# 2.2: Introduction to Truth Tables

- Page ID
- 92956

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To evaluate a logic statement, we first must learn how to evaluate results of the basic logic operations discussed in the previous section – *negation, conjunction, disjunction, conditional* and *biconditional. *Then, we will learn to use the order of operations for connectors in more complex statements.

The situation in evaluating logic statement is similar to simplifying an arithmetic statement, where one first learns how to perform addition, subtraction, multiplication, and division and then to evaluate any arithmetic expression using the proper order of operations.

According to our previous definition, a *statement* in logic is either true or false, but never both. This means that a simple statement \(p\) can only have two values: 'True' noted as \(p=T\), or 'False' noted \(p = F\). To evaluate basic logic operations, we need to determine truth value outcomes for all possible truth values for the simple statements involved in the operation.

Let us consider how *negation* works.

## Negation

Recall from the previous section of this chapter that the definition of the **negation** of statement \(p\) is another statement that is exactly the opposite of statement \(p\). The negation of \(p\) is "not \(p\)."The symbol used to indicate the negation is \(\sim\), and the negation of statement \(p\) is written \(\sim p\).

Suppose we let \(p\) represent the statement "*I like ice cream*." Then, \(\sim p\) represents the statement "*I do not like ice cream*" or "*It is not true that I like ice cream*."

Statement \(p\) can be true or statement \(p\) can be false as presented in the first column of the table:

\(\begin{array}{|c|c|}

\hline p & \sim p \\

\hline \mathrm{T} & \mathrm{ } \\

\hline \mathrm{F} & \mathrm{ } \\

\hline

\end{array}\)

Obviously, when statement \(p\) is true, statement \(\sim p\) is false. When statement \(p\) is false, statement \(\sim p\) is true. A statement and its negation will always have opposite truth values. In short, the operation of negation works in following way:

- \(\sim T = F\)
- \(\sim F = T\).

The truth values for negation can also be summarized in a type of table called a *truth table*. A **truth table** is a table that shows the resulting truth value of a compound statement for all possible truth values of the simple statements. Because many logic statements can get tricky to think about, we often create a truth table to keep track of what truth values for the simple statements make a compound statement true or false.

\(\begin{array}{|c|c|}

\hline p & \sim p \\

\hline \mathrm{T} & \mathrm{F} \\

\hline \mathrm{F} & \mathrm{T} \\

\hline

\end{array}\)

Next, we review the logical connector '*and*' and build a truth table for the *conjunction* of two simple statements.

## Conjunction

If \(p\) and \(q\) are simple statements, their **conjunction** is “\(p\) and \(q\)” noted as \(p \wedge q\).

Because each of statements \(p\) and \(q\) can be true or false, we must consider all possible combinations of their truth values. There are four of these possible combinations:

- \(p\) and \(q\) are both true,
- \(p\) is be true but \(q\) is false
- \(p\) is false but \(q\) is true,
- \(p\) and \(q\) are both false.

These four combinations are presented as the first two columns in the following table:

\(\begin{array}{| c | c | c |}

\hline p & q & p \wedge q \\

\hline \mathrm{ T } & \mathrm{ T } & \mathrm{ } \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{ } \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{ } \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{ } \\

\hline

\end{array}\)

Recall from the previous section of this chapter that for the conjunction of two simple statements to be true, both simple statements must be true. To illustrate how the outcome of a conjunction is determined, let us consider the following example:

A company is offering a job you are interested in. To apply, an applicant must have an associates degree in information technology (IT) ** and** at least two years of experience in the IT field. These application requirements can be translated into symbolic form:

\(p\): must have associates degree in IT \(q\): must have at least two years of experience in IT field

Then, the requirement to apply is \(p \wedge q\). That is, to qualify for the job, \(p \wedge q\) must be true.

The only way a candidate qualifies for the job \((p \wedge q = T)\) is to have both an associates degree \((p=T)\) and at least two years of experience \((q=T)\). In all other cases, a candidate does not qualify.

