Donna Kirk

Your Turn

2.1

1.

A logical statement is a sentence that can be identified as being true or false. This sentence is a logical statement because it can be identified as false.

Logical statement, false.

2.

A logical statement is a sentence that can be identified as being true or false. This sentence is a logical statement because it can be identified as true.

Logical statement, true.

3.

A logical statement is a sentence that can be identified as being true or false. This sentence is a not a logical statement because questions cannot be identified as either true or false.

Not a logical statement, questions cannot be determined to be either true or false.

2.2

1.

Choose a lowercase letter. The most common choices are $p$ , $q$ , and $r$ . You are not being asked to determine the truth value.

$p$ : The movie Gandhi won the Academy Award for Best Picture in 1982.

p{\text{:}}

The movie Gandhi won the Academy Award for Best Picture in 1982.

2.

Choose a lowercase letter. The most common choices are $p$ , $q$ , and $r$ . You are not being asked to determine the truth value.

$q$ : Soccer is the most popular sport in the world.

q{\text{:}}

Soccer is the most popular sport in the world.

3.

Choose a lowercase letter. The most common choices are $p$ , $q$ , and $r$ . You are not being asked to determine the truth value.

$r$ : All oranges are citrus fruits.

r{\text{:}}

All oranges are citrus fruits.

2.3

1.

The negation of a logical statement has the opposite truth value of the original statement.

Original statement: Ted Cruz was not born in Texas.

Remove “not.”

Negation: Ted Cruz was born in Texas.

Ted Cruz was born in Texas.

2.

The negation of a logical statement has the opposite truth value of the original statement.

Original statement: Adele has a lovely singing voice.

Add “not.”

Negation: Adele does not have a lovely singing voice.

Adele does not have a beautiful voice.

3.

The negation of a logical statement has the opposite truth value of the original statement.

Original statement: Leaves convert carbon dioxide to oxygen during the process of photosynthesis.

Add “not.”

Negation: Leaves do not convert carbon dioxide to oxygen during the process of photosynthesis.

Leaves do not convert carbon dioxide to oxygen during the process of photosynthesis.

2.4

1.

The negation of a logical statement has the opposite truth value of the original statement.

Original statement: ${\sim}p$

To negate a negative statement, remove the tilde, since the opposite of a negative is a positive.

Negation: ${\sim}({\sim}p)=p$

p

2.

The negation of a logical statement has the opposite truth value of the original statement.

Original statement: $q$

To negate an affirmative statement, add a tilde.

Negation: ${\sim}q$

\text~q

3.

The negation of a logical statement has the opposite truth value of the original statement.

Original statement: $r$

To negate an affirmative statement, add a tilde.

Negation: ${\sim}r$

\text~r

2.5

1.

“Wonder Woman is stronger than Captain Marvel” is the negation of “Wonder Woman is not stronger than Captain Marvel.”

Thus, symbolically you write ${\sim}({\sim}p)=p$ .

p

2.

The statement $r$ represents “Woody and Buzz Lightyear are best friends.”

The negation ${\sim}r$ is translated “Woody and Buzz Lightyear are not best friends.”

Woody and Buzz Lightyear are not best friends.

2.6

1.

It would be incorrect to say that the sum of all consecutive integers results in a prime number because 0 and

1 = 1

, which is not prime. You can say that “The sum of some consecutive integers results in a prime number.”

The sum of some consecutive integers results in a prime number.

2.

Based on the premises provided, no birds give live birth to their young. This statement, however, is false, as some birds do give birth to live young.

No birds give live birth to their young.

3.

All squares are rectangles. All rectangles are parallelograms, so a square is a parallelogram. All parallelograms have four sides, so a square has four sides. Thus, all squares have four sides.

You could draw a Venn diagram that shows this. Squares are a subset of rectangles. Rectangles are a subset of parallelograms. Parallelograms are a subset of things with four sides. Thus, squares are a subset of things with four sides.

All squares are parallelograms and have four sides.

2.7

1.

You negate a “Some…not” statement by replacing “Some…not” with “All.”

Original statement: Some apples are not sweet.

Negation: All apples are sweet.

All apples are sweet.

2.

You negate a “No” statement by replacing “No” with “Some.”

Original statement: No triangles are squares.

Negation: Some triangles are squares.

Some triangles are squares.

3.

You negate a “some” statement by replacing “Some” with “No.”

Original statement: Some vegetables are green.

Negation: No vegetables are green.

No vegetables are green.

2.8

1.

Negation;

\sim

.

2.

Conjunction;

\wedge

.

3.

Biconditional;

\leftrightarrow

.

2.9

1.

Today we will go skiing, but last night it did not snow.

Both “and” and “but” use the conjunction connective.

The given statements:

$p$ : Last night it snowed.

$q$ : Today we will go skiing.

Break down the compound statement into its components.

Today we will go skiing,	but	not	last night it snowed
$q$	$\wedge$	$\sim$	$p$

$q \wedge {\text{~}}p$

q \wedge \text{~}p

2.

q \leftrightarrow p

Today we will go skiing if and only if it snowed last night.

The given statements:

$p$ : Last night it snowed.

$q$ : Today we will go skiing.

Break down the compound statement into its components.

Today we will go skiing	if and only if	it snowed last night
$q$	$\leftrightarrow$	$p$

$q$ $\leftrightarrow$ $p$

3.

p \vee \text{~}q

Last night it snowed or today we will not go skiing.

The given statements:

$p$ : Last night it snowed.

$q$ : Today we will go skiing.

Break down the compound statement into its components.

Last night it snowed	or	not	today we will go skiing
$p$	$\vee$	$\sim$	$q$

$p \vee {\text{~}}q$

4.

p \to q

If it snowed last night, then today we will go skiing.

The given statements:

$p$ : Last night it snowed.

$q$ : Today we will go skiing.

Break down the compound statement into its components.

Last night it snowed	conditional	today we will go skiing
$p$	$\to$	$q$

$p$ $\to$ $p$

2.10

1.

$\text{~}r \to (p \vee q)$

Given statements:

$p$ : My roommates ordered pizza.

$q$ : I ordered wings.

$r$ : Our friends came over to watch the game.

Translate ${\text{~}}r$ as :Our friends did not come over to watch the game.

$\to$ is the conditional, which will be translated “if…then.”

$\vee$ is the inclusive or.

The comma before “then” shows that the statements in parentheses belong together.

If our friends did not come over to watch the game, then my roommates ordered pizza or I ordered wings.

2.

$(p \wedge q) \to r$

$\wedge$ is the conjunction, which will be translated “and.”

$\to$ is the conditional, which will be translated “if…then.”

The comma before “then” shows that the statements in parentheses belong together.

Given statements:

$p$ : My roommates ordered pizza.

$q$ : I ordered wings.

$r$ : Our friends came over to watch the game.

If my roommates ordered pizza and I ordered wings, then our friends came over to watch the game.

3.

${\text{~}}(p \vee r)$

$\sim$ is the negation symbol. Replace the symbol with “It is not the case that…”

$\vee$ is the inclusive or.

Given statements:

$p$ : My roommates ordered pizza.

$q$ : I ordered wings.

$r$ : Our friends came over to watch the game.

It is not the case that my roommates ordered pizza or our friends came over to watch the game.

2.11

1.

((p \vee q) \wedge (\text{~}r))

The order of logical operations is

Parentheses
Negation
Disjunction, conjunction
Conditional
Biconditional

For same order operations, work left to right.

The highest order operation in this expression is negation, so add parentheses around the negation.

$p \vee q\; \wedge \left( {{\text{~}}r} \right)$

Disjunction and conjunction are the same order, so work left to right. First evaluate the exclusive or.

$(p \vee q) \wedge \left( {{\text{~}}r} \right)$

Then finish by evaluating the conjunction.

$((p \vee q) \wedge \left( {{\text{~}}r} \right))$

2.

((\text{~}p) \to (q \vee r))

The order of logical operations is

Parentheses
Negation
Disjunction, conjunction
Conditional
Biconditional

For same order operations, work left to right.

${\text{~}}p{\text{ }} \to {\text{ }}q \vee r$

The highest order operation in this expression is negation, so add parentheses around the negation.

