Chapter 1.10-1.13

Indicate whether the argument is valid or invalid. For valid arguments, prove that the argument is valid using a truth table. For invalid arguments give truth values for the variables showing that the argument is not valid. p ∨ q ¬q ∴ p ↔ q

Not valid. p = T and q = F.

Translate each of the following English statements into logical expressions. The domain of discourse is the set of all real numbers. The reciprocal of every positive number is also positive.

∀x ((x > 0) → (1/x > 0))

Use the rules of inference and the laws of propositional logic to prove that each argument is valid. Number each line of your argument and label each line of your proof "Hypothesis" or with the name of the rule of inference used at that line. If a rule of inference is used, then include the numbers of the previous lines to which the rule is applied. (p ∨ q) → r p ∴ r

1. (p ∨ q) → r Hypothesis 2. p Hypothesis 3. p ∨ q Addition, 2. 4. r Modus ponens, 1, 3.

Use the rules of inference and the laws of propositional logic to prove that each argument is valid. Number each line of your argument and label each line of your proof "Hypothesis" or with the name of the rule of inference used at that line. If a rule of inference is used, then include the numbers of the previous lines to which the rule is applied. p → q r → u p ∧ r ∴ q ∧ u

1. p → q Hypothesis 2. p ∧ r Hypothesis 3. p Simplification, 2 4. q Modus ponens, 1, 3. 5. r → u Hypothesis 6. r Simplification, 2 7. u Modus ponens, 5, 6 8. q ∧ u Conjunction, 4, 7.

Some of the rules of inference can be proven using the other rules of inference and the laws of propositional logic. One of the rules of inference is Modus ponens: p → q p ∴ q Prove that Modus ponens is valid using the laws of propositional logic and any of the other rules of inference besides Modus ponens. (Hint: you will need one of the conditional identities from the laws of propositional logic).

1. p → q Hypothesis 2. ¬p ∨ q Conditional identity, 1 3. p Hypothesis 4. ¬¬p Double negation, 3 5. q Disjunctive syllogism, 2, 4

Some of the rules of inference can be proven using the other rules of inference and the laws of propositional logic. One of the rules of inference is Resolution: p ∨ q ¬p ∨ r ∴ q ∨ r Prove that Resolution is valid using the laws of propositional logic and any of the other rules of inference besides Resolution. (Hint: you will need one of the conditional identities from the laws of propositional logic).

1. p ∨ q Hypothesis 2. q ∨ p Commutative law, 1 3. ¬¬q ∨ p Double negation, 2 4. ¬q → p Conditional identity, 3 5. ¬p ∨ r Hypothesis 6. p → r Conditional identity, 5 7. ¬q → r Hypothetical syllogism, 4, 6 8. ¬¬q ∨ r Conditional identity, 7 9. q ∨ r Double negation, 8

Use the rules of inference and the laws of propositional logic to prove that each argument is valid. Number each line of your argument and label each line of your proof "Hypothesis" or with the name of the rule of inference used at that line. If a rule of inference is used, then include the numbers of the previous lines to which the rule is applied. p → (q ∧ r) ¬q ∴ ¬p

1. ¬q Hypothesis 2. ¬q ∨ ¬r Addition, 1 3. ¬(q ∧ r) De Morgan's law, 2 4. p → (q ∧ r) Hypothesis 5. ¬p Modus tollens, 3, 4

hypotheses

An argument is a sequence of propositions

conclusion

An argument is a sequence of propositions, called hypotheses, followed by a final proposition

Prove that the given argument is valid. First find the form of the argument by defining predicates and expressing the hypotheses and the conclusion using the predicates. Then use the rules of inference to prove that the form is valid. The domain of discourse is the set of people who live in a city. Linda lives in the city. Linda lives in the city. Linda owns a Ferrari. Everyone who owns a Ferrari has gotten a speeding ticket. ∴ Linda has gotten a speeding ticket..

F(x): x owns a Ferrari S(x): x has gotten a speeding ticket The form of the argument is Linda lives in the city F(Linda) ∀x (F(x) → S(x)) ∴ S(Linda) 1. Linda lives in the city Hypothesis 2. ∀x (F(x) → S(x)) Hypothesis 3. F(Linda) → S(Linda) Universal instantiation, 1, 2 4. F(Linda) Hypothesis 5. S(Linda) Modus ponens, 3, 4

The domain of discourse for this problem is a group of three people who are working on a project. To make notation easier, the people are numbered 1, 2, 3. The predicate M(x, y) indicates whether x has sent an email to y, so M(2, 3) is read "Person 2 has sent an email to person 3." The table below shows the value of the predicate M(x,y) for each (x,y) pair. The truth value in row x and column y gives the truth value for M(x,y). M 1 2 3 1 T T T 2 T F T 3 T T F Indicate whether the quantified statement is true or false. Justify your answer. ∃x ∃y ((x ≠ y) ∧ ¬M(x,y))

False. For every x and y such that x ≠ y, M(x, y) is true.