That is, when a candidate

- has an associates degree and has at least two years of experience, they qualify: \(T \wedge T = T\)
- has an associates degree and does not have at least two years of experience, they do not qualify: \(T \wedge F = F\)
- does not have an associates degree and has at least two years of experience, they do not qualify: \(F \wedge T = F\)
- does not have an associates degree and does not have at least two years of experience, they do not qualify: \(F \wedge F = F\).

These results can be presented as the truth table for conjunction:

\(\begin{array}{| c | c | c |}

\hline p & q & p \wedge q \\

\hline \mathrm{ T } & \mathrm{ T } & \mathrm{ T } \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{F} \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{F} \\

\hline

\end{array}\)

Next, we review the logical connector '*or*' and build a truth table for the *disjunction* of two simple statements.

## Disjunction

If \(p\) and \(q\) are simple statements, their **disjunction** is “\(p\) or \(q\)” noted as \(p \vee q\).

Just as with a conjunction, each of \(p\) and \(q\) can be either true or false. We can present all four possible combinations of truth values in the first two columns of a truth table:

\(\begin{array}{| c | c | c |}

\hline p & q & p \vee q \\

\hline \mathrm{ T } & \mathrm{ T } & \mathrm{ } \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{ } \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{ } \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{ } \\

\hline

\end{array}\)

Recall from the previous section of this chapter that for the disjunction of two simple statements to be true, at least one of the simple statements must be true. The only time that a disjunction of two simple statements is false is when both simple statements are false. To illustrate how the outcome of a disjunction is determined, let us consider the following example:

A company is offering a job you are interested in. To apply, an applicant must have an associates degree in information technology (IT) ** or** at least two years of experience in the IT field. These application requirements can be translated into symbolic form:

\(p\): must have associates degree in IT \(q\): must have at least two years of experience in IT field

Then, the requirement to apply is \(p \vee q\). That is, to qualify for the job, \(p \vee q\) must be true.

Obviously, any candidate who has either an associates degree \((p=T)\) or two years of experience \((q=T)\) qualifies for the job \((p \vee q = T)\). Only a person who has neither the degree \((p=F)\) nor the experience \((q=F)\) will not qualify \((p \vee q = F)\).

That is, when a candidate

- has an associates degree and has at least two years of experience, they qualify: \(T \vee T = T\)
- has an associates degree and does not have at least two years of experience, they still qualify: \(T \vee F = T\)
- does not have an associates degree and has at least two years of experience, they still qualify: \(F \vee T = T\)
- does not have an associates degree and does not have at least two years of experience, they do not qualify: \(F \vee F = F\).

These results can be presented as the truth table for disjunction:

\(\begin{array}{| c | c | c |}

\hline p & q & p \vee q \\

\hline \mathrm{ T } & \mathrm{ T } & \mathrm{ T } \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{T} \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{T} \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{F} \\

\hline

\end{array}\)

Now, we review the logical connector '*if...then*' and build a truth table for a *conditional statement*.

## Conditional

Recall that a **conditional statement** consists of two simple statements \(p\) and \(q\) joined as “If \(p\), then \(q\).” We write this as \(p \rightarrow q\). The simple statement used in the "*if*" part is the *antecedent* and the simple statement used in the "*then*" part is called the *consequent*. Sometimes a conditional statement is called an *implication, *and we say that "\(p\) implies \(q\).

Just as with conjunctions and disjunctions, each of statements \(p\) and \(q\) can be either true or false. We can present all four possible combinations of truth values in the first two columns of a truth table:

\(\begin{array}{| c | c | c |}

\hline p & q & p \rightarrow q \\

\hline \mathrm{ T } & \mathrm{ T } & \mathrm{ } \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{ } \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{ } \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{ } \\

\hline

\end{array}\)

We must determine the truth value for each of the four outcomes in the table much the same way we did for the conjunction and disjunction of statements \(p\) and \(q\). To illustrate how the outcome of a conditional statement is determined, let us consider an example.

A parent makes this statement to a child: "*If you finish your homework, I will buy you ice cream."*

The two simple statements can be translated into symbolic form:

\(p\): Y*ou ** finish your homework.* \(q\):

*I will buy you ice cream.*

Then, the promise made by the parent can be written as \(p \rightarrow q\).