$\left( {{\text{~}}p} \right) \to {\text{ }}q \vee r$

The next highest order operation is the disjunction. Add parentheses around the disjunction.

$\left( {{\text{~}}p} \right) \to (q \vee r)$

Finally, you can evaluate the conditional.

$(\left( {{\text{~}}p} \right) \to (q \vee r))$

3.

((\text{~}p) \vee (\text{~}q)) \leftrightarrow (\text{~}(p \wedge q))

; this is another example of De Morgan’s Laws and it is always true.

The order of logical operations is

Parentheses
Negation
Disjunction, conjunction
Conditional
Biconditional

For same order operations, work left to right.

${\text{~}}p \vee {\text{~}}q{\text{ }} \leftrightarrow {\text{~}}(p \wedge q)$

The right side of the biconditional already has parentheses. The next step in the order of operation is negation, so add parentheses around the three negations.

$\left( {{\text{~}}p} \right) \vee ({\text{~}}q) \leftrightarrow ({\text{~}}(p \wedge q))$

The next step in the order of operations is disjunction/conjunction. Add parentheses around the disjunction.

$(\left( {{\text{~}}p} \right) \vee ({\text{~}}q)) \leftrightarrow ({\text{~}}(p \wedge q))$

The lowest order operation is the biconditional. Finish by evaluating the biconditional.

$((\left( {{\text{~}}p} \right) \vee ({\text{~}}q)) \leftrightarrow ({\text{~}}(p \wedge q)))$

2.12

1.

The negation is $p{\text :}\ 3 \times 5 \ne 14$

Since $3 \times 5 = 15$ , the negation is true.

p{\text{:}}\,3 \times 5 \ne 14

, true

2.

The negation is $q{\text :}$ No house are built with bricks.

The negation is false since some houses are built of bricks.

q

: No houses are built with bricks; false

3.

The negation is ${\sim}r{\text :}$ Abuja is not the capital of Nigeria.

The negation is a false statement since Abuja is the capital of Nigeria.

\text~r

: Abuja is not the capital of Nigeria; false

2.13

1.

Given statements:

$p$ : Yellow is a primary color.

$p$ is true.

$q$ : Blue is a primary color.

$q$ is true.

The $\wedge$ symbol represents the conjunction and is translated “and” or “but.”

A conjunction is only true when both statements are true.

$p \wedge q$

${\rm True} \wedge {\rm True}={\rm True}$

True

2.

Given statements:

$p$ : Yellow is a primary color.

$p$ is true.

$r$ : Green is a primary color.

$r$ is false.

${\sim}r$ is true.

${\sim}r \wedge p$

A conjunction is only true when both statements are true.

$\rm True \wedge True = True$

False

3.

Given statements:

$q$ : Blue is a primary color.

$q$ is true.

$r$ : Green is a primary color.

$r$ is false.

$q \wedge r$

A conjunction is only true when both statements are true.

$\rm True \wedge False = False$

True

2.14

1.

Given statements:

$p$ : Yellow is a primary color.

$p$ is true.

$q$ : Blue is a primary color.

$q$ is true.

$p \vee q$

A disjunction is only false when both statements are false.

$\rm True \vee True = True$

True

2.

Given statements:

$p$ : Yellow is a primary color.

$p$ is true.

${\sim}p$ is false.

$r$ : Green is a primary color.

$r$ is false.

${\sim}p \vee r$

A disjunction is only false when both statements are false.

$\rm False \vee False = False$

False

3.

Given statements:

$q$ : Blue is a primary color.

$q$ is true.

$r$ : Green is a primary color.

$r$ is false.

$q \vee r$

A disjunction is only false when both statements are false.

$\rm True \vee False = True$

True

2.15

1.

Given statements:

$p$ : Yellow is a primary color.

$p$ is true.

$q$ : Blue is a primary color.

$q$ is true.

$r$ : Green is a primary color.

$r$ is false.

$p$	$q$	$r$	${\sim}q$	${\sim}q \wedge p$	$(q \wedge p) \vee r$
T	T	F	F	F	F

$p$	$q$	$r$	$\text{~}q$	$\text{~}q \wedge p$	$(\text{~}q \wedge p) \vee r$
T	T	F	F	F	F

false

2.

Given statements:

$p$ : Yellow is a primary color.

$p$ is true.

$q$ : Blue is a primary color.

$q$ is true.

$r$ : Green is a primary color.

$r$ is false.

$p$	$q$	$r$	${\sim}r$	$p \vee q$	$(p \vee q) \wedge {\sim}r$
T	T	F	T	T	T

$p$	$q$	$r$	$\text{~}r$	$p \vee q$	$(p \vee q) \wedge (\text{~}r)$
T	T	F	T	T	T

true

3.

Given statements:

$p$ : Yellow is a primary color.

$p$ is true.

$q$ : Blue is a primary color.

$q$ is true.

$r$ : Green is a primary color.

$r$ is false.

$p$	$q$	$r$	$p \wedge r$	${\sim}(p \wedge r$ )	${\sim}(p \wedge r)\wedge q$
T	T	F	F	T	T

$p$	$q$	$r$	$(p \wedge r)$	$\text{~}(p \wedge r)$	$\text{~}(p \wedge r) \wedge q$
T	T	F	F	T	T

true

2.16

1.

$p$	$q$	${\sim}q$	$p \wedge {\sim}q$
T	T	F	F
T	F	T	T
F	T	F	F
F	F	T	F

$p$	$q$	$\text~q$	$p \wedge \text{~}q$
T	T	F	F
T	F	T	T
F	T	F	F
F	F	T	F

2.

$p$	$q$	$p \vee q$	${\sim}(p \vee q)$
T	T	T	F
T	F	T	F
F	T	T	F
F	F	F	T

$p$	$q$	$p \vee q$	$\text{~}(p \vee q)$
T	T	T	F
T	F	T	F
F	T	T	F
F	F	F	T

3.

$p$	$q$	$r$	${\sim}q$	$p \wedge {\sim}q$	$(p \wedge {\sim}q) \vee r$
T	T	T	F	F	T
T	T	F	F	F	F
T	F	T	T	T	T
T	F	F	T	T	T
F	T	T	F	F	T
F	T	F	F	F	F
F	F	T	T	F	T
F	F	F	T	F	F

$p$	$q$	$r$	$\text{~}q$	$p\, \wedge \text{~}q$	$(p \wedge \text{~}q) \vee r$
T	T	T	F	F	T
T	T	F	F	F	F
T	F	T	T	T	T
T	F	F	T	T	T
F	T	T	F	F	T
F	T	F	F	F	F
F	F	T	T	F	T
F	F	F	T	F	F

2.17

1.

$p$	${\sim}p$	$p \vee {\sim}p$
T	F	T
F	T	T

Valid

$p$	$\text{~}{p}$	${p} \vee \text{~}{p}$
T	F	T
F	T	T

2.

$p$	$q$	${\sim}p$	${\sim}q$	${\sim}p \vee {\sim}q$
T	T	F	F	F
T	F	F	T	T
F	T	T	F	T
F	F	T	T	T

Not valid

$p$	$q$	$\text{~}{p}$	$\text{~}{q}$	$\text{~}p \vee ~q$
T	T	F	F	F
T	F	F	T	T
F	T	T	F	T
F	F	T	T	T

2.18

1.

False

$p$	$q$	${p} \to {q}$
T	F	F

Since you are told to assume $p$ is true and $q$ is false, you only need one row of possible truth values in your truth table.

$p$	$q$	$p \to q$
T	F

$p \to q$

The conditional is only false when the hypothesis is true, and the conclusion is false. Using the provided truth values, you have.

If True, then False

The result is False, so write F in the last column.

$p$	$q$	$p \to q$
T	F	F

2.

True

$p$	$q$	$\text{~}{q}$	${p} \to \text{~}{q}$
T	F	T	T

Since you are told to assume $p$ is true and $q$ is false, you only need one row of possible truth values in your truth table.

$p$	$q$	${\text{~}}q$	$p \to {\text{~}}q$
T	F

${\text{~}}q$

The negation of false is true, so write T under ${\text{~}}q$ .

