3.4: Verifying Arguments
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)An argument is a series of statements called premises followed by a single statement called the conclusion.
An argument is valid if whenever all the premises are true, then the conclusion must also be true. Otherwise the argument is invalid.
Arguments can also be analyzed using truth tables, although this can be a lot of work.
To analyze an argument with a truth table:
- Make a truth table with a seperate column for each premise and the conclusion.
- Examine only the lines in the table which all of the premises are true.
- If the conclusion is also true for the lines you examined in step 2, the argument is valid.
- If the conclusion is false for even one of the lines you examined in step 2, the argument is invalid.
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’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 create a truth table and a separate column for both premises and the conclusion:
\(\begin{array}{|c|c|c|}
\hline b & s & b \rightarrow s & b & s\\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \\
\hline
\end{array}\)
We only examine the first line since that is the only time both premises are true. We see that the conclusion is also true. This argument is valid.
Determine whether the argument is valid:
\(\begin{array} {ll} \text{Premise:} & \text{If I have a shovel, I can dig a hole.} \\ \text{Premise:} & \text{I dug a hole.} \\ \text{Conclusion:} & \text{Therefore, I had a shovel.} \end{array}\)
Solution
Let \(S=\) have a shovel, \(D=\operatorname{dig}\) a hole. The argument becomes:
\(\begin{array} {ll} \text{Premise:} & S \rightarrow D \\ \text{Premise:} & D \\ \text{Conclusion:} & S \end{array}\)
We can construct a truth table with a seperate column for each premise and the conclusion:
\(\begin{array}{|c|c|c|c|c|c|c|c|}
\hline S & D & S \rightarrow D & D & S \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T}\\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F}\\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \\
\hline
\end{array}\)
We need to examine both Line 1 and Line 3 since both premises are true in both cases. Since the conclusion is false in Line 3, the argument is invalid.
Rather than making a truth table for every argument, we may be able to recognize certain common forms of arguments that are valid (or invalid). If we can determine that an argument fits one of the common forms, we can immediately state whether it is valid or invalid.
The law of detachment is:
\(\begin{array} {ll} \text{Premise:} & p \rightarrow q \\ \text{Premise:} & p \\ \text{Conclusion:} & q \end{array}\)
Recall this argument from an earlier example:
\(\begin{array} {ll} \text{Premise:} & \text{If you bought bread, then you went to the store.} \\ \text{Premise:} & \text{You bought bread.} \\ \text{Conclusion:} & \text{You went to the store.} \end{array}\)
In symbolic form:
\(\begin{array} {ll} \text{Premise:} & b \rightarrow s \\ \text{Premise:} & b \\ \text{Conclusion:} & s \end{array}\)
This argument has the structure described by the law of detachment. (The second premise and the conclusion are simply the two parts of the first premise detached from each other.) Instead of making a truth table, we can say that this argument is valid by stating that it satisfies the law of detachment.
The law of contraposition is:
\(\begin{array} {ll} \text{Premise:} & p \rightarrow q \\ \text{Premise:} & \sim q \\ \text{Conclusion:} & \sim p \end{array}\)
Notice that the second premise and the conclusion look like the contrapositive of the first premise, \(\sim q \rightarrow \sim p\), but they have been detached. You can think of the law of contraposition as a combination of the law of detachment and the fact that the contrapositive is logically equivalent to the original statement.
\(\begin{array} {ll} \text{Premise:} & \text{If I drop my phone into the swimming pool, my phone will be ruined.} \\ \text{Premise:} & \text{My phone isn’t ruined.} \\ \text{Conclusion:} & \text{I didn’t drop my phone into the swimming pool.} \end{array}\)
If we let \(d=\mathrm{I}\) drop the phone in the pool and \(r=\) the phone is ruined, then we can represent the argument this way:
\(\begin{array} {ll} \text{Premise:} & d \rightarrow r \\ \text{Premise:} & \sim r \\ \text{Conclusion:} & \sim d \end{array}\)
The form of this argument matches what we need to invoke the law of contraposition, so it is a valid argument.
Is this argument valid?
\(\begin{array} {ll} \text{Premise:} & \text{If you brushed your teeth before bed, then your toothbrush will be wet.} \\ \text{Premise:} & \text{Your toothbrush is dry.} \\ \text{Conclusion:} & \text{You didn’t brush your teeth before bed.} \end{array}\)
- Answer
-
Let \(b=\) brushed teeth and \(w=\) toothbrush is wet.
\(\begin{array} {ll} \text{Premise:} & b \rightarrow w \\ \text{Premise:} & \sim w \\ \text{Conclusion:} & \sim b \end{array}\)
This argument is valid by the Law of Contraposition.