Let's consider the outcome of the promise on a case-by-case basis for each of the four combinations of truth values for statement \(p\) and statement \(q\). It may be helpful to think about the truth value of the conditional statement \(p \rightarrow q\) in the following way: \(p \rightarrow q\) will be true only when the parent's promise is not broken.

- If the child finishes the homework \((p=T)\) and the parent buys ice cream \((q=T)\), then statement \(p \rightarrow q\) is true as the promise was fulfilled and not broken: \(T \rightarrow T = T\).
- If the child finishes the homework \((p=T)\) and the parent does not buy ice cream \((q=T)\), then the parent's promise \(p \rightarrow q\) was a lie. The parent broke the promise, and so the conditional statement is false: \(T \rightarrow F = F\).
- If the child does not finish the homework \((p=F)\) and the parent buys ice cream \((q=T)\), then statement \(p \rightarrow q\) is still true as again the parent did not break the promise. The parent's promise is in effect only when the child finishes the homework. The parent didn't say what would happen if the child failed to finish the homework. With the '
*if*' part being false, we haven't said anything about what the parent will do so the promise was not broken: \(F \rightarrow T = T\). - If the child does not finish the homework \((p=F)\) and the parent does not buy ice cream \((q=T)\), then statement \(p \rightarrow q\) is true as the parent did not break the promise because the child did not fulfill the required action to finish the homework: \(F \rightarrow F = T\).

These results can be presented as the truth table for a conditional statement.

\(\begin{array}{| c | c | c |}

\hline p & q & p \rightarrow q \\

\hline \mathrm{ T } & \mathrm{ T } & \mathrm{ T } \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{T} \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\

\hline

\end{array}\)

Notice that a conditional statement is false only in one case -- when the antecedent is true and the consequent is false. In our example of the homework and ice cream promise, that is the only time the parent's promise is broken.

Finally, we review the logical connector '*if and only if*' and build a truth table for a *biconditional statement*.

## Biconditional

A **biconditional statement** consists of two simple statements \(p\) and \(q\) joined as “\(p\) if and only if \(q\).” A biconditional statement is a conjunction of the two conditional statements "If \(p\), then \(q\)" and "If \(q\), then \(p\)." We write this as \(p \leftrightarrow q\).

We can use this definition of a biconditional statement to construct the truth table for it. For each of the four possible combination of values for \(p\) and \(q\), we evaluate \(p \rightarrow q\) in the third column and \(q \rightarrow p\) in the fourth column using the truth table for the conditional statement we created previously. Next, we evaluate the conjunction using the truth values in the third and fourth columns to determine the truth value of \((p \rightarrow q) \wedge (q \rightarrow p)\) in the fifth column using the truth table for the conjunction.

\(\begin{array}{| c | c | c | c | c |}

\hline p & q & p \rightarrow q & q \rightarrow p & p \leftrightarrow q = (p \rightarrow q) \wedge (q \rightarrow p)\\

\hline \mathrm{ T } & \mathrm{ T } & \mathrm{ T \rightarrow T = T } & \mathrm{ T \rightarrow T = T } & \mathrm{ T \wedge T = T }\\

\hline \mathrm{T} & \mathrm{F} & \mathrm{T \rightarrow F = F} & \mathrm{ F \rightarrow T = T} & \mathrm{ F \wedge T = F }\\

\hline \mathrm{F} & \mathrm{T} & \mathrm{F \rightarrow T = T} & \mathrm{ T \rightarrow F = F } & \mathrm{ T \wedge F = F}\\

\hline \mathrm{F} & \mathrm{F} & \mathrm{F \rightarrow F = T} & \mathrm{ F\rightarrow F = T } & \mathrm{ T \wedge T = T }\\

\hline

\end{array}\)

As you can see from the truth table, a biconditional statement is only true when both simple statements have the same truth values (both \(p\) and \(q\) are true or both \(p\) and \(q\) are false.) When the simple statements in a biconditional have different truth values, a biconditional is false.

Now that we have explored truth tables for five basic operations used to form compound logical statements, here is a summary chart of truth values for easy reference.