$p$	$q$	${\text{~}}q$	$p \to {\text{~}}q$
T	F	T

$p \to {\text{~}}q$

The conditional is only false when the hypothesis is true, and the conclusion is false. Using $p$ and the previous column, you have.

If True, then True

The result is True, so write T in the last column.

$p$	$q$	${\text{~}}q$	$p \to {\text{~}}q$
T	F	T	T

3.

Since you are told to assume $p$ is true and $q$ is false, you only need one row of possible truth values in your truth table.

$p$	$q$	${\text{~}}p$	${\text{~}}p \to q$
T	F

${\text{~}}p$

The negation of true is false, so write F under ${\text{~}}p$ .

$p$	$q$	${\text{~}}p$	${\text{~}}p \to q$
T	F	F

${\text{~}}p \to q$

The conditional is only false when the hypothesis is true, and the conclusion is false. Using the previous two columns, you have.

If False, then False

The result is True, so write T in the last column.

$p$	$q$	${\text{~}}p$	${\text{~}}p \to q$
T	F	F	T

True

$p$	$q$	$\text{~}{p}$	$\text{~}{p} \to \text{~}q$
T	F	F	T

2.19

1.

Valid

$p$	$q$	$\text{~}{p}$	$\text{~}{p} \vee {q}$	${q} \to \left( {\text{~}{p} \vee {q}} \right)$
T	T	F	T	T
T	F	F	F	T
F	T	T	T	T
F	F	T	T	T

Since there are two statements, you need four rows for all possible combinations.

$p$	$q$	${\text{~}}p$	${\text{~}}p \vee q$	$q \to {\text{ }}({\text{~}}p \vee q)$
T	T
T	F
F	T
F	F

${\text{~}}p$

The negation of a statement has the opposite truth value. Write the opposite value of $p$ in this column.

$p$	$q$	${\text{~}}p$	${\text{~}}p \vee q$	$q \to {\text{ }}({\text{~}}p \vee q)$
T	T	F
T	F	F
F	T	T
F	F	T

${\text{~}}p \vee q$

A disjunction is only false when both statements are false. Write F if both the ${\text{~}}p$ column and $q$ column have an F. Otherwise, write T.

$p$	$q$	${\text{~}}p$	${\text{~}}p \vee q$	$q \to {\text{ }}({\text{~}}p \vee q)$
T	T	F	T
T	F	F	F
F	T	T	T
F	F	T	T

$q \to {\text{ }}({\text{~}}p \vee q)$

A conditional is only false when the premise is true, and the conclusion is false. Write F if the $q$ column is T and the previous column is F. Otherwise, write T.

The statement is valid because the last column is always true.

$p$	$q$	${\text{~}}p$	${\text{~}}p \vee q$	$q \to {\text{ }}({\text{~}}p \vee q)$
T	T	F	T	T
T	F	F	F	T
F	T	T	T	T
F	F	T	T	T

2.

Not valid

$p$	$q$	$\text{~}{p}$	${q} \wedge {p}$	$\text{~}{p} \to \left( {{q} \wedge {p}} \right)$
T	T	F	T	T
T	F	F	F	T
F	T	T	F	F
F	F	T	F	F

Since there are two statements, you need four rows for all possible combinations.

$p$	$q$	${\text{~}}p$	$q \wedge p$	${\text{~}}p \to {\text{ }}(q \wedge p)$
T	T
T	F
F	T
F	F

${\text{~}}p$

The negation of a statement has the opposite truth value. Write the opposite value of $p$ in this column.

$p$	$q$	${\text{~}}p$	$q \wedge p$	${\text{~}}p \to {\text{ }}(q \wedge p)$
T	T	F
T	F	F
F	T	T
F	F	T

$q \wedge p$

A conjunction is true only when both statements are true. Otherwise, it is false. Write T if both the $q$ and $p$ columns have a T. Otherwise, write an F.

$p$	$q$	${\text{~}}p$	$q \wedge p$	${\text{~}}p \to {\text{ }}(q \wedge p)$
T	T	F	T
T	F	F	F
F	T	T	F
F	F	T	F

${\text{~}}p \to {\text{ }}(q \wedge p)$

A conditional is only false when the premise is true, and the conclusion is false. Write F if the ${\text{~}}p$ column has an T and the $q \wedge p$ column has an F. Otherwise, write T.

$p$	$q$	${\text{~}}p$	$q \wedge p$	${\text{~}}p \to {\text{ }}(q \wedge p)$
T	T	F	T	T
T	F	F	F	T
F	T	T	F	F
F	F	T	F	F

The statement is not valid because the last column is not all true.

2.20

1.

Since you are told to assume $p$ is true and $q$ is false, you only need one row of possible truth values in your truth table.

$p$	$q$	$p \leftrightarrow q$
T	F

$p \leftrightarrow q$

A biconditional is true when the two statements have the same truth value. Otherwise, it is false.

Since $p$ and $q$ have different truth values, the biconditional is false. Write an F in the last column.

$p$	$q$	$p \leftrightarrow q$
T	F	F

The statement is false because there is an F in the last column.

False

$p$	$q$	${p} \leftrightarrow {q}$
T	F	F

2.

Since you are told to assume $p$ is true and $q$ is false, you only need one row of possible truth values in your truth table.

$p$	$q$	${\text{~}}q$	$p \leftrightarrow {\text{~}}q$
T	F

${\text{~}}q$

Since $q$ is false, ${\text{~}}q$ is true. Write T under ${\text{~}}q$ .

$p$	$q$	${\text{~}}q$	$p \leftrightarrow {\text{~}}q$
T	F	T

$p \leftrightarrow {\text{~}}q$

A biconditional is true when the two statements have the same truth value. Otherwise, it is false.

Since $p$ and ${\text{~}}q$ are both true, the biconditional is true. Write a T in the last column.

$p$	$q$	${\text{~}}q$	$p \leftrightarrow {\text{~}}q$
T	F	T	T

The statement is true because there is a T in the last column.

True

$p$	$q$	$\text{~}{q}$	${p} \leftrightarrow \text{~}{q}$
T	F	T	T

3.

Since you are told to assume $p$ is true and $q$ is false, you only need one row of possible truth values in your truth table.

$p$	$q$	${\text{~}}p$	${\text{~}}p \leftrightarrow q$
T	F

${\text{~}}p$

Since $p$ is true, ${\text{~}}p$ is false. Write F under ${\text{~}}p$ .

$p$	$q$	${\text{~}}p$	${\text{~}}p \leftrightarrow q$
T	F	F

${\text{~}}p \leftrightarrow q$

A biconditional is true only when the statements have the same truth value. Otherwise, it is false. The premise is false, ${\text{~}}p$ and $q$ have the same truth value, so write T in the last column.

$p$	$q$	${\text{~}}p$	${\text{~}}p \leftrightarrow q$
T	F	F	T

The statement is true because there is a T in the last column.

True

$p$	$q$	$\text{~}{p}$	$\text{~}{p} \leftrightarrow {q}$
T	F	F	T

2.21

1.

Since there are two statements, you need four rows for all possible combinations.

$p$	$q$	$p \wedge q$	${\text{~}}(p \wedge q)$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	${\text{~}}(p \wedge q) \leftrightarrow {\text{ }}({\text{~}}p \vee {\text{~}}q)$
T	T
T	F
F	T
F	F

$p \wedge q$

A conjunction is true only when both statements are true. Write T when both $p$ and $q$ are true. Otherwise, write F.

$p$	$q$	$p \wedge q$	${\text{~}}(p \wedge q)$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	${\text{~}}(p \wedge q) \leftrightarrow {\text{ }}({\text{~}}p \vee {\text{~}}q)$
T	T	T
T	F	F
F	T	F
F	F	F

${\text{~}}(p \wedge q)$

The negation of a statement has the opposite truth value. Write the opposite of $p \wedge q$ in this column.

$p$	$q$	$p \wedge q$	${\text{~}}(p \wedge q)$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	${\text{~}}(p \wedge q) \leftrightarrow {\text{ }}({\text{~}}p \vee {\text{~}}q)$
T	T	T	F
T	F	F	T
F	T	F	T
F	F	F	T

${\text{~}}p$

The negation of a statement has the opposite truth value. Write the opposite of $p$ in this column.

