5.6: Evaluating Deductive Arguments with Truth Tables
- Page ID
- 152064
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- Determine if a deductive argument is valid using a truth table.
Arguments can also be analyzed using truth tables, although this can be a lot of work.
To analyze an argument with a truth table:
- Represent each of the premises symbolically
- Create a conditional statement, joining all the premises to form the antecedent, and using the conclusion as the consequent.
- Create a truth table for the statement. If it is always true, then the argument is valid.
Consider the argument
\(\begin{array} {ll} \text{Premise:} & \text{If you bought bread, then you went to the store.} \\ \text{Premise:} & \text{You bought bread.} \\ \text{Conclusion:} & \text{You went to the store.} \end{array}\)
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:
\(\begin{array} {ll} \text{Premise:} & b \rightarrow s \\ \text{Premise:} & b \\ \text{Conclusion:} & s \end{array}\)
To test the validity, we look at whether the combination of both premises implies the conclusion; is it true that \([(b \rightarrow s) \wedge b] \rightarrow s ?\)
\(\begin{array}{|c|c|c|}
\hline b & s & b \rightarrow s \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)
\(\begin{array}{|c|c|c|c|}
\hline b & s & b \rightarrow s & (b \rightarrow s) \wedge b \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline
\end{array}\)
\(\begin{array}{|c|c|c|c|c|}
\hline b & s & b \rightarrow s & (b \rightarrow s) \wedge b & {[(b \rightarrow s) \wedge b] \rightarrow s} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)
Since the truth table for \([(b \rightarrow s) \wedge b] \rightarrow s\) is always true, this is a valid argument.
Determine whether the argument is valid:
\(\begin{array} {ll} \text{Premise:} & \text{If I have a shovel, I can dig a hole.} \\ \text{Premise:} & \text{I dug a hole.} \\ \text{Conclusion:} & \text{Therefore, I had a shovel.} \end{array}\)
- Answer
-
Let \(S=\) have a shovel, \(D=\operatorname{dig}\) a hole. The first premise is equivalent to \(S \rightarrow D\). The second premise 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, so this is an invalid argument.
\(\begin{array} {ll} \text{Premise:} & \text{If I go to the mall, then I’ll buy new jeans.} \\ \text{Premise:} & \text{If I buy new jeans, I’ll buy a shirt to go with it.} \\ \text{Conclusion:} & \text{If I go to the mall, I’ll buy a shirt.} \end{array}\)
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:
\(\begin{array} {ll} \text{Premise:} & m \rightarrow j \\ \text{Premise:} & j \rightarrow s \\ \text{Conclusion:} & m \rightarrow s \end{array}\)
We can construct a truth table for \([(m \rightarrow j) \wedge(j \rightarrow s)] \rightarrow(m \rightarrow s) .\) Try to recreate each step and see 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, we can see this is a valid argument.