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Chapter 1: Problem 55

Find a compound proposition logically equivalent to \(p \rightarrow q\) using only the logical operator \(\downarrow .\)

Short Answer

Expert verified
\[(p \rightarrow q) \equiv ((p \downarrow p) \downarrow (q \downarrow q))\]

Step by step solution

01

Express the implication

Start by recalling that the implication \(p \rightarrow q\) can be rewritten using other logical operators. One common way to express it is: \(p \rightarrow q \equiv eg p \lor q\).
02

Rewrite using NAND

Next, remember that for any propositions \(a\) and \(b\), the logical NAND operator, \(a \downarrow b\), is equivalent to eg(a \land b). We need to express both \(eg p\) and \(eg p \lor q\) using only NAND.
03

Express NOT using NAND

The negation of a proposition \(a\), i.e., \(eg a\), can be expressed using NAND as follows: \(eg a = a \downarrow a\). So, \(eg p = p \downarrow p\).
04

Express OR using NAND

To express \(eg p \lor q\) using only NAND, use De Morgan's law: \(eg p \lor q = eg(eg(eg p) \land eg q)\). Rewriting using NAND: \(eg p \lor q = eg((p \downarrow p) \downarrow (q \downarrow q))\).
05

Simplify the expression

Simplify the expression in step 4 further: \(eg p \lor q = (p \downarrow p) \downarrow (q \downarrow q)\).

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Key Concepts

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

Implication
In logic, 'implication' refers to a conditional statement, typically in the form of 'if p, then q' and denoted as \( p \rightarrow q \). When we talk about implications, we're essentially stating that if the first proposition (p) is true, then the second proposition (q) must also be true. This relationship is fundamental in logical reasoning. For better understanding, remember that this can be rewritten using other logical operators as \( eg p \vee q \). This means that either p is false, or q is true. Understanding implications helps in breaking down complex logical statements into simpler parts.
NAND Operator
A NAND operator is a combination of AND and NOT operations. It is denoted by \( \downarrow \). For any propositions \(a\) and \(b\), the NAND operation is expressed as \( a \downarrow b \), which means \( eg (a \land b) \). Simply put, NAND produces true unless both operands are true. In simpler terms:
  • If both \(a\) and \(b\) are true, the result is false.
  • If either \(a\) or \(b\) is false, the result is true.
NAND is powerful because any logical operation, including AND, OR, and NOT, can be constructed using NAND alone.
De Morgan's Laws
De Morgan's Laws are rules for working with negations in logic. They provide a way to simplify expressions involving NOT, AND, and OR. These laws are:
1. The negation of a conjunction is the disjunction of the negations: \( eg (p \land q) \equiv (eg p \vee eg q) \).
2. The negation of a disjunction is the conjunction of the negations: \( eg (p \vee q) \equiv (eg p \land eg q) \).
These laws come in handy, especially in the context of expressing logical statements using different operators like NAND.
Negation
Negation is a fundamental logical operation that inverts the truth value of a proposition. For a given proposition \(p\), its negation is represented as \( eg p \), which reads 'not p'. If \(p\) is true, then \( eg p \) is false, and vice versa. Negation plays a crucial role in the transformation of logical expressions, especially when using NAND to represent other logical operations. For example, \( eg a \) can be expressed as \( a \downarrow a \).
Logical Operators
Logical operators are symbols or words used to connect two or more propositions in a logical expression, thus forming a compound proposition. The most common logical operators include:
  • AND (\( \land \)): True if both operands are true.
  • OR (\( \vee \)): True if at least one operand is true.
  • NOT (\( eg \)): Inverts the truth value of the operand.
  • IMPLICATION (\( \rightarrow \)): True if the first operand implies the truth of the second operand.
  • NAND (\( \downarrow \)): NOT AND; true unless both operands are true.
Understanding these operators is key to mastering logical equivalence and transforming logical expressions accurately.

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Most popular questions from this chapter

Express each of these statements using predicates and quantifiers. a) A passenger on an airline qualifies as an elite flyer if the passenger flies more than 25,000 miles in a year or takes more than 25 flights during that year. b) A man qualifies for the marathon if his best previous time is less than 3 hours and a woman qualifies for the marathon if her best previous time is less than 3.5 hours. c) A student must take at least 60 course hours, or at least 45 course hours and write a master鈥檚 thesis, and receive a grade no lower than a B in all required courses, to receive a master鈥檚 degree. d) There is a student who has taken more than 21 credit hours in a semester and received all A鈥檚.

Prove that there are infinitely many solutions in positive integers \(x, y,\) and \(z\) to the equation \(x^{2}+y^{2}=\) \(z^{2} .\left[\text { Hint: Let } x=m^{2}-n^{2}, y=2 m n, \text { and } z=m^{2}+n^{2}\right.\) where \(m\) and \(n\) are integers. \(]\)

Express the negation of each of these statements in terms of quantifiers without using the negation symbol. a) \(\forall x(x>1)\) b) \(\forall x(x \leq 2)\) \(\begin{array}{ll}\text { c) } & \exists x(x \geq 4)\end{array}\) d) \(\exists x(x<0)\) e) \(\forall x((x<-1) \vee(x>2))\) f) \(\exists x((x<4) \vee(x>7))\)

Use a proof by exhaustion to show that a tiling using dominoes of a \(4 \times 4\) checkerboard with opposite corners removed does not exist. [Hint: First show that you can assume that the squares in the upper left and lower right corners are removed. Number the squares of the original checkerboard from 1 to \(16,\) starting in the first row, moving right in this row, then starting in the leftmost square in the second row and moving right, and so on. Remove squares 1 and \(16 .\) To begin the proof, note that square 2 is covered either by a domino laid horizontally, which covers squares 2 and \(3,\) or vertically, which covers squares 2 and \(6 .\) Consider each of these cases separately, and work through all the subcases that arise. \(]\)

Identify the error or errors in this argument that supposedly shows that if \(\exists x P(x) \wedge \exists x Q(x)\) is true then \(\exists x(P(x) \wedge Q(x))\) is true. $$ \begin{array}{l}{\text { 1. } \exists x P(x) \vee \exists x Q(x)} \\ {\text { 2. } \exists x P(x)} \\ {\text { 3. } P(c)} \\ {\text { 4. } \exists x Q(x)} \\\ {\text { 5. } Q(c)} \\ {\text { 6. } P(c) \wedge Q(c)} \\ {\text { 7. } \exists x(P(x) \wedge Q(x))}\end{array} $$
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