2.4: Analyzing Symbolic Arguments

Our previous work with statements and truth tables allows us to analyze and to evaluate arguments using logic. Here, the word argument is not used in the sense that two or more people disagree with each other. Rather, argument is used in the sense of trying to make a convincing case that something must be true using ideas of logic.

The everyday notion of an argument is that it is used to convince us to believe something. An argument is formed with two components: a set of premises and a conclusion. The thing that we are being encouraged to believe is the conclusion, while the premises are the statements offered as supporting evidence for the conclusion that we want to make.

Here are some examples of arguments.

Example 1

\(\begin{array} {ll} \text{Premise 1:} & \text{If I do not have to go to summer school, then I will get an internship.} \\ \text{Premise 2:} & \text{I have to go to summer school.} \\ \text{Conclusion:} & \text{I won’t get an internship.} \end{array}\)

Example 2

\(\begin{array} {ll} \text{Premise 1:} & \text{I studied or I failed the class.} \\ \text{Premise 2:} & \text{I did not fail the class.} \\ \text{Conclusion:} & \text{I studied.} \end{array}\)

To give an argument is to give some premises in support of a conclusion. But suppose that you are given an argument for some conclusion such as the in the previous examples. How can you tell whether that argument is a good argument or bad argument? Our goal in this section will be to analyze an argument to determine whether or not it is a "good" argument.

For an argument to be "good," the truth of an argument’s premises must guarantee the conclusion occurs. If this is the case, we will say that the argument is valid. When the truth of an argument’s premises fails to guarantee the conclusion, we will say that the argument is invalid.

Valid Argument

An argument is valid if its conclusion necessarily follows from the premises. Otherwise, the argument is invalid.

For the conclusion of an argument to necessarily follow from the premises, it means that the symbolic statement

[ Premise 1 \(\wedge\) Premise 2 ] \(\rightarrow\) Conclusion

is always true for all possible truth values of the simple statements involved. You might recall that these types of statements are known as tautologies. If at least one truth value of this symbolic statement is false, then the argument is invalid. Sometimes, this type of argument is called a fallacy.

It is important to note here that if an argument is valid, it only means that conclusion necessarily follows from the premises. It does not mean that conclusion is true.

Arguments can be analyzed using truth tables.

Analyzing Arguments Using Truth Tables

To analyze an argument with a truth table:

1. Represent each of the premises symbolically
2. Create a conditional statement, joining all the premises to form the antecedent, and using the conclusion as the consequent.
3. Create a truth table for the statement. If it is always true, then the argument is valid.

 

Example 3

Consider the argument

\(\begin{array} {ll} \text{Premise 1:} & \text{If you bought bread, then you went to the store.} \\ \text{Premise 2:} & \text{You bought bread.} \\ \text{Conclusion:} & \text{You went to the store.} \end{array}\)

Solution

While this example is fairly obviously a valid argument, we can analyze it using a truth table by representing each of the premises symbolically. We can then form a conditional statement showing that the premises together imply the conclusion. If the truth table is a tautology (always true), then the argument is valid.

We’ll let \(b\) represent “you bought bread” and \(s\) represent “you went to the store”. Then, the argument becomes

\(\begin{array} {ll} \text{Premise 1:} & b \rightarrow s \\ \text{Premise 2:} & b \\ \text{Conclusion:} & s \end{array}\)

To test the validity, we look at whether the combination of both premises implies the conclusion. Is it true that \([(b \rightarrow s) \wedge b] \rightarrow s ?\)

\(\begin{array}{|c|c|c|c|c|}
\hline b & s & b \rightarrow s & (b \rightarrow s) \wedge b & {[(b \rightarrow s) \wedge b] \rightarrow s} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T \rightarrow T = T} & \mathrm{T \wedge T = T} & \mathrm{T \rightarrow T = T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T \rightarrow F = F} & \mathrm{F \wedge T = F} & \mathrm{F \rightarrow F = T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F \rightarrow T = T} & \mathrm{T \wedge F = F} & \mathrm{F \rightarrow T = T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F \rightarrow F = T} & \mathrm{T \wedge F = F} & \mathrm{F \rightarrow F = T} \\
\hline
\end{array}\)

The last column of the truth table shows that \([(b \rightarrow s) \wedge b] \rightarrow s\) is always true. This is a tautology, and so the argument is valid. Based on these premises, it is correct to conclude "You went to the store."

The next two examples revisit the arguments presented in Example 1 and Example 2. Let's determine whether the arguments presented there were examples of valid or invalid reasoning.

Example 4

Determine whether the argument is valid or invalid:

\(\begin{array} {ll} \text{Premise 1:} & \text{If I do not have to go to summer school, then I will get an internship.} \\ \text{Premise 2:} & \text{I have to go to summer school.} \\ \text{Conclusion:} & \text{I won’t get an internship.} \end{array}\)

Solution

We’ll let \(p\) represent “I go to summer school” and \(q\) represent “I will get an internship”. In symbolic form, the argument becomes

\(\begin{array} {ll} \text{Premise 1:} & \sim p \rightarrow q \\ \text{Premise 2:} & p \\ \text{Conclusion:} & \sim q \end{array}\)

Form a truth table to find the truth values for \([( \sim p \rightarrow q) \wedge p] \rightarrow \; \sim q \).

