee eg q) \).
To break it down:
  • The left subtree begins with the NOT operator (\( eg \)), followed by an AND node (\( \wedge \)) which links to \( p \) and \( q \).
  • The right subtree is comprised of an OR node (\( \vee \)) with children negating both \( p \) and \( q \).
This organizational structure allows for evaluating the overall tree from the bottom up, calculating the truth values systematically.
Logical Operators
Logical operators form the backbone of logical expressions in mathematics and computer science. In the formula \( eg(p \wedge q) \leftrightarrow(eg p \vee eg q) \), we encounter four core operators:
  • NOT (\( eg \)): Reverses the truth value.

  • AND (\( \wedge \)): True if both operands are True.

  • OR (\( \vee \)): True if at least one operand is True.

  • BICONDITIONAL (\( \leftrightarrow \)): True if both sides have the same truth value.
Understanding these operators is crucial, as they define how expressions are evaluated and enable us to manipulate logical statements.
Truth Values
Truth values form the binary foundation for logical statements, typically denoted as True or False. When evaluating logical expressions, each variable in the expression can take on these values. For the exercise given, we analyze all possible combinations of \( p \) and \( q \):
  • \( p = \text{True}, q = \text{True} \)
  • \( p = \text{True}, q = \text{False} \)
  • \( p = \text{False}, q = \text{True} \)
  • \( p = \text{False}, q = \text{False} \)
Each combination offers a scenario where the expression needs to be evaluated independently. This approach ensures that all potential outcomes are considered and demonstrates the expression's complete behavior.
Logical Equivalence
Logical equivalence is when two statements yield the same truth value in all possible scenarios. Demonstrating logical equivalence shows that two expressions, although constructed differently, represent the same truth. In this task, our focus is the formula \( eg(p \wedge q) \leftrightarrow (eg p \vee eg q) \).
Through evaluation, each combination of \( p \) and \( q \) results in the biconditional evaluating to True, indicating both parts of the formula are logically equivalent. This consistency across all combinations labels the expression a tautology, which means it is always true regardless of the truth values assigned to its variables, effectively reinforcing the concept of logical equivalence.

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Most popular questions from this chapter

Prove that a combinatorial network for $$ (x \wedge y \wedge z) \vee(\neg x \wedge y \wedge z) \vee(x \wedge \neg y \wedge z) \vee(x \wedge y \wedge \neg z) $$ can be simplified to a combinatorial network representing $$ (x \wedge y) \vee(x \wedge z) \vee(y \wedge z) $$

Describe in words the difference between \(\emptyset\) and \(19 .\)

The length of a clause is the number of literals in the clause. The length of a CNF formula is the sum of the length of its clauses. The number of excess literals in a CNF formula is the length of the formula minus the number of clauses in the formula. (a) Show that if an unsatisfiable set \(S\) of clauses contains only clauses of length 0 and 1 , it has a resolution refutation. (Hint: Prove the following: If \(S\) contains a clause of length 0 , it has [trivially] a resolution refutation. If, for some proposition letter \(p, S\) contains both \(p\) and \(\neg p,\) then \(S\) has a resolution refutation. Otherwise, \(S\) is satisfiable.) (b) Show that if a set \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k} \vee \lambda_{k+1}+\cup S(k \geq 1)\right.\) of clauses is un- satisfiable, so are \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \lambda_{k}\right\\} \cup S\) and \(\left\\{\lambda_{k+1}\right\\} \cup S\). (Hint: For the first half, prove that if an interpretation \(I\) satisfies \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k}\right\\} \cup S,\) it also satisfies \(\left.\left\\{\lambda_{1} \vee \lambda_{2} \vee \cdots \vee \lambda_{k} \vee \lambda_{k+1}\right\\} \cup S_{.}\right)\) (c) Show that for \(k \geq 1,\) the number of excess literals in \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \cdots \vee \lambda_{k}\right\\} \cup S\) and the number of excess literals in \(\left\\{\lambda_{k+1}\right\\} \cup S\) are both less than the number of excess literals in \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k} \vee \lambda_{k+1}\right\\} \cup S\). (d) A resolution derivation of a clause \(r_{k}\) from a set \(S\) of clauses is a sequence \(r_{0}, r_{1}, r_{2}, \ldots, r_{k}\) of clauses where each \(r_{l}\) is either an element of \(S\) or a resolvant of two previous \(r\) 's. (Thus, resolution refutation of \(S\) is just a resolution derivation of \(F\) from \(S .\) ) Show that if there is a resolution derivation of \(\lambda\) from \(S\) and a resolution refutation of \(S \cup\\{\lambda\\},\) then there is a resolution refutation of \(S .\) (e) Prove that if there is a resolution refutation \(\rho\) of \(\left\\{\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k}\right\\} \cup S,\) then either (i) there is a resolution refutation of \(\left.\mid \lambda_{1} \vee \lambda_{2} \vee \cdots \vee \lambda_{k} \vee \lambda_{k+1}\right\\} \cup S\) or (ii) there is a resolution derivation of \(\lambda_{k+1}\) from \(\lambda_{1} \vee \lambda_{2} \vee \ldots \vee \lambda_{k} \vee \lambda_{k+1} \cup S\). (Hint: Prove this by induction on the length \(\rho\). You will have to add \(\lambda_{k+1}\) as a disjunct to some of the clauses in \(\rho\). It is not true in general that if \(S \models \lambda\), then there is a resolution derivation of \(\lambda\) from \(S .\) ) (f) Prove that resolution refutation is complete.

For any two integers \(m\) and \(n,\) we say \(m\) divides \(n\) if there is an integer \(k\) such that \(n=\) \(m k\). (Many programming languages give easy ways to say that, such as \(n \% m=0\) or \(n\) div \(m=0 .\) ) Define \(D i v(m, n)\) to be \(m\) divides \(n\). Translate each of the following propositions and quantified formulas into a clear English sentence. Label each as being true or false, with the universe as the set \(\mathbb{Z}\). (a) \(\operatorname{Div}(5,7)\) (b) \(\operatorname{Div}(4,16)\) (c) \(\operatorname{Div}(16,4)\) (d) \(\operatorname{Div}(-8,0)\) (e) \(\forall m(\forall n(\operatorname{Div}(m, n)))\) (f) \(\forall n(\operatorname{Div}(1, n))\) (g) \(\forall m(\operatorname{Div}(m, 0))\) (h) \(\forall m(\forall n(\operatorname{Div}(m, n) \rightarrow \operatorname{Div}(n, m)))\) (i) \(\forall m(\forall n(\forall p((D i v(m, n) \wedge \operatorname{Div}(n, p)) \rightarrow \operatorname{Div}(m, p))))\) (j) \(\forall m(\forall n((D i v(m, n) \wedge \operatorname{Div}(n, m)) \rightarrow m=n))\)

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.)
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