Operation |
Notation |
Summary of truth values |
---|---|---|

Negation | \(\sim p\) | The opposite truth value of \(p\) |

Conjunction | \(p \wedge q\) | True only when both \(p\) and \(q\) are true |

Disjunction | \(p \vee q\) | False only when both \(p\) and \(q\) are false |

Conditional | \(p \rightarrow q\) | False only when \(p\) is true and \(q\) is false |

Biconditional | \(p \leftrightarrow q\) | True only when both \(p\) and \(q\) are true or are both are false |

Consider the following compound statements. Use appropriate entries from truth tables to determine whether each statement is true or false.

- There are 50 states in the United States or Santa Claus is vice president.
- There are 50 states in the United States and Santa Claus is vice president.
- If there are 50 states in the United States, then Santa Claus is vice president.
- If Santa Claus is vice president, then there are 50 states in the United States.
- There are 50 states in the United States if and only if Santa Claus is vice president.

### Solution

First, identify and determine the truth value for each simple statement:

*There are 50 states in the United States. *(True) *Santa Claus is vice president.* (False)

- True. This statement can be represented as \(T \vee F\) because the word "
*or*" connects a true statement and a false statement. According to the truth table for a disjunction, \(T \vee F = T\). - False. This statement can be represented as \(T \wedge F\) because the word "
*and*" connects a true statement and a false statement. According to the truth table for a conjunction, \(T \wedge F = F\). - False. This statement can be represented as \(T \rightarrow F\) because "
*if ... then...*" connects a true antecedent statement to a false consequent statement. According to the truth table for a conditional, \(T \rightarrow F = F\). This is a broken promise situation. - True. This statement can be represented as \(F \rightarrow T\) because "
*if ... then...*" connects a false antecedent statement to a true consequent statement. According to the truth table for a conditional, \(F \rightarrow T = T\). - False. This statement can be represented as \(T \leftrightarrow F\) because "
*...if and only if...*" connects a true statement and a false statement. According to the truth table for a biconditional, \(T \leftrightarrow F = F\) because the two simple statements have different truth values.

Try it Now 1

Consider the following compound statements. Use appropriate entries from truth tables to determine whether each statement is true or false.

- 8 is odd and 6 is even.
- 8 is odd or 6 is even.
- If 8 is odd, then 6 is even.
- If 6 is even, then 8 is odd.
- 8 is odd if and only if 6 is even.

**Answer**-
- False because (\(F \wedge T = F\))
- True because (\(F \vee T = T\))
- True because (\(F \rightarrow T = T\))
- False because (\(T \rightarrow F = F\))
- False because (\(F \leftrightarrow T = F\))

Compound statements in logic can be quite complex and can have more than one operation or connector. It is important to follow the correct order of performing operations when evaluating compound statements for truth values just as we follow a prescribed order of operating when evaluating an expression in arithmetic or algebra.

To evaluate any compound statement that includes grouping symbols or more than one connective we need to adhere to the following order of operations:

- Any expression inside parentheses
- Negation (\(\sim\))
- Conjunction or disjunction \(( \wedge, \vee\))
- Conditional \(( \rightarrow )\)
- Biconditional \(( \leftrightarrow )\)

Let us now practice constructing truth tables for various compound statements that are slightly more complex.

Create a truth table for \(\sim p \rightarrow \; \sim q\).

### Solution

Recall that all four possible combinations of truth values for \(p\) and \(q\) can be presented in the first two columns of the truth table. Then, following the order of operations, we evaluate \(\sim p\) in the third column and \(\sim q\) in the fourth column using the truth table for negation determined in the beginning of this section. Finally, we evaluate the conditional \(\sim p \rightarrow \; \sim q\) using truth values of \(\sim p\) found in the third column and truth values of \(\sim q\) found in the fourth column, applying the rules for a conditional connector.

\(\begin{array}{|c|c|c|c|c|}

\hline p & q & \sim p & \sim q & \sim p \rightarrow \; \sim q \\

\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F \rightarrow F = T} \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F \rightarrow T = T} \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T \rightarrow F = F} \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T \rightarrow T = T} \\

\hline

\end{array}\)

Create a truth table for the statement \(p \; \vee \sim q\).