\(\begin{array}{|c|c|c|c|c|c|c|}
\hline p & q & \sim p & \sim q & \sim p \rightarrow q & (\sim p \rightarrow q) \wedge p & {[( \sim p \rightarrow q) \wedge p] \rightarrow \; \sim q } \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F \rightarrow T = T} & \mathrm{T \wedge T = T} & \mathrm{T \rightarrow F = F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F \rightarrow F = T} & \mathrm{T \wedge T = T} & \mathrm{T \rightarrow T = T}\\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T \rightarrow T = T} & \mathrm{T \wedge F = F} & \mathrm{F \rightarrow F = T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T \rightarrow F = F} & \mathrm{F \wedge F = F} & \mathrm{F \rightarrow T = T}\\
\hline
\end{array}\)

One truth value for \([( \sim p \rightarrow q) \wedge p] \rightarrow \; \sim q \) in the last column of the table is False, while the remaining truth values are True. This is not a tautology, and so the argument is invalid. Based on these premises, it is not correct to conclude "I won't get an internship."

Example 5

Determine whether the argument is valid or invalid:

\(\begin{array} {ll} \text{Premise 1:} & \text{I studied or I failed the class.} \\ \text{Premise 2:} & \text{I did not fail the class.} \\ \text{Conclusion:} & \text{I studied.} \end{array}\)

Solution

Let \(s=\) I studied and \(f=\) I failed the class.

The premises and conclusion can be stated as:

\(\begin{array} {ll} \text{Premise 1:} & s \vee f \\ \text{Premise 2:} & \sim f \\ \text{Conclusion:} & s \end{array}\)

Form a truth table to find the truth values for \([(s \vee f) \wedge \sim f] \rightarrow s .\)

\(\begin{array}{|c|c|c|c|c|}
\hline s & f & s \vee f & \sim f & (s \vee f) \; \wedge \sim f & {[(s \vee f) \; \wedge \sim f] \rightarrow s} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T \vee T =T} & \mathrm{F} & \mathrm{T \wedge F = F} & \mathrm{F \rightarrow T = T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T \vee F =T} & \mathrm{T} & \mathrm{T \wedge T = T} & \mathrm{T \rightarrow T = T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F \vee T =T} & \mathrm{F} & \mathrm{T \wedge F = F} & \mathrm{F \rightarrow F = T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F \vee F = F} & \mathrm{T} & \mathrm{F \wedge T = F} & \mathrm{F \rightarrow F = T} \\
\hline
\end{array}\)

All truth values for \([(s \vee f) \wedge \sim f] \rightarrow s \) in the last column of the table are True. This is a tautology, and so the argument is valid. Based on these premises, it is correct to conclude "I studied."

Try it Now 1

Determine whether the argument is valid:

\(\begin{array} {ll} \text{Premise 1:} & \text{If I have a shovel, I can dig a hole.} \\ \text{Premise 2:} & \text{I dug a hole.} \\ \text{Conclusion:} & \text{Therefore, I had a shovel.} \end{array}\)

Answer

Let \(S=\) have a shovel and \(D=\) dig a hole. Premise 1 is equivalent to \(S \rightarrow D\). Premise 2 is \(D\). The conclusion is \(S\). We are testing \([(S \rightarrow D) \wedge D] \rightarrow S\).

\(\begin{array}{|c|c|c|c|c|}
\hline S & D & S \rightarrow D & (S \rightarrow D) \wedge D & {[(S \rightarrow D) \wedge D] \rightarrow S} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)

This is not a tautology, and so this is an invalid argument. It is not correct to conclude "I had a shovel."

The final example follows the same process of using a truth table to analyze an argument for validity. However, this argument contains 3 simple statements, and its truth table is a bit more complex.

 

Example 6

\(\begin{array} {ll} \text{Premise 1:} & \text{If I go to the mall, then I’ll buy new jeans.} \\ \text{Premise 2:} & \text{If I buy new jeans, I’ll buy a shirt to go with it.} \\ \text{Conclusion:} & \text{If I go to the mall, I’ll buy a shirt.} \end{array}\)

Solution

Let \(m=\) I go to the mall, \(j=\) I buy jeans, and \(s=\) I buy a shirt.

The premises and conclusion can be stated as:

\(\begin{array} {ll} \text{Premise 1:} & m \rightarrow j \\ \text{Premise 2:} & j \rightarrow s \\ \text{Conclusion:} & m \rightarrow s \end{array}\)

We can construct a truth table for \([(m \rightarrow j) \wedge(j \rightarrow s)] \rightarrow(m \rightarrow s) .\) Try to recreate each step to verify how the truth table was constructed.

\(\begin{array}{|c|c|c|c|c|c|c|c|}
\hline m & j & s & m \rightarrow j & j \rightarrow s & (m \rightarrow j) \wedge(j \rightarrow s) & m \rightarrow s & {[(m \rightarrow j) \wedge(j \rightarrow s)] \rightarrow(m \rightarrow s)} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline
\end{array}\)

From the final column of the truth table, all truth values for \([(m \rightarrow j) \wedge(j \rightarrow s)] \rightarrow(m \rightarrow s)\) are True. This is a tautology, and so the argument is valid. Based on these premises, it is correct to conclude "If I go to the mall, I'll buy a shirt."