$p$	$q$	$p \wedge q$	${\text{~}}(p \wedge q)$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	${\text{~}}(p \wedge q) \leftrightarrow {\text{ }}({\text{~}}p \vee {\text{~}}q)$
T	T	T	F	F
T	F	F	T	F
F	T	F	T	T
F	F	F	T	T

${\text{~}}q$

Write the opposite of $q$ in this column.

$p$	$q$	$p \wedge q$	${\text{~}}(p \wedge q)$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	${\text{~}}(p \wedge q) \leftrightarrow {\text{ }}({\text{~}}p \vee {\text{~}}q)$
T	T	T	F	F	F
T	F	F	T	F	T
F	T	F	T	T	F
F	F	F	T	T	T

${\text{~}}p \vee {\text{~}}q$

A disjunction is only false when both statements are false. Write an F where both ${\text{~}}p$ and ${\text{~}}q$ are false. Otherwise, write T.

$p$	$q$	$p \wedge q$	${\text{~}}(p \wedge q)$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	${\text{~}}(p \wedge q) \leftrightarrow {\text{ }}({\text{~}}p \vee {\text{~}}q)$
T	T	T	F	F	F	F
T	F	F	T	F	T	T
F	T	F	T	T	F	T
F	F	F	T	T	T	T

${\text{~}}(p \wedge q) \leftrightarrow {\text{ }}({\text{~}}p \vee {\text{~}}q)$

A biconditional is true only when both statements have the same truth value. Write T where ${\text{~}}(p \wedge q)$ and ${\text{~}}p \vee {\text{~}}q$ have the same truth value. Write F where they do not.

$p$	$q$	$p \wedge q$	${\text{~}}(p \wedge q)$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	${\text{~}}(p \wedge q) \leftrightarrow {\text{ }}({\text{~}}p \vee {\text{~}}q)$
T	T	T	F	F	F	F	T
T	F	F	T	F	T	T	T
F	T	F	T	T	F	T	T
F	F	F	T	T	T	T	T

This statement is valid since the last column is always true.

Valid

$p$	$q$	${p} \wedge {q}$	$\text{~}\left( {{p} \wedge {q}} \right)$	$\text{~}{p}$	$\text{~}{q}$	$\text{~}{p} \vee \text{~}{q}$	$\text{~}\left( {{p} \wedge {q}} \right) \leftrightarrow \left( {\text{~}{p} \vee \text{~}{q}} \right)$
T	T	T	F	F	F	F	T
T	F	F	T	F	T	T	T
F	T	F	T	T	F	T	T
F	F	F	T	T	T	T	T

2.

Since there are two statements, you need four rows for all possible combinations.

$p$	$q$	${\text{~}}p$	$q \wedge p$	${\text{~}}p \leftrightarrow {\text{ }}(q \wedge p)$
T	T
T	F
F	T
F	F

${\text{~}}p$

Write the opposite of $p$ in this column.

$p$	$q$	${\text{~}}p$	$q \wedge p$	${\text{~}}p \leftrightarrow {\text{ }}(q \wedge p)$
T	T	F
T	F	F
F	T	T
F	F	T

$q \wedge p$

A conjunction is true only when both statements are true. Write a T if both $q$ and $p$ are true. Otherwise, write an F.

$p$	$q$	${\text{~}}p$	$q \wedge p$	${\text{~}}p \leftrightarrow {\text{ }}(q \wedge p)$
T	T	F	T
T	F	F	F
F	T	T	F
F	F	T	F

${\text{~}}p \leftrightarrow {\text{ }}(q \wedge p)$

A biconditional is true only when both statements have the same truth value. Write T where ${\text{~}}p$ and $q \wedge p$ have the same truth value. Write F where they do not.

$p$	$q$	${\text{~}}p$	$q \wedge p$	${\text{~}}p \leftrightarrow {\text{ }}(q \wedge p)$
T	T	F	T	F
T	F	F	F	T
F	T	T	F	F
F	F	T	F	F

The statement is not valid since the last column is not always true.

Not Valid

$p$	$q$	$\text{~}{p}$	${q} \wedge {p}$	$\text{~}{p} \leftrightarrow \left( {{q} \wedge {p}} \right)$
T	T	F	T	F
T	F	F	F	T
F	T	T	F	F
F	F	T	F	F

3.

Valid

$p$	$q$	${p} \to {q}$	$\text{~}{p}$	$\text{~}{p} \vee {q}$	$\left( {p \to q} \right) \leftrightarrow \left( {\text{~}p \vee q} \right)$
T	T	T	F	T	T
T	F	F	F	F	T
F	T	T	T	T	T
F	F	T	T	T	T

Since there are two statements, you need four rows for all possible combinations.

$p$	$q$	$p \to q$	${\text{~}}p$	${\text{~}}p \vee q$	$(p \to q) \leftrightarrow {\text{ }}({\text{~}}p \vee q)$
T	T
T	F
F	T
F	F

$p \to q$

A conditional is only false when the premise is true and the hypotheses is false. Write an F if $p$ is true and $q$ is false. Otherwise, write a T.

$p$	$q$	$p \to q$	${\text{~}}p$	${\text{~}}p \vee q$	$(p \to q) \leftrightarrow {\text{ }}({\text{~}}p \vee q)$
T	T	T
T	F	F
F	T	T
F	F	T

${\text{~}}p$

Write the opposite of $p$ in this column.

$p$	$q$	$p \to q$	${\text{~}}p$	${\text{~}}p \vee q$	$(p \to q) \leftrightarrow {\text{ }}({\text{~}}p \vee q)$
T	T	T	F
T	F	F	F
F	T	T	T
F	F	T	T

${\text{~}}p \vee q$

A disjunction is only false when both statements are false. Write an F when both ${\text{~}}p$ and $q$ are false. Otherwise, write a T.

$p$	$q$	$p \to q$	${\text{~}}p$	${\text{~}}p \vee q$	$(p \to q) \leftrightarrow {\text{ }}({\text{~}}p \vee q)$
T	T	T	F	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

$(p \to q) \leftrightarrow {\text{ }}({\text{~}}p \vee q)$

A biconditional is true only when both statements have the same truth value. Write a T where both $p \to q$ and ${\text{~}}p \vee q$ are true. Otherwise, write an F.

$p$	$q$	$p \to q$	${\text{~}}p$	${\text{~}}p \vee q$	$(p \to q) \leftrightarrow {\text{ }}({\text{~}}p \vee q)$
T	T	T	F	T	T
T	F	F	F	F	T
F	T	T	T	T	T
F	F	T	T	T	T

The statement is valid since the last column is always true.

4.

Since there are three statements, you need eight rows for all possible combinations.

$p$	$q$	$r$	$p \wedge q$	$(p \wedge q) \to r$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	$({\text{~}}p \vee {\text{~}}q) \vee r$	$((p \wedge q) \to r) \leftrightarrow {\text{ }}(({\text{~}}p \vee {\text{~}}q) \vee r)$
T	T	T
T	T	F
T	F	T
T	F	F
F	T	T
F	T	F
F	F	T
F	F	F

$p \wedge q$

A conjunction is true only when both statements are true. Otherwise, it is false. Write a T where both $p$ and $q$ are true. Otherwise, write an F.

$p$	$q$	$r$	$p \wedge q$	$(p \wedge q) \to r$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	$({\text{~}}p \vee {\text{~}}q) \vee r$	$((p \wedge q) \to r) \leftrightarrow {\text{ }}(({\text{~}}p \vee {\text{~}}q) \vee r)$
T	T	T	T
T	T	F	T
T	F	T	F
T	F	F	F
F	T	T	F
F	T	F	F
F	F	T	F
F	F	F	F

$(p \wedge q){\text{ }} \to r$

A conditional is only false when the premise is true and the hypotheses is false. Write an F if $p \wedge q$ is true and $r$ is false. Otherwise, write a T.

$p$	$q$	$r$	$p \wedge q$	$(p \wedge q) \to r$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	$({\text{~}}p \vee {\text{~}}q) \vee r$	$((p \wedge q) \to r) \leftrightarrow {\text{ }}(({\text{~}}p \vee {\text{~}}q) \vee r)$
T	T	T	T	T
T	T	F	T	F
T	F	T	F	T
T	F	F	F	T
F	T	T	F	T
F	T	F	F	T
F	F	T	F	T
F	F	F	F	T

${\text{~}}p$

Write the opposite of $p$ in this column.