Show that the two quantified statements in each problem are not logically equivalent by filling in a table so that, for the domain of discourse {a, b, c}, the values of the predicate P you select for the table causes one of the statements to be true and the other to be false. For example, the table below shows that ∀x ∀ y P(x, y) and ∃x ∃y P(x, y) are not logically equivalent because for the given values of the predicate P, ∀x ∀ y P(x, y) is false and ∃x ∃y P(x, y) is true. P a b c a T T T b T F T c T T F ∀x ∃y ((x ≠ y) ∧ P(x, y)) and ∀x ∃y P(x, y)

For the table below, ∀x ∃y P(x, y) is true and ∀x ∃y ((x ≠ y) ∧ P(x, y)) is false. P a b c a T F T b F T F c F T F

Determine whether each argument is valid. If the argument is valid, give a proof using the laws of logic. If the argument is invalid, give values for the predicates P and Q over the domain {a, b} that demonstrate the argument is invalid. ∀x (P(x) ∨ Q(x)) ∴ ∀x Q(x) ∨ ∀x P(x)

Not valid. P Q a F T b T F P(a) ∨ Q(a) and P(b) ∨ Q(b) are both true, so ∀x (P(x) ∨ Q(x)) is true. However, neither ∀x P(x) nor ∀x Q(x) is true. Therefore the hypothesis is true and the conclusion is false.

Determine whether each argument is valid. If the argument is valid, give a proof using the laws of logic. If the argument is invalid, give values for the predicates P and Q over the domain {a, b} that demonstrate the argument is invalid. ∃x Q(x) ∧ ∃x P(x) ∴ ∃x (P(x) ∧ Q(x))

Not valid. P Q a F T b T F ∃x Q(x) is true because Q(a) is true. ∃x P(x) is true because P(b) is true. However, neither P(a) ∧ Q(a) nor P(b) ∧ Q(b) is true, so ∃x (P(x) ∧ Q(x)) is false. Therefore the hypothesis is true and the conclusion is false.

Which of the following arguments are valid? Explain your reasoning. Every girl scout who sells at least 50 boxes of cookies will get a prize. Suzy, a girl scout, got a prize. Therefore Suzy sold 50 boxes of cookies.

Not valid. The first hypothesis is that ∀x (S(x) → P(x)), where the predicate S(x) means that x has sold at least 50 boxes of cookies and P(x) means that x got a prize. Consider a scenario in which there is a girl scout troop for which ∀x (S(x) → P(x)) is true. Furthermore, there is a girl named Suzy in the troop such that P(Suzy) is true and S(Suzy) is false. Then all the hypotheses are true but the conclusion is false. Therefore the argument is not valid.

Show that the given argument is invalid by giving values for the predicates P and Q over the domain {a, b}. ∃x (P(x) ∨ Q(x)) ∃x ¬Q(x) ∴ ∃x P(x)

P Q a F T b F F ∃x (P(x) ∨ Q(x)) is true because P(a) ∨ Q(a) is true. ∃x ¬Q(x) is true since ¬Q(b) is true. However, since P(a) = P(b) = F, ∃x P(x) is false. Therefore both hypotheses are true and the conclusion is false.

Prove that the given argument is valid. First find the form of the argument by defining predicates and expressing the hypotheses and the conclusion using the predicates. Then use the rules of inference to prove that the form is valid. The domain is the set of students at an elementary school. Every student who has a permission slip can go on the field trip. Every student has a permission slip. ∴ Every student can go on the field trip.

P(x): x has a permission slip F(x): x can go on the field trip The form of the argument is: ∀x (P(x) → F(x)) ∀x P(x) ∴ ∀x F(x) 1. c is an arbitrary student at the school Element definition 2. ∀x (P(x) → F(x)) Hypothesis 3. P(c) → F(c) Universal instantiation, 1, 2 4. ∀x P(x) Hypothesis 5. P(c) Universal instantiation, 1, 4 6. F(c) Modus ponens, 3, 5 7. ∀x F(x) Universal generalization, 1, 6

The domain of discourse for this problem is a group of three people who are working on a project. To make notation easier, the people are numbered 1, 2, 3. The predicate M(x, y) indicates whether x has sent an email to y, so M(2, 3) is read "Person 2 has sent an email to person 3." The table below shows the value of the predicate M(x,y) for each (x,y) pair. The truth value in row x and column y gives the truth value for M(x,y). M 1 2 3 1 T T T 2 T F T 3 T T F Indicate whether the quantified statement is true or false. Justify your answer. ∀x ∀y ((x ≠ y) → M(x,y))

True. For every x and y such that x ≠ y, M(x, y) is true.