### Solution

All possible combinations of truth values for simple statements \(p\) and \(q\) are presented in the first two columns of the table. Next, we evaluate \(\sim q\) in the third column using the truth values for statement \(q\) in the second column. Now, we can use the first and third columns to find the truth values of \( p \; \vee \sim q\).

\(\begin{array}{|c|c|c|c|}

\hline p & q & \sim q & p \; \vee \sim q \\

\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T \vee F = T} \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T \vee T = T} \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F \vee F = F} \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F \vee T = T} \\

\hline

\end{array}\)

The truth table shows that \(p \; \vee \sim q\) is true in three cases and false in one case.

Try it Now 2

Create a truth table for the statement \(\sim p \wedge \; q\).

**Answer**-
\(\begin{array}{|c|c|c|c|}

\hline p & q & \sim p & \sim p \wedge \; q \\

\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F \wedge T = F} \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F \wedge F = F} \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T \wedge T = T} \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T \wedge F = F} \\

\hline

\end{array}\)

In the previous examples, each compound statement was formed using two simple statements, \(p\) and \(q\). In these cases, there were only \(2 \times 2 = 4\) possible combinations of truth values for the two statements. Sometimes, however, compound statements have more than two simple statements. For example, a compound statement might include three simple statements \(p\), \(q\), and \(r\). When there are three simple statements, then there are \(2 \times 2 \times 2 = 8\) possible combinations of truth values for the three statements. The next example is an illustration.

Create a truth table for the statement \(p \; \wedge \sim(q \vee r)\).

### Solution

The eight combinations of truth values for three simple statements obviously can be listed in any order, but the standard way is presented in the first two columns of the truth table. This list of combinations is created by first letting \(p\) = T and then creating the standard four combinations of truth values for \(q\) and \(r\). Then, the rest of the list of combinations is created by letting \(p\) = F and repeating the combinations of truth values for \(q\) and \(r\).

Following the order of operations rules, we first use the rules for disjunction to evaluate the statement in parentheses \((q \vee r)\) using the truth values for \(q\) and \(r\) in the second and third columns. These results for \((q \vee r)\) are placed in the fourth column. Next, we negate the results in this fourth column and write the results for \(\sim(q \vee r)\) in the fifth column. The last step is to evaluate \(p \; \wedge \sim(q \vee r)\), the conjunction between truth values in the first and fifth columns.

\(\begin{array}{|c|c|c|c|c|c|}

\hline p & q & r & q \vee r & \sim \; (q \vee r) & p \; \wedge \sim (q \vee r \ { ) } \\

\hline \mathrm { T } & \mathrm{ T } & \mathrm{ T } & \mathrm { T \vee T = T } & \mathrm { \sim T = F } & \mathrm { T \wedge F = F } \\

\hline \mathrm { T } & \mathrm { T } & \mathrm { F } & \mathrm { T \vee F = T } & \mathrm { \sim T = F } & \mathrm { T \wedge F = F } \\

\hline \mathrm { T } & \mathrm { F } & \mathrm{ T } & \mathrm { F \vee T = T } & \mathrm { \sim T = F } & \mathrm { T \wedge F = F } \\

\hline \mathrm { T } & \mathrm { F } & \mathrm { F } & \mathrm { F \vee F = F} & \mathrm { \sim F = T } & \mathrm { T \wedge T = T } \\

\hline \mathrm { F } & \mathrm { T } & \mathrm { T } & \mathrm { T \vee T = T } & \mathrm { \sim T = F } & \mathrm { F \wedge F = F } \\

\hline \mathrm { F } & \mathrm { T } & \mathrm { F } & \mathrm { T \vee F = T } & \mathrm { \sim T = F } & \mathrm { F \wedge F = F } \\

\hline \mathrm { F } & \mathrm { F } & \mathrm { T } & \mathrm { F \vee T = T } & \mathrm { \sim T = F } & \mathrm { F \wedge F = F } \\

\hline \mathrm { F } & \mathrm { F } & \mathrm { F } & \mathrm { F \vee F = F } & \mathrm { \sim F = T } & \mathrm { F \wedge T = F } \\

\hline

\end{array}\)

It turns out that this complex expression is true in only one case -- when \(p\) is true, \(q\) is false, and \(r\) is false.