$p$	$q$	$r$	$p \wedge q$	$(p \wedge q) \to r$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	$({\text{~}}p \vee {\text{~}}q) \vee r$	$((p \wedge q) \to r) \leftrightarrow {\text{ }}(({\text{~}}p \vee {\text{~}}q) \vee r)$
T	T	T	T	T	F
T	T	F	T	F	F
T	F	T	F	T	F
T	F	F	F	T	F
F	T	T	F	T	T
F	T	F	F	T	T
F	F	T	F	T	T
F	F	F	F	T	T

${\text{~}}q$

Write the opposite of $q$ in this column.

$p$	$q$	$r$	$p \wedge q$	$(p \wedge q) \to r$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	$({\text{~}}p \vee {\text{~}}q) \vee r$	$((p \wedge q) \to r) \leftrightarrow {\text{ }}(({\text{~}}p \vee {\text{~}}q) \vee r)$
T	T	T	T	T	F	F
T	T	F	T	F	F	F
T	F	T	F	T	F	T
T	F	F	F	T	F	T
F	T	T	F	T	T	F
F	T	F	F	T	T	F
F	F	T	F	T	T	T
F	F	F	F	T	T	T

${\text{~}}p \vee {\text{~}}q$

A disjunction is only false when both statements are false. Write an F in this column of both ${\text{~}}p$ and ${\text{~}}q$ are false. Otherwise, write a T.

$p$	$q$	$r$	$p \wedge q$	$(p \wedge q) \to r$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	$({\text{~}}p \vee {\text{~}}q) \vee r$	$((p \wedge q) \to r) \leftrightarrow {\text{ }}(({\text{~}}p \vee {\text{~}}q) \vee r)$
T	T	T	T	T	F	F	F
T	T	F	T	F	F	F	F
T	F	T	F	T	F	T	T
T	F	F	F	T	F	T	T
F	T	T	F	T	T	F	T
F	T	F	F	T	T	F	T
F	F	T	F	T	T	T	T
F	F	F	F	T	T	T	T

$({\text{~}}p \vee {\text{~}}q) \vee r$

A disjunction is only false when both statements are false. Write an F in this column of both ${\text{~}}p \vee {\text{~}}q$ and $r$ are false. Otherwise, write a T.

$p$	$q$	$r$	$p \wedge q$	$(p \wedge q) \to r$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	$({\text{~}}p \vee {\text{~}}q) \vee r$	$((p \wedge q) \to r) \leftrightarrow {\text{ }}(({\text{~}}p \vee {\text{~}}q) \vee r)$
T	T	T	T	T	F	F	F	T
T	T	F	T	F	F	F	F	F
T	F	T	F	T	F	T	T	T
T	F	F	F	T	F	T	T	T
F	T	T	F	T	T	F	T	T
F	T	F	F	T	T	F	T	T
F	F	T	F	T	T	T	T	T
F	F	F	F	T	T	T	T	T

$((p \wedge q){\text{ }} \to r) \leftrightarrow {\text{ }}(({\text{~}}p \vee {\text{~}}q) \vee r)$

A biconditional is true only when both statements have the same truth value. Write T where $(p \wedge q){\text{ }} \to r$ and $({\text{~}}p \vee {\text{~}}q) \vee r$ have the same truth value. Write F where they do not.

$p$	$q$	$r$	$p \wedge q$	$(p \wedge q) \to r$	${\text{~}}p$	${\text{~}}q$	${\text{~}}p \vee {\text{~}}q$	$({\text{~}}p \vee {\text{~}}q) \vee r$	$((p \wedge q) \to r) \leftrightarrow {\text{ }}(({\text{~}}p \vee {\text{~}}q) \vee r)$
T	T	T	T	T	F	F	F	T	T
T	T	F	T	F	F	F	F	F	T
T	F	T	F	T	F	T	T	T	T
T	F	F	F	T	F	T	T	T	T
F	T	T	F	T	T	F	T	T	T
F	T	F	F	T	T	F	T	T	T
F	F	T	F	T	T	T	T	T	T
F	F	F	F	T	T	T	T	T	T

The statement is valid because the last column is always true.

Valid

$p$	$q$	$r$	$\text{~}{p}$	$\text{~}{q}$	${p} \wedge {q}$	$\left( {{p} \wedge {q}} \right) \to {r}$	$\text{~}{p} \vee \text{~}{q}$	$\left( {{\text{~}p} \vee {\text{~}q}} \right) \vee {r}$	$\left( {{p} \wedge {q} \to {r}} \right) \leftrightarrow \left( {{\text{~}p} \vee {\text{~}q} \vee {r}} \right)$
T	T	T	F	F	T	T	F	T	T
T	T	F	F	F	T	F	F	F	T
T	F	T	F	T	F	T	T	T	T
T	F	F	F	T	F	T	T	T	T
F	T	T	T	F	F	T	T	T	T
F	T	F	T	F	F	T	T	T	T
F	F	T	T	T	F	T	T	T	T
F	F	F	T	T	F	T	T	T	T

2.22

1.

Construct a truth table for a biconditional using the first statement as the hypothesis and the second statement as the conclusion.

$(p \rightarrow q) \leftrightarrow (q \rightarrow {\sim}p)$

If the last column in the truth table is true for every entry, the statements are logically equivalent. Since you have two statements, you need four rows of possible truth values.

$p$	$q$	$p \rightarrow q$	${\sim}p$	$q \rightarrow {\sim}p$	$(p \rightarrow q) \leftrightarrow (q \rightarrow {\sim}p)$
T	T
T	F
F	T
F	F

$p \rightarrow q$

A conditional is only false when the premise is true, and the conclusion is false. Otherwise, it is true. Write an F when $p$ is true and $q$ is false. Otherwise, write a T.

$p$	$q$	$p \rightarrow q$	${\sim}p$	$q \rightarrow {\sim}p$	$(p \rightarrow q) \leftrightarrow (q \rightarrow {\sim}p)$
T	T	T
T	F	F
F	T	T
F	F	T

${\sim}p$

Write the opposite of $p$ .

$p$	$q$	$p \rightarrow q$	${\sim}p$	$q \rightarrow {\sim}p$	$(p \rightarrow q) \leftrightarrow (q \rightarrow {\sim}p)$
T	T	T	F
T	F	F	F
F	T	T	T
F	F	T	T

$q \rightarrow {\sim}p$

A conditional is only false when the premise is true, and the conclusion is false. Otherwise, it is true. Write an F when $q$ is true and ${\sim}p$ is false. Otherwise, write a T.

$p$	$q$	$p \rightarrow q$	${\sim}p$	$q \rightarrow {\sim}p$	$(p \rightarrow q) \leftrightarrow (q \rightarrow {\sim}p)$
T	T	T	F	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

$(p \rightarrow q) \leftrightarrow (q \rightarrow {\sim}p)$

A biconditional is true only with both statements have the same truth value. Otherwise, it is false. Write a T if $p \rightarrow q$ and $q \rightarrow {\sim}p$ have the same truth value. Otherwise, write an F.

$p$	$q$	$p \rightarrow q$	${\sim}p$	$q \rightarrow {\sim}p$	$(p \rightarrow q) \leftrightarrow (q \rightarrow {\sim}p)$
T	T	T	F	T	T
T	F	F	F	F	T
F	T	T	T	T	T
F	F	T	T	T	T

The statements are equivalent since the last column is always true.

p \to q

is logically equivalent to

\text{~}q \to\text{~}p.

${p}$	${q}$	${p} \to {q}$	$\text{~}{q}$	$\text{~}{p}$	$\text{~}{q} \to \text{~}{p}$	$\left( {p} \to {q} \right) \leftrightarrow \left( {\text{~}{q} \to \text{~}{p}} \right)$
T	T	T	F	F	T	T
T	F	F	T	F	F	T
F	T	T	F	T	T	T
F	F	T	T	T	T	T

2.

Construct a truth table for a biconditional using the first statement as the hypothesis and the second statement as the conclusion.