The domain of discourse for this problem is a group of three people who are working on a project. To make notation easier, the people are numbered 1, 2, 3. The predicate M(x, y) indicates whether x has sent an email to y, so M(2, 3) is read "Person 2 has sent an email to person 3." The table below shows the value of the predicate M(x,y) for each (x,y) pair. The truth value in row x and column y gives the truth value for M(x,y). M 1 2 3 1 T T T 2 T F T 3 T T F Indicate whether the quantified statement is true or false. Justify your answer. ∃x ∀y M(x,y)

True. If x = 1, then M(x, y) is true for all y.

Which of the following arguments are invalid and which are valid? Prove your answer by replacing each proposition with a variable to obtain the form of the argument. Then prove that the form is valid or invalid. He studied for the test or he failed the test or both. He passed the test. ∴ He studied for the test.

Valid. Assign: s: he studied for his test f: he failed the test The form of the argument is s ∨ f ¬f ∴ s The hypotheses s ∨ f and ¬f are both true only in the second line of the truth table below. The conclusion s is true in the second row. s f s ∨ f ¬f T T T F T F T T F T T F F F F T

Indicate whether the argument is valid or invalid. For valid arguments, prove that the argument is valid using a truth table. For invalid arguments give truth values for the variables showing that the argument is not valid. (p ∨ q) → r ∴ (p ∧ q) → r

Valid. For every row in which (p ∨ q) → r is true (rows 1, 3, 5, 7, and 8), (p ∧ q) → r is also true. p q r (p ∨ q) → r (p ∧ q) → r T T T T T T T F F F T F T T T T F F F T F T T T T F T F F T F F T T T F F F T T

Indicate whether the argument is valid or invalid. For valid arguments, prove that the argument is valid using a truth table. For invalid arguments give truth values for the variables showing that the argument is not valid. p ↔ q p ∨ q ∴ p

Valid. The only row in which the hypotheses p ↔ q and p ∨ q are both true is the first row, and the conclusion p is true in the first row. p q p ↔ q p ∨ q T T T T T F F T F T F T F F T F

Universal

c is an arbitrary element P(c) ∴ ∀x P(x) example Let c be an arbitrary integer. c ≤ c2 Therefore, every integer is less than or equal to its square.

Existential generalization

c is an element (arbitrary or particular) P(c) ∴ ∃x P(x) example Sam is a particular student in the class. Sam completed the assignment. Therefore, there is a student in the class who completed the assignment.

Universal instantiation

c is an element (arbitrary or particular) ∀x P(x) ∴ P(c) example Sam is a student in the class. Every student in the class completed the assignment. Therefore, Sam completed his assignment.

Prove that each argument is valid by replacing each proposition with a variable to obtain the form of the argument. Then use the rules of inference to prove that the form is valid. If it was not foggy or it didn't rain (or both), then the race was held and there was a trophy ceremony. The trophy ceremony was not held. ∴ It rained.

f: it was foggy r: it rained h: the race was held t: the trophy ceremony was held The form of the argument is (¬f ∨ ¬r) → (h ∧ t) ¬t ∴ r 1. ¬t Hypothesis 2. ¬t ∨ ¬h Addition, 1 3. ¬(t ∧ h) De Morgan's law, 2 4. ¬(h ∧ t) Commutative law, 3 5. (¬f ∨ ¬r) → (h ∧ t) Hypothesis 6. ¬(¬f ∨ ¬r) Modus tollens, 4, 5 7. ¬¬f ∧ ¬¬r De Morgan's law, 6 8. ¬¬r Simplification, 7 9. r Double negation, 8

argument

is a sequence of propositions, called hypotheses

The validity of an argument can be established by applying the rules of inference and laws of propositional logic in a

logical proof

Addition

p ∴ p ∨ q

Modus ponens

p p → q ∴ q

Conjunction

p q ∴ p ∧ q

Hypothetical syllogism

p → q q → r ∴ p → r

Simplification

p ∧ q ∴ p

Disjunctive syllogism

p ∨ q ¬p ∴ q

Resolution

p ∨ q ¬p ∨ r ∴ q ∨ r

Modus tollens

¬q p → q ∴ ¬p

Translate each of the following English statements into logical expressions. The domain of discourse is the set of all real numbers. There is no smallest number.

