6.3: Truth Tables- Conditional, Biconditional

Conditional

A conditional is a logical compound statement in which a statement \(p\), called the hypothesis, implies a statement \(q\), called the conclusion.

A conditional is written as \(p \rightarrow q\) and is translated as "if \(p\), then \(q\)".

Example 1

The English statement “If it is raining, then there are clouds is the sky” is a conditional statement. It makes sense because if the hypothesis “it is raining” is true, then the conclusion “there are clouds in the sky” must also be true.

Notice that the statement tells us nothing of what to expect if it is not raining; there might be clouds in the sky, or there might not. If the hypothesis is false, then the conclusion becomes irrelevant.

Example 2

Suppose you order a team jersey online on Tuesday and want to receive it by Friday so you can wear it to Saturday’s game. The website says that if you pay for expedited shipping, you will receive the jersey by Friday. In what situation is the website telling a lie?

There are four possible outcomes:

1. You pay for expedited shipping and receive the jersey by Friday
2. You pay for expedited shipping and don’t receive the jersey by Friday
3. You don’t pay for expedited shipping and receive the jersey by Friday
4. You don’t pay for expedited shipping and don’t receive the jersey by Friday

Only one of these outcomes proves that the website was lying: the second outcome in which you pay for expedited shipping but don’t receive the jersey by Friday. The first outcome is exactly what was promised, so there’s no problem with that. The third outcome is not a lie because the website never said what would happen if you didn’t pay for expedited shipping; maybe the jersey would arrive by Friday whether you paid for expedited shipping or not. The fourth outcome is not a lie because, again, the website didn’t make any promises about when the jersey would arrive if you didn’t pay for expedited shipping.

It may seem strange that the third outcome in the previous example, in which the first part is false but the second part is true, is not a lie. Remember, though, that if the hypothesis is false, we cannot make any judgment about the conclusion. The website never said that paying for expedited shipping was the only way to receive the jersey by Friday.

Example 3

A friend tells you “If you upload that picture to Facebook, you’ll lose your job.” Under what conditions can you say that your friend was wrong?

There are four possible outcomes:

1. You upload the picture and lose your job
2. You upload the picture and don’t lose your job
3. You don’t upload the picture and lose your job
4. You don’t upload the picture and don’t lose your job

There is only one possible case in which you can say your friend was wrong: the second outcome in which you upload the picture but still keep your job. In the last two cases, your friend didn’t say anything about what would happen if you didn’t upload the picture, so you can’t say that their statement was wrong. Even if you didn’t upload the picture and lost your job anyway, your friend never said that you were guaranteed to keep your job if you didn’t upload the picture; you might lose your job for missing a shift or punching your boss instead.

Truth Table for the Conditional

\(\begin{array}{|c|c|c|}
\hline p & q & p \rightarrow q \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)

Again, if the hypothesis \(p\) is false, we cannot prove that the statement is a lie, so the result of the third and fourth rows is true.

Example 4

Construct a truth table for the statement \((m \wedge \sim p) \rightarrow r\)

Solution

We start by constructing a truth table with 8 rows to cover all possible scenarios. Next, we can focus on the hypothesis, \(m \wedge \sim p\).

\(\begin{array}{|c|c|c|}
\hline m & p & r \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} \\ \hline
\end{array}\)

\(\begin{array}{|c|c|c|c|}
\hline m & p & r & \sim p \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)

\(\begin{array}{|c|c|c|c|c|}
\hline m & p & r & \sim p & m \wedge \sim p \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline
\end{array}\)

Now we can create a column for the conditional. Because it can be confusing to keep track of all the Ts and \(\mathrm{Fs}\), why don't we copy the column for \(r\) to the right of the column for \(m \wedge \sim p\) ? This makes it a lot easier to read the conditional from left to right.

