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Chapter 1: Problem 7

Translate these statements into English, where \(C(x)\) is " \(x\) is a comedian" and \(F(x)\) is " \(x\) is funny" and the domain consists of all people. $$ \begin{array}{ll}{\text { a) } \forall x(C(x) \rightarrow F(x))} & {\text { b) } \forall x(C(x) \wedge F(x))} \\ {\text { c) } \quad \exists x(C(x) \rightarrow F(x))} & {\text { d) } \exists x(C(x) \wedge F(x))}\end{array} $$

Short Answer

Expert verified
a) All comedians are funny. b) Everyone is a funny comedian. c) There is at least one person who, if a comedian, is funny. d) There is at least one funny comedian.

Step by step solution

01

Translate statement a

The statement \(\forall x (C(x) \rightarrow F(x))\) means 'For all people x, if x is a comedian, then x is funny.' In simpler terms, 'All comedians are funny.'
02

Translate statement b

The statement \(\forall x (C(x) \rightarrow F(x))\) means 'For all people x, x is a comedian and x is funny.' In other words, 'Everyone is a funny comedian.'
03

Translate statement c

The statement \(\forall x (C(x) \rightarrow F(x))\) means 'There exists some person x such that if x is a comedian, then x is funny.' This translates to 'There is at least one person who, if they are a comedian, they are funny.'
04

Translate statement d

The statement \(\forall x (C(x) \rightarrow F(x))\) means 'There exists some person x such that x is a comedian and x is funny.' In simpler terms, 'There is at least one funny comedian.'

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Key Concepts

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

Predicate Logic
In predicate logic, a predicate is a statement or expression that involves variables. Predicates can be true or false depending on the values of their variables. In our exercise, the predicates are defined as follows:
  • C(x): 'x is a comedian'
  • F(x): 'x is funny'
By combining these predicates with logical expressions and quantifiers, we can form statements about a domain, which, in this case, is 'all people.'
Quantifiers
Quantifiers are symbols used in predicate logic to indicate the scope of the variables within a statement.
There are two main types of quantifiers:
  • Universal quantifier (\(\forall\)): This symbol means 'for all' or 'every' in the context of the domain. In our exercise, the universal quantifier is used in statements like \(\forall x (C(x) \rightarrow F(x))\), translating to 'For all people x, if x is a comedian, then x is funny.'
  • Existential quantifier (\(\forallx\)): This symbol means 'there exists' or 'for some' in the domain. An example from the exercise is \(\forall x (C(x) \rightarrow F(x))\), meaning 'There exists some person x such that if x is a comedian, then x is funny.'
Logical Equivalence
Logical equivalence refers to the property where two statements are true in the same contexts or conditions.
In our exercise, logical equivalence is important for accurate translations. For instance, the statements: 'For all people x, if x is a comedian, then x is funny' and 'All comedians are funny' are logically equivalent.
Understanding logical equivalence helps us ensure that statements maintain the same truth conditions when translated from symbolic logic to plain English.
Translation of Logical Statements
Translating logical statements into plain English requires a clear understanding of the logical structure and meaning of each part of the statement.
The steps from the exercise illustrate how different logical constructs鈥攍ike implication, conjunction, and quantifiers鈥攖ranslate into meaningful English sentences.
For example:
  • \(\forall x (C(x) \rightarrow F(x))\): 'All comedians are funny.'
  • \(\forall x (C(x) \rightarrow F(x))\): 'Everyone is a funny comedian.'
  • \(\forall x (C(x) \rightarrow F(x))\): 'There is at least one person who, if they are a comedian, they are funny.'
  • \(\forall x (C(x) \rightarrow F(x))\): 'There is at least one funny comedian.'
Breaking down the statements into smaller parts can make the translation easier to understand.