$(p \rightarrow q) \leftrightarrow (p \vee {\sim}q)$

If the last column in the truth table is true for every entry, the statements are logically equivalent.

$p$	$q$	$p \rightarrow q$	${\sim}q$	$p \vee {\sim}q$	$(p \rightarrow q) \leftrightarrow (p \vee {\sim}q)$
T	T
T	F
F	T
F	F

$p \rightarrow q$

A conditional is only false when the premise is true, and the conclusion is false. Otherwise, it is true. Write an F when $p$ is true and $q$ is false. Otherwise, write a T.

$p$	$q$	$p \rightarrow q$	${\sim}q$	$p \vee {\sim}q$	$(p \rightarrow q) \leftrightarrow (p \vee ~q)$
T	T	T
T	F	F
F	T	T
F	F	T

${\sim}q$

Write the opposite of $q$ .

$p$	$q$	$p \rightarrow q$	${\sim}q$	$p \vee {\sim}q$	$(p \rightarrow q) \leftrightarrow (p \vee {\sim}q)$
T	T	T	F
T	F	F	T
F	T	T	F
F	F	T	T

$p \vee {\sim}q$

A disjunction is only false when both statements are false. Otherwise, it is true. Write an F if both $p$ and ${\sim}q$ are both false. Otherwise, write a T. The columns are bold to help you compare.

p	$q$	$p \rightarrow q$	${\sim}q$	$p \vee {\sim}q$	$(p \rightarrow q) \leftrightarrow (p \vee {\sim}q)$
T	T	T	F	T
T	F	F	T	T
F	T	T	F	F
F	F	T	T	T

$(p \rightarrow q) \leftrightarrow (p \vee {\sim}q)$

A biconditional is true only if both statements have the same truth value. Otherwise, it is false. Write a T if both $p \rightarrow q$ and $p \vee {\sim}q$ are true. If they are different, write an F. The columns are bold to help you compare.

p	$q$	$p \rightarrow q$	${\sim}q$	$p \vee {\sim}q$	$(p \rightarrow q) \leftrightarrow (p \vee {\sim}q)$
T	T	T	F	T	T
T	F	F	T	T	F
F	T	T	F	F	F
F	F	T	T	T	T

The statements are not equivalent since the last column is not always true.

p \to q

is not logically equivalent to

p \vee \text{~}q.

$p$	$q$	${p} \to {q}$	$\text{~}{q}$	${p}\rm \vee \text{~}{q}$	$\left( {{p} \to {q}} \right) \leftrightarrow \left( {{p}{\rm{ }} \vee \text{~}{q}} \right)$
T	T	T	F	T	T
T	F	F	T	T	F
F	T	T	F	F	F
F	F	T	T	T	T

2.23

1.

Given statements:

$p$ : Elvis Presley wore capes.

$q$ : Some superheroes wear capes.

The symbol is the conditional and is translated “if…then…” or “implies.”

	$p$	$\rightarrow$	$q$
If	Elvis Presley wore capes,	then	some superheroes wear capes.

If Elvis Presley wore capes, then some superheroes wear capes.

2.

Given statements:

$p$ : Elvis Presley wore capes.

$q$ : Some superheroes wear capes.

The converse of $p \rightarrow q$ is $q \rightarrow p$ .

The symbol is the conditional and is translated “if…then…” or “implies.”

	$q$	$\rightarrow$	$p$
If	some superheroes wear capes,	then	Elvis Presley wore capes.

If some superheroes wear capes, then Elvis Presley wore capes.

3.

Given statements:

$p$ : Elvis Presley wore capes.

$q$ : Some superheroes wear capes.

The inverse of $p \rightarrow q$ is ${\sim}p \rightarrow {\sim}q$ .

The symbol is the conditional and is translated “if…then…” or “implies.”

The negation symbol $({\sim})$ is translated “not.”

A “some…” statement is negated with “no….”

	${\sim}p$	$\rightarrow$	${\sim}q$
If	Elvis Presley did not wear capes,	then	no superheroes wear capes.

If Elvis Presley did not wear capes, then no superheroes wear capes.

4.

Given statements:

$p$ : Elvis Presley wore capes.

$q$ : Some superheroes wear capes.

The contrapositive of $p \rightarrow q$ is ${\sim}q \rightarrow {\sim}p$ .

The symbol is the conditional and is translated “if…then…” or “implies.”

The negation symbol $({\sim})$ is translated “not.”

A “some…” statement is negated with “no….”

	$- q$	$\rightarrow$	$- p$
If	no superheroes wear capes,	then	Elvis Presley did not wear capes.

If no superheroes wear capes, then Elvis Presley did not wear capes.

2.24

1.

p

: Dora is an explorer.

2.

q

: Boots is a monkey.

3.

The original statement:

If Dora is an explorer, then Boots is a monkey.

$p$ : Dora is an explorer.

$q$ : Boots is a monkey.

In symbols, the original statement is $p \rightarrow q$ .

The new statement:

If Dora is not an explorer, then Boots is not a monkey.

In symbols, the new statement is ${\sim}p \rightarrow {\sim}q$ .

This is the inverse of the original statement. The inverse negates each statement but does not change the order.

Inverse

4.

The original statement:

If Dora is an explorer, then Boots is a monkey.

$p$ : Dora is an explorer.

$q$ : Boots is a monkey.

In symbols, the original statement is $p \rightarrow q$ .

The new statement:

If Boots is a monkey, then Dora is an explorer.

In symbols, the new statement is $q \rightarrow p$ .

This is the converse of the original statement. A converse changes the order of the original statement.

Converse

5.

The inverse and the converse are equivalent. To show that two statements are logically equivalent, construct a truth table for a biconditional using the first statement as the hypothesis (the inverse, ${\sim}p \rightarrow {\sim}q)$ and the second statement (the converse, $q \rightarrow p$ ) as the conclusion. If the last column holds true only, the statements are equivalent.

$p$	$q$	${\sim}p$	${\sim}q$	${\sim}p \rightarrow {\sim}q$	$q \rightarrow p$	$({\sim}p \rightarrow {\sim}q) \leftrightarrow (q \rightarrow p)$
T	T	F	F	T	T	T
T	F	F	T	T	T	T
F	T	T	F	F	F	T
F	F	T	T	T	T	T

The inverse and converse are equivalent since the last column is always true.

Converse

2.25

1.

The converse of $p \rightarrow q$ is $q \rightarrow p$ .

The original statement is “If my friend lives in San Francisco, then my friend does not live in California.

The converse switches the position.

Converse in words: If my friend does not live in California, then my friend lives in San Francisco.

A conditional is only false if the premise is true, and the conclusion is false.

For the original conditional statement $p \rightarrow q$ to be false, you know that $p$ is true, and $q$ is false.

Now consider the converse, $q \rightarrow p$ . The premise, $q$ , is false, so the converse is true.

If my friend does not live in California, then my friend lives in San Francisco. True.

2.

The inverse of $p \rightarrow q$ is ${\sim}p \rightarrow {\sim}q$ .

The original statement is “If my friend lives in San Francisco, then my friend does not live in California.

The inverse negates each part without changing the order.

Inverse in words: If my friend does not live in San Francisco, then my friend lives in California.

A conditional is only false if the premise is true, and the conclusion is false.

For the original conditional statement $p \rightarrow q$ to be false, you know that $p$ is true, and $q$ is false.

Now consider the inverse, ${\sim}p \rightarrow {\sim}q$ . The premise, ${\sim}p$ , is false, so the inverse is true.

If my friend does not live in San Francisco, then my friend lives in California. True.

3.

The contrapositive of $p \rightarrow q$ is ${\sim}q \rightarrow {\sim}p$ .

The original statement is “If my friend lives in San Francisco, then my friend does not live in California.

The contrapositive switches the position and negates each part.

Contrapositive in words: If my friend lives in California, then my friend does not live in San Francisco.

A conditional is only false if the premise is true, and the conclusion is false.

For the original conditional statement $p \rightarrow q$ to be false, you know that $p$ is true, and $q$ is false.

Now consider the contrapositive, ${\sim}q \rightarrow {\sim}p$ . The premise, ${\sim}q$ , is true and the conclusion, ${\sim}p$ is false. This makes the contrapositive false.