¬∃x ∀y (x ≤ y)

The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some point in time. Give a logical expression equivalent to the following English statements. You can assume that it is possible for a person to beat himself or herself. Josephine has beaten everyone else.

∀x ((x ≠ Josephine) → B(Josephine, x))

Translate each of the following English statements into logical expressions. The domain of discourse is the set of all real numbers. The ratio of every two positive numbers is also positive.

∀x ∀y (((x > 0) ∧ (y > 0)) → (x/y > 0))

The domain for the variables x and y are the set of musicians in an orchestra. The predicates S, B, and P are defined as: S(x): x plays a string instrument B(x): x plays a brass instrument P(x, y): x practices more than y Give a quantified expression that is equivalent to the following English statements: All the string players practice more than all the brass players.

∀x ∀y ((S(x) ∧ B(y)) → P(x, y))

Translate each of the following English statements into logical expressions. The domain of discourse is the set of all real numbers. Every number besides 0 has a unique multiplicative inverse.

∀x ∃y (((x ≠ 0) → (xy = 1)) ∧ ∀z ((z ≠ y) → (xz ≠ 1))) which is equivalent to ∀x ∃y ∀z (((x ≠ 0) → (xy = 1)) ∧ ((z ≠ y) → (xz ≠ 1)))

The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some point in time. Give a logical expression equivalent to the following English statements. You can assume that it is possible for a person to beat himself or herself. Everyone has won at least one game.

∀x ∃y B(x, y)

The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some point in time. Give a logical expression equivalent to the following English statements. You can assume that it is possible for a person to beat himself or herself. Everyone has been beaten before.

∀x ∃y B(y, x) which is the same as ∀y ∃x B(x, y)

The domain for the first input variable to predicate T is a set of students at a university. The domain for the second input variable to predicate T is the set of Math classes offered at that university. The predicate T(x, y) indicates that student x has taken class y. Sam is a student at the university and Math 101 is one of the courses offered at the university. Give a logical expression for each sentence. Every student has taken at least one math class.

∀x ∃y T(x, y)

The domain for the variables x and y are the set of musicians in an orchestra. The predicates S, B, and P are defined as: S(x): x plays a string instrument B(x): x plays a brass instrument P(x, y): x practices more than y Give a quantified expression that is equivalent to the following English statements: Sam practices more than anyone else in the orchestra.

∀y ((y ≠ Sam) → P(Sam, y))

The domain for the variables x and y are the set of musicians in an orchestra. The predicates S, B, and P are defined as: S(x): x plays a string instrument B(x): x plays a brass instrument P(x, y): x practices more than y Give a quantified expression that is equivalent to the following English statements: Someone in the orchestra plays a string instrument and a brass instrument.

∃x (B(x) ∧ S(x))

Existential instantiation* *Note: each use of Existential instantiation must define a new element with its own name (e.g., "c" or "d").

∃x P(x) ∴ (c is a particular element) ∧ P(c) example There is an integer that is equal to its square. Therefore, c2 = c, for some integer c.

The domain for the first input variable to predicate T is a set of students at a university. The domain for the second input variable to predicate T is the set of Math classes offered at that university. The predicate T(x, y) indicates that student x has taken class y. Sam is a student at the university and Math 101 is one of the courses offered at the university. Give a logical expression for each sentence. There is a student who has taken every math class besides Math 101.

∃x ∀y ((y ≠ Math 101) → T(x, y))

The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some point in time. Give a logical expression equivalent to the following English statements. You can assume that it is possible for a person to beat himself or herself. There are at least two members who have never been beaten.

∃x ∃y ((x ≠ y) ∧ ∀z (¬B(z, x) ∧ ¬B(z, y))) or ∃x ∃y ∀z ((x ≠ y) ∧ (¬B(z, x) ∧ ¬B(z, y)))

The domain for the first input variable to predicate T is a set of students at a university. The domain for the second input variable to predicate T is the set of Math classes offered at that university. The predicate T(x, y) indicates that student x has taken class y. Sam is a student at the university and Math 101 is one of the courses offered at the university. Give a logical expression for each sentence. Sam has taken exactly two math classes.

∃y ∃z ∀w ((z ≠ y) ∧ T(Sam, y) ∧ T(Sam, z) ∧ ((w ≠ y ∧ w ≠ z) → ¬T(Sam, w)))