\(\begin{array}{|c|c|c|c|c|c|c|}
\hline m & p & r & \sim p & m \wedge \sim p & r & (m \wedge \sim p) \rightarrow r \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)

When \(m\) is true, \(p\) is false, and \(r\) is false- -the fourth row of the table-then the hypothesis \(m \wedge \sim p\) will be true but the conclusion false, resulting in an invalid conditional; every other case gives a valid conditional.

If you want a real-life situation that could be modeled by \((m \wedge \sim p) \rightarrow r\), consider this: let \(m=\) we order meatballs, \(p=\) we order pasta, and \(r=\) Rob is happy. The statement \((m \wedge \sim p) \rightarrow r\) is "if we order meatballs and don't order pasta, then Rob is happy". If \(m\) is true (we order meatballs), \(p\) is false (we don't order pasta), and \(r\) is false (Rob is not happy), then the statement is false, because we satisfied the hypothesis but Rob did not satisfy the conclusion.

For any conditional, there are three related statements, the converse, the inverse, and the contrapositive.

Derived Forms of a Conditional

The original conditional is \(\quad\) "if \(p,\) then \(q^{\prime \prime} \quad p \rightarrow q\)

The converse is \(\quad\) "if \(q,\) then \(p^{\prime \prime} \quad q \rightarrow p\)

The inverse is \(\quad\) "if not \(p,\) then not \(q^{\prime \prime} \quad \sim p \rightarrow \sim q\)

The contrapositive is "if not \(q,\) then not \(p^{\prime \prime} \quad \sim q \rightarrow \sim p\)

Example 5

Consider again the conditional “If it is raining, then there are clouds in the sky.” It seems reasonable to assume that this is true.

The converse would be “If there are clouds in the sky, then it is raining.” This is not always true.

The inverse would be “If it is not raining, then there are not clouds in the sky.” Likewise, this is not always true.

The contrapositive would be “If there are not clouds in the sky, then it is not raining.” This statement is true, and is equivalent to the original conditional.

Looking at truth tables, we can see that the original conditional and the contrapositive are logically equivalent, and that the converse and inverse are logically equivalent.

Equivalence

A conditional statement and its contrapositive are logically equivalent.

The converse and inverse of a conditional statement are logically equivalent.

In other words, the original statement and the contrapositive must agree with each other; they must both be true, or they must both be false. Similarly, the converse and the inverse must agree with each other; they must both be true, or they must both be false.

We typically represent the conditional using the words, "if ..., then ...," but there are other ways this logical connective can be represented in English. Consider the conditional from Example 5: "If it is raining, then there are clouds in the sky." We could equivalently write, "It is raining only if there are clouds in the sky." You can probably imagine how these two statements are saying the same thing - whenever it's raining outside, it is a safe conclusion there are clouds in the sky as well. Some other wordings that communicate the same information use either "sufficient" or "necessary." For example, "Raining is a sufficient condition for it to be cloudy," and "Being cloudy is a necessary condition for it to be raining." Here is a table summarizing the different wordings.

Different Wordings of the Conditional

The following statements are equivalent:

• If \(p\), then \(q\).
• \(q\) only if \(p\).
• \(p\) is sufficient for \(q\).
• \(q\) is necessary for \(p\).

In everyday life, we often have a stronger meaning in mind when we use a conditional statement. Consider “If you submit your hours today, then you will be paid next Friday.” What the payroll rep really means is “If you submit your hours today, then you will be paid next Friday, and if you don’t submit your hours today, then you won’t be paid next Friday.” The conditional statement if t, then p also includes the inverse of the statement: if not t, then not p. A more compact way to express this statement is “You will be paid next Friday if and only if you submit your timesheet today.” A statement of this form is called a biconditional.

Biconditional

A biconditional is a logical conditional statement in which the hypothesis and conclusion are interchangeable.

A biconditional is written as \(p \leftrightarrow q\) and is translated as " \(p\) if and only if \(q^{\prime \prime}\).