If my friend lives in California, then my friend does not live in San Francisco. False.

2.26

1.

$p \vee q\; \equiv$ Jackie played softball or she ran track

$p :$ Jackie played softball.

$q :$ Jackie ran track.

Negations:

$\text{~}p:$ Jackie did not play softball.

$\text{~}q:$ Jackie did not run track.

Use De Morgan’s Law:

$\text{~}(p \vee q) \equiv \;\text{~}p \wedge \text{~}q$

The negation $\text{~}(p \vee q)$ is equivalent to $\text{~}p \wedge \text{~}q$ .

Jackie did not play softball and she did not run track.

2.

$p \wedge q =$ Brandon studied for his certification exam, and he passed his exam.

$p :$ Brandon studied for his certification exam.

$q :$ Brandon passed his exam.

Negations:

$\text{~}p:$ Brandon did not study for his certification exam.

$\text{~}q:$ Brandon did not pass his exam.

Use De Morgan’s Law:

$\text{~}(p \wedge q) \equiv \;\text{~}p \vee \text{~}q$

Brandon did not study for his certification exam, or he did not pass his exam.

2.27

1.

$p \to q:$ If Edna Mode makes a new superhero costume, then it will not include a Cape.

$p :$ Edna Mode makes a new superhero costume.

$q :$ The costume does not include a cape.

Negation:

$\text{~}q:$ The costume does include a cape.

The negation of a conditional is the conjunction of $p$ and not $q$ .

$\text{~}(p \to q){\rm{ }} \equiv \;p \wedge \text{~}q$

Edna Mode makes a new superhero costume, and the costume does include a cape.

Edna Mode made a new superhero costume, and it includes a cape.

2.

$p \to q:$ If I had pancakes for breakfast, then I used maple syrup.

$p :$ I had pancakes for breakfast.

$q :$ I used maple syrup.

Negation:

$\text{~}q:$ I did not use maple syrup.

The negation of a conditional is the conjunction of $p$ and not $q$ .

$\text{~}(p \to q){\rm{ }} \equiv \;p \wedge \text{~}q$

I had pancakes for breakfast, and I did not use maple syrup.

2.28

1.

$p \to q:$ If some people like ice cream, then ice cream makers will make a profit.

$p :$ Some people like ice cream.

$q :$ Ice cream makers will make a profit.

Negation:

$\text{~}q:$ Ice cream makers will not make a profit.

The negation of a conditional is the conjunction of $p$ and not $q$ .

$\text{~}(p \to q){\rm{ }} \equiv \;p \wedge \text{~}q$

Some people like ice cream, and ice cream makers will not make a profit.

Some people like ice cream, but ice cream makers will not make a profit.

2.

$p \to q:$ If Racquel cannot play video games, then nobody will play video games.

$p :$ Racquel cannot play video games.

$q :$ Nobody will play video games.

Negation:

The negation of a “No…” statement is a “Some…” statement.

$\text{~}q:$ Somebody will play video games.

The negation of a conditional is the conjunction of $p$ and not $q$ .

$\text{~}(p \to q){\rm{ }} \equiv \;p \wedge \text{~}q$

Racquel cannot play video games, but somebody will play video games.

Raquel cannot play video games, but somebody will play video games.

2.29

1.

Eric needs to replace the light bulb, and Marcos did not leave the light bulb on all night, and Dan did not break the light bulb.

$p \to q:$ If Eric needs to replace the light bulb, then Marcus left the light on all night or Dan broke the bulb.

$p :$ Eric needs to replace the light bulb.

$q :$ Marcus left the light on all night or Dan broke the bulb.

Negation:

Use De Morgan’s Law to write the negation of $q:\;\text{~}(p \vee q){\rm{ }} \equiv \;\text{~}p \wedge \text{~}q$

$\text{~}q:$ Marcus did not leave the light on all night and Dan did not break the bulb.

The negation of a conditional is the conjunction of $p$ and not $q$ .

$\text{~}(p \to q){\rm{ }} \equiv \;p \wedge \text{~}q$

Eric needs to replace the light bulb but Marcus did not leave the light on all night and Dan did not break the bulb.

2.

Trenton went to school, and Regina went to work, and Merika did not clean the house.

$p \to q:$ If Trenton went to school and Regina went to work, then Merika cleaned the house.

$p :$ Trenton went to school and Regina went to work.

$q :$ Merika cleaned the house.

Negation:

$\text{~}q:$ Merika did not clean the house.

The original sentence is a conditional.

The negation of a conditional is the conjunction of $p$ and not $q$ .

You do not need to use De Morgan’s Law since $p$ is not negated.

$\text{~}(p \to q){\rm{ }} \equiv \;p \wedge \text{~}q$

Trenton went to school and Regina went to work, and Merika did not clean the house.

2.30

1.

The equivalence is valid if the last column is all true.

There are two statements, so you need four rows of possible truth values.

$p$	$q$	$p \vee q$	$\text{~}(p \vee q)$	$\text{~}p$	$\text{~}q$	$\text{~}p \wedge \text{~}q$	$\text{~}(p \vee q) \leftrightarrow (\text{~}p \wedge \text{~}q)$
T	T
T	F
F	T
F	F

Columns are bold to help you make comparisons.

$p \vee q$

A disjunction is only false if both statements are false. Write an F if both $p$ and $q$ are false. Otherwise, write a T.

$p$	$q$	$p \vee q$	$\text{~}(p \vee q)$	$\text{~}p$	$\text{~}q$	$\text{~}p \wedge \text{~}q$	$\text{~}(p \vee q) \leftrightarrow (\text{~}p \wedge \text{~}q)$
T	T	T
T	F	T
F	T	T
F	F	F

$\text{~}(p \vee q)$

Write the opposite of the previous column.

$p$	$q$	$p \vee q$	$\text{~}(p \vee q)$	$\text{~}p$	$\text{~}q$	$\text{~}p \wedge \text{~}q$	$\text{~}(p \vee q) \leftrightarrow (\text{~}p \wedge \text{~}q)$
T	T	T	F
T	F	T	F
F	T	T	F
F	F	F	T

$\text{~}p$

Write the opposite of $p$ .

$p$	$q$	$p \vee q$	$\text{~}(p \vee q)$	$\text{~}p$	$\text{~}q$	$\text{~}p \wedge \text{~}q$	$\text{~}(p \vee q) \leftrightarrow (\text{~}p \wedge \text{~}q)$
T	T	T	F	F
T	F	T	F	F
F	T	T	F	T
F	F	F	T	T

$\text{~}q$

Write the opposite of $q$ .

$p$	$q$	$p \vee q$	$\text{~}(p \vee q)$	$\text{~}p$	$\text{~}q$	$\text{~}p \wedge \text{~}q$	$\text{~}(p \vee q) \leftrightarrow (\text{~}p \wedge \text{~}q)$
T	T	T	F	F	F
T	F	T	F	F	T
F	T	T	F	T	F
F	F	F	T	T	T

$\text{~}p \wedge \text{~}q$

A conjunction is true only if both statements are true. Write an F if both $\text{~}p$ and $\text{~}q$ are true. Otherwise, write a T.

$p$	$q$	$p \vee q$	$\text{~}(p \vee q)$	$\text{~}p$	$\text{~}q$	$\text{~}p \wedge \text{~}q$	$\text{~}(p \vee q) \leftrightarrow (\text{~}p \wedge \text{~}q)$
T	T	T	F	F	F	F
T	F	T	F	F	T	F
F	T	T	F	T	F	F
F	F	F	T	T	T	T

$\text{~}(p \vee q) \leftrightarrow (\text{~}p \wedge \text{~}q)$

A biconditional is true with both statements have the same truth value. Write a T if both $\text{~}(p \vee q)$ and $(\text{~}p \wedge \text{~}q)$ are true. Write an F if they have different truth values.

$p$	$q$	$p \vee q$	$\text{~}(p \vee q)$	$\text{~}p$	$\text{~}q$	$\text{~}p \wedge \text{~}q$	$\text{~}(p \vee q) \leftrightarrow (\text{~}p \wedge \text{~}q)$
T	T	T	F	F	F	F	T
T	F	T	F	F	T	F	T
F	T	T	F	T	F	F	T
F	F	F	T	T	T	T	T

The statements are equivalent since the last column is always true.