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Most popular questions from this chapter

A statement is in prenex normal form (PNF) if and only if it is of the form $$ Q_{1} x_{1} Q_{2} x_{2} \cdots Q_{k} x_{k} P\left(x_{1}, x_{2}, \ldots, x_{k}\right) $$ where each \(Q_{i}, i=1,2, \ldots, k,\) is either the existential quantifier or the universal quantifier, and \(P\left(x_{1}, \ldots, x_{k}\right)\) is a predicate involving no quantifiers. For example, \(\exists x \forall y(P(x, y) \wedge Q(y))\) is in prenex normal form, whereas \(\exists x P(x) \vee \forall x Q(x)\) is not (because the quantifiers do not all occur first). Every statement formed from propositional variables, predicates, \(\mathbf{T},\) and \(\mathbf{F}\) using logical connectives and quantifiers is equivalent to a statement in prenex normal form. Exercise 51 asks for a proof of this fact. Show how to transform an arbitrary statement to a statement in prenex normal form that is equivalent to the given statement. (Note: A formal solution of this exercise requires use of structural induction, covered in Section \(5.3 . )\)

Determine the truth value of each of these statements if the domain of each variable consists of all real numbers. $$ \begin{array}{ll}{\text { a) } \forall x \exists y\left(x^{2}=y\right)} & {\text { b) } \forall x \exists y\left(x=y^{2}\right)} \\ {\text { c) } \exists x \forall y(x y=0)} & {\text { d) } \exists x \exists y(x+y \neq y+x)}\end{array} $$ $$ \begin{array}{l}{\text { e) } \forall x(x \neq 0 \rightarrow \exists y(x y=1))} \\ {\text { f) } \exists x \forall y(y \neq 0 \rightarrow x y=1)} \\\ {\text { g) } \forall x \exists y(x+y=1)} \\ {\text { h) } \exists x \exists y(x+2 y=2 \wedge 2 x+4 y=5)} \\ {\text { i) } \forall x \exists y(x+y=2 \wedge 2 x-y=1)} \\ {\text { j) } \forall x \forall y \exists z(z=(x+y) / 2)}\end{array} $$

Construct a truth table for each of these compound propositions. a) \(p \rightarrow(\neg q \vee r)\) b) \(\neg p \rightarrow(q \rightarrow r)\) c) \((p \rightarrow q) \vee(\neg p \rightarrow r)\) d) \((p \rightarrow q) \wedge(\neg p \rightarrow r)\) e) \((p \rightarrow q) \vee(\neg q \rightarrow r)\) f) \((\neg p \leftrightarrow \neg q) \leftrightarrow(q \leftrightarrow r)\)

Construct a truth table for each of these compound propositions. a. \((p \vee q) \vee r\) b) \((p \vee q) \wedge r\) c) \((p \wedge q) \vee r\) d) \((p \wedge q) \wedge r\) e) \((p \vee q) \wedge \neg r\) f) \((p \wedge q) \vee \neg r\)

For each of these collections of premises, what relevant conclusion or conclusions can be drawn? Explain the rules of inference used to obtain each conclusion from the premises. a) 鈥淚f I take the day off, it either rains or snows.鈥� 鈥淚 took Tuesday off or I took Thursday off.鈥� 鈥淚t was sunny on Tuesday.鈥� 鈥淚t did not snow on Thursday.鈥� b) 鈥淚f I eat spicy foods, then I have strange dreams.鈥� 鈥淚 have strange dreams if there is thunder while I sleep.鈥� 鈥淚 did not have strange dreams.鈥� c) 鈥淚 am either clever or lucky.鈥� 鈥淚 am not lucky.鈥� 鈥淚f I am lucky, then I will win the lottery.鈥� d) 鈥淓very computer science major has a personal computer.鈥� 鈥淩alph does not have a personal computer.鈥� 鈥淎nn has a personal computer.鈥� e) 鈥淲hat is good for corporations is good for the United States.鈥� 鈥淲hat is good for the United States is good for you.鈥� 鈥淲hat is good for corporations is for you to buy lots of stuff.鈥� f ) 鈥淎ll rodents gnaw their food.鈥� 鈥淢ice are rodents.鈥� 鈥淩abbits do not gnaw their food.鈥� 鈥淏ats are not ro- dents.鈥�
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