Create a truth table for the statement \((p \vee q) \leftrightarrow \; \sim r\).

### Solution

There are 3 simple statements so start by listing all the possible truth value combinations for \(p, q,\) and \(r \) in the first three columns. After creating the 8 combinations, use the truth values for \(p\) and \(q\) to write the results for \(p \vee q\) in the fourth column. In the fifth column, write the truth values for \(\sim r\), the negation of the truth values in the third column. Finally, find the truth values of \((p \vee q) \leftrightarrow \; \sim r\). Remember, a biconditional is true when the truth value of the two parts match, but it is false when the truth values do not match.

\(\begin{array}{|c|c|c|c|c|c|}

\hline p & q & r & p \vee q & \sim r & (p \vee q) \leftrightarrow \; \sim r \\

\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T \vee T = T} & \mathrm{\sim T =F} & \mathrm{T \leftrightarrow F = F} \\

\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T \vee T = T} & \mathrm{\sim F =T} & \mathrm{T \leftrightarrow T = T} \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T \vee F = T} & \mathrm{\sim T =F} & \mathrm{T \leftrightarrow F = F} \\

\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T \vee F = T} & \mathrm{\sim F =T} & \mathrm{T \leftrightarrow T = T} \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F \vee T = T} & \mathrm{\sim T =F} & \mathrm{T \leftrightarrow F = F} \\

\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F \vee T = T} & \mathrm{\sim F =T} & \mathrm{T \leftrightarrow T = T} \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F \vee F = F} & \mathrm{\sim T =F} & \mathrm{F \leftrightarrow F = T} \\

\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F \vee F = F} & \mathrm{\sim F =T} & \mathrm{F \leftrightarrow T = F} \\

\hline

\end{array}\)

Try it Now 3

Create a truth table for the statement \((p \; \wedge \sim q) \rightarrow r\).

**Answer**-
\(\begin{array}{|c|c|c|c|c|c|c|}

\hline p & q & r & \sim q & p \; \wedge \sim q & r & (p \; \wedge \sim q) \rightarrow r\ \\

\hline \mathrm { T } & \mathrm{ T } & \mathrm{ T } & \mathrm { \sim T = F } & \mathrm { T \wedge F = F} & \mathrm{ T } & \mathrm { F \rightarrow T = T } \\

\hline \mathrm { T } & \mathrm { T } & \mathrm { F } & \mathrm { \sim T = F } & \mathrm { T \wedge F = F } & \mathrm{ F } & \mathrm { F \rightarrow F = T } \\

\hline \mathrm { T } & \mathrm { F } & \mathrm{ T } & \mathrm { \sim F = T } & \mathrm { T \wedge T = T } & \mathrm{ T } & \mathrm { T \rightarrow T = T} \\

\hline \mathrm { T } & \mathrm { F } & \mathrm { F } & \mathrm { \sim F = T} & \mathrm { T \wedge T = T } & \mathrm{ F } & \mathrm { T \rightarrow F = F } \\

\hline \mathrm { F } & \mathrm { T } & \mathrm { T } & \mathrm { \sim T = F } & \mathrm { F \wedge F = F } & \mathrm{ T }& \mathrm { F \rightarrow T = T } \\

\hline \mathrm { F } & \mathrm { T } & \mathrm { F } & \mathrm { \sim T = F } & \mathrm {F \wedge F = F } & \mathrm{ F } & \mathrm { F \rightarrow F = T} \\

\hline \mathrm { F } & \mathrm { F } & \mathrm { T } & \mathrm { \sim F = T } & \mathrm { F \wedge F = F } & \mathrm{ T } & \mathrm { F \rightarrow T = T } \\

\hline \mathrm { F } & \mathrm { F } & \mathrm { F } & \mathrm { \sim F = T } & \mathrm { F \wedge T = F } & \mathrm{ F } & \mathrm { F \rightarrow F = T } \\

\hline

\end{array}\)