$p$	$q$	${p} \vee {q}$	$\text{~}\left( {{p} \vee {q}} \right)$	$\text{~}p$	$\text{~}q$	$\text{~}p \wedge \text{~}q$	$\text{~}({p} \vee {q}) \leftrightarrow \left( {\text{~}p \wedge \text{~}q} \right)$
T	T	T	F	F	F	F	T
T	F	T	F	F	T	F	T
F	T	T	F	T	F	F	T
F	F	F	T	T	T	T	T

2.31

1.

The Law of Detachment
Premise:	$p \to q$	If my classmates like history, then some people like history.
Premise:	$p$	My classmates like history.
Conclusion:	$\therefore q$	Some people like history.

Some people like history.

2.

The Law of Detachment
Premise:	$p \to q$	If you do not like to read, then some people do not like reading.
Premise:	$p$	You do not like to read.
Conclusion:	$\therefore q$	Some people do not like reading.

Some people do not like reading.

3.

The Law of Detachment
Premise:	$p \to q$	If the polygon has five sides, then it is not an octagon.
Premise:	$p$	The polygon has five sides.
Conclusion:	$\therefore q$	It is not an octagon.

The polygon is not an octagon.

2.32

1.

The negation of a “Some…” statement is a “No…” statement.

The Law of Denying the Consequent
Premise:	$p \to q$	If my classmate likes history, then some people like history.
Premise:	$\text{~}q$	Nobody likes history.
Conclusion:	$\therefore \text{~}p$	My classmate does not like history.

My classmate does not like history.

2.

The negation of an “All…” statement is a “Some….not…” statement.

The Law of Denying the Consequent
Premise:	$p \to q$	If Homer does not like to read, then some people do not like reading.
Premise:	$\text{~}q$	All people like reading.
Conclusion:	$\therefore \text{~}p$	Homer does like to read.

Homer likes to read.

3.

The Law of Denying the Consequent
Premise:	$p \to q$	If the polygon has five sides, then it is not an octagon.
Premise:	$\text{~}q$	The polygon is an octagon.
Conclusion:	$\therefore \text{~}p$	The polygon does not have five sides.

The polygon does not have five sides.

2.33

1.

The Chain Rule for Conditional Arguments
Premise:	$p \to q$	If my roommate does not go to work, then my roommate will not get paid.
Premise:	$q \to r$	If my roommate does not get paid, then they will not be able to pay their bills.
Conclusion:	$\therefore p \to r$	If my roommate does not go to work, then they will not be able to pay their bills.

If my roommate does not go to work, then they will not be able to pay their bills.

2.

The Chain Rule for Conditional Arguments
Premise:	$p \to q$	If penguins cannot fly, then some birds cannot fly.
Premise:	$q \to r$	If some birds cannot fly, then we will watch the news.
Conclusion:	$\therefore p \to r$	If penguins cannot fly, then we will watch the news.

If penguins cannot fly, then we will watch the news.

3.

The Chain Rule for Conditional Arguments
Premise:	$p \to q$	If Marcy goes to the movies, then Marcy will buy popcorn.
Premise:	$q \to r$	If Marcy buys popcorn, then she will buy water.
Conclusion:	$\therefore p \to r$	If Marcy goes to the movies, Marcy will buy water.

If Marcy goes to the movies, then she will buy water.

Check Your Understanding

1.

A logical statement is a complete sentence that makes a claim that may be either true or false. Questions and commands cannot be logical statements. Opinions are not logical statements.

logical statement

2.

The negation of a logical statement has the opposite truth value. If the original statement is true, the negation is false. If the original statement is false, the negation is true.

negation

3.

\text{~}p

4.

${\sim}({\sim}p) = p$

The negation of a negation is the original statement.

$p$

5.

A logical statement used to support a conclusion of an argument is called a premise. You can have any number of premise statements.

premises

6.

Inductive arguments attempt to draw a general conclusion from specific premises. They can be either strong or weak, depending on the relevance of the premise statements.

Inductive

7.

quantifiers

8.

Some giraffes are not tall.

9.

compound statement

10.

connective

11.

biconditional, $\leftrightarrow$

12.

Parentheses, $(\,)$

13.

Conjunction, $\wedge$ ; disjunction, $\vee$ (in any order)

14.

A logical statement is valid if its conclusion follows from its premises. The validity is not affected by whether those premises are true or false.

valid

15.

true

16.

truth table

17.

A truth table for two statements has four rows of truth values.

The general rule is that for n statements there are ${2^n}$ rows in the truth table.

${2^2}=4$

four

18.

The truth table for one statement has two rows of truth values.

The general rule is that for n statements there are ${2^n}$ rows in the truth table.

${2^1}=2$

two

19.

one-way contract

20.

conclusion

21.

If the hypothesis of a conditional statement is false, the conditional statement is true. A false hypothesis can have either a true or a false conclusion. In both situations, the conditional is true.

hypothesis

22.

biconditional

23.

biconditional

24.

true

25.

always true, valid, or a tautology.

26.

conditional

27.

logically equivalent

28.

inverse

29.

converse, inverse

30.

\text{~}p \vee \text{~}q

31.

\text{~}p \wedge \text{~}q

32.

p \wedge \text{~}q

33.

De Morgan’s Laws

34.

premise

35.

A logical argument is valid if its conclusion follows from the premises. The validity is not affected by whether the premises are true or not.

valid

36.

inductive

37.

deductive

38.

fallacy

39.

sound

$p$	$q$	$r$	$\text{~}{p}$	$\text{~}{q}$	${p} \wedge {q}$	$\left( {{p} \wedge {q}} \right) \to {r}$	$\text{~}{p} \vee \text{~}{q}$	$\left( {{\text{~}p} \vee {\text{~}q}} \right) \vee {r}$	$\left( {{p} \wedge {q} \to {r}} \right) \leftrightarrow \left( {{\text{~}p} \vee {\text{~}q} \vee {r}} \right)$
T	T	T	F	F	T	T	F	T	T
T	T	F	F	F	T	F	F	F	T
T	F	T	F	T	F	T	T	T	T
T	F	F	F	T	F	T	T	T	T
F	T	T	T	F	F	T	T	T	T
F	T	F	T	F	F	T	T	T	T
F	F	T	T	T	F	T	T	T	T
F	F	F	T	T	F	T	T	T	T

$p$	$q$	$r$	$\text{~}{p}$	$\text{~}{q}$	${p} \wedge {q}$	$\left( {{p} \wedge {q}} \right) \to {r}$	$\text{~}{p} \vee \text{~}{q}$	$\left( {{\text{~}p} \vee {\text{~}q}} \right) \vee {r}$	$\left( {{p} \wedge {q} \to {r}} \right) \leftrightarrow \left( {{\text{~}p} \vee {\text{~}q} \vee {r}} \right)$
T	T	T	F	F	T	T	F	T	T
T	T	F	F	F	T	F	F	F	T
T	F	T	F	T	F	T	T	T	T
T	F	F	F	T	F	T	T	T	T
F	T	T	T	F	F	T	T	T	T
F	T	F	T	F	F	T	T	T	T
F	F	T	T	T	F	T	T	T	T
F	F	F	T	T	F	T	T	T	T

Chapter 2

Your Turn

Check Your Understanding

$p$	$q$	$r$	$\text{~}{p}$	$\text{~}{q}$	${p} \wedge {q}$	$\left( {{p} \wedge {q}} \right) \to {r}$	$\text{~}{p} \vee \text{~}{q}$	$\left( {{\text{~}p} \vee {\text{~}q}} \right) \vee {r}$	$\left( {{p} \wedge {q} \to {r}} \right) \leftrightarrow \left( {{\text{~}p} \vee {\text{~}q} \vee {r}} \right)$
T	T	T	F	F	T	T	F	T	T
T	T	F	F	F	T	F	F	F	T
T	F	T	F	T	F	T	T	T	T
T	F	F	F	T	F	T	T	T	T
F	T	T	T	F	F	T	T	T	T
F	T	F	T	F	F	T	T	T	T
F	F	T	T	T	F	T	T	T	T
F	F	F	T	T	F	T	T	T	T