Because a biconditional statement \(p \leftrightarrow q\) is equivalent to \((p \rightarrow q) \wedge(q \rightarrow p),\) we may think of it as a conditional statement combined with its converse: if \(p\), then \(q\) and if \(q\), then \(p\). The double-headed arrow shows that the conditional statement goes from left to right and from right to left. A biconditional is considered true as long as the hypothesis and the conclusion have the same truth value; that is, they are either both true or both false.

Truth Table for the Biconditional

\(\begin{array}{|c|c|c|}
\hline p & q & p \leftrightarrow q \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)

Notice that the fourth row, where both components are false, is true; if you don’t submit your timesheet and you don’t get paid, the person from payroll told you the truth.

Example 6

Suppose this statement is true: “The garbage truck comes down my street if and only if it is Thursday morning.” Which of the following statements could be true?

1. It is noon on Thursday and the garbage truck did not come down my street this morning.
2. It is Monday and the garbage truck is coming down my street.
3. It is Wednesday at 11:59PM and the garbage truck did not come down my street today.

Solution

1. This cannot be true. This is like the second row of the truth table; it is true that I just experienced Thursday morning, but it is false that the garbage truck came.
2. This cannot be true. This is like the third row of the truth table; it is false that it is Thursday, but it is true that the garbage truck came.
3. This could be true. This is like the fourth row of the truth table; it is false that it is Thursday, but it is also false that the garbage truck came, so everything worked out like it should.

Try it Now 1

Suppose this statement is true: “I wear my running shoes if and only if I am exercising.” Determine whether each of the following statements must be true or false.

1. I am exercising and I am not wearing my running shoes.
2. I am wearing my running shoes and I am not exercising.
3. I am not exercising and I am not wearing my running shoes.

Answer

Choices a & b are false; c is true.

Example 7

Create a truth table for the statement \((A \vee B) \leftrightarrow \sim C\)

Solution

Whenever we have three component statements, we start by listing all the possible truth value combinations for \(A, B,\) and \(C .\) After creating those three columns, we can create a fourth column for the hypothesis, \(A \vee B\). Now we will temporarily ignore the column for \(C\) and focus on \(A\) and \(B\), writing the truth values for \(A \vee B\).

\(\begin{array}{|c|c|c|}
\hline A & B & C \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}\)

\(\begin{array}{|c|c|c|c|}
\hline A & B & C & A \vee B \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}\)

Next we can create a column for the negation of \(C\). (Ignore the \(A \vee B\) column and simply negate the values in the \(C\) column.)

\(\begin{array}{|c|c|c|c|c|}
\hline A & B & C & A \vee B & \sim C \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)

Finally, we find the truth values of \((A \vee B) \leftrightarrow \sim C\). 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.

\(\begin{array}{|c|c|c|c|c|c|}
\hline A & B & C & A \vee B & \sim C & (A \vee B) \leftrightarrow \sim C \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline
\end{array}\)

To illustrate this situation, suppose your boss needs you to do either project \(A\) or project \(B\) (or both, if you have the time). If you do one of the projects, you will not get a crummy review ( \(C\) is for crummy). So \((A \vee B) \leftrightarrow \sim C\) means "You will not get a crummy review if and only if you do project \(A\) or project \(B\)." Looking at a few of the rows of the truth table, we can see how this works out. In the first row, \(A, B,\) and \(C\) are all true: you did both projects and got a crummy review, which is not what your boss told you would happen! That is why the final result of the first row is false. In the fourth row, \(A\) is true, \(B\) is false, and \(C\) is false: you did project \(A\) and did not get a crummy review. This is what your boss said would happen, so the final result of this row is true. And in the eighth row, \(A, B\), and \(C\) are all false: you didn't do either project and did not get a crummy review. This is not what your boss said would happen, so the final result of this row is false. (Even though you may be happy that your boss didn't follow through on the threat, the truth table shows that your boss lied about what would happen.)