The Law of Syllogism is:
\(\begin{array} {ll} \text{Premise:} & p \rightarrow q \\ \text{Premise:} & q \rightarrow r \\ \text{Conclusion:} & p \rightarrow r \end{array}\)
The earlier example about buying a shirt at the mall is an example illustrating the transitive property. It describes a chain reaction: if the first thing happens, then the second thing happens, and if the second thing happens, then the third thing happens. Therefore, if we want to ignore the second thing, we can say that if the first thing happens, then we know the third thing will happen. We don’t have to mention the part about buying jeans; we can simply say that the first event leads to the final event. We could even have more than two premises; as long as they form a chain reaction, the transitive property will give us a valid argument.
\(\begin{array} {ll} \text{Premise:} & \text{If a soccer player commits a reckless foul, she will receive a yellow card.} \\ \text{Premise:} & \text{If Hayley receives a yellow card, she will be suspended for the next match.} \\ \text{Conclusion:} & \text{If Hayley commits a reckless foul, she will be suspended for the next match.} \end{array}\)
If we let \(r=\) committing a reckless foul, \(y=\) receiving a yellow card, and \(s=\) being suspended, then our argument looks like this:
\(\begin{array} {ll} \text{Premise:} & r \rightarrow y \\ \text{Premise:} & y \rightarrow s \\ \text{Conclusion:} & r \rightarrow s \end{array}\)
This argument has the exact structure required to use the transitive property, so it is a valid argument.
Is this argument valid?
\(\begin{array} {ll} \text{Premise:} & \text{If the old lady swallows a fly, she will swallow a spider.} \\ \text{Premise:} & \text{If the old lady swallows a spider, she will swallow a bird.} \\ \text{Premise:} & \text{Premise:} & \text{If the old lady swallows a bird, she will swallow a cat.} \\ \text{Premise:} & \text{If the old lady swallows a cat, she will swallow a dog.} \\\text{Premise:} & \text{If the old lady swallows a dog, she will swallow a goat.} \\ \text{Premise:} & \text{If the old lady swallows a goat, she will swallow a cow.} \\ \text{Premise:} & \text{If the old lady swallows a cow, she will swallow a horse.} \\ \text{Premise:} & \text{If the old lady swallows a horse, she will die, of course.} \\ \text{Conclusion:} & \text{If the old lady swallows a fly, she will die, of course.} \end{array}\)
- Answer
-
This argument is valid by the Law of Syllogism, which can involve more than two premises, as long as they continue the chain reaction. The premises \(f \rightarrow s, s \rightarrow b, b \rightarrow c, c \rightarrow d\) \(d \rightarrow g, g \rightarrow w, w \rightarrow h, h \rightarrow x\) can be reduced to \(f \rightarrow x. \) (Because we had already used \(c\) and \(d\) we decided to use \(w\) for cow and \(x\) for death. If the old lady swallows the fly, she will eventually eat a horse and die.
Disjunctive syllogism is:
\(\begin{array} {ll} \text{Premise:} & p \vee q \\ \text{Premise:} & \sim p \\ \text{Conclusion:} & q \end{array}\)
The order of the two parts of the disjunction isn't important. In other words, we could have the premises \(p \vee q\) and \(\sim q,\) and the conclusion \(p\)
\(\begin{array} {ll} \text{Premise:} & \text{I can either drive or take the train.} \\ \text{Premise:} & \text{I refuse to drive.} \\ \text{Conclusion:} & \text{I will take the train.} \end{array}\)
If we let \(d=I\) drive and \(t=I\) take the train, then the symbolic representation of the argument is:
\(\begin{array} {ll} \text{Premise:} & d \vee t \\ \text{Premise:} & \sim d \\ \text{Conclusion:} & t \end{array}\)
This argument is valid because it has the form of a disjunctive syllogism. I have two choices, and one of them is not going to happen, so the other one must happen.
Is this argument valid?
\(\begin{array} {ll} \text{Premise:} & \text{Alison was required to write a 10-page paper or give a 5-minute speech.} \\ \text{Premise:} & \text{Alison did not give a 5-minute speech.} \\ \text{Conclusion:} & \text{Alison wrote a 10-page paper.} \end{array}\)
- Answer
-
Let \(p=\) wrote a paper and \(s=\) gave a speech.
\(\begin{array} {ll} \text{Premise:} & p \vee s \\ \text{Premise:} & -s \\ \text{Conclusion:} & p \end{array}\)
This argument is valid by Disjunctive Syllogism. Alison had to do one or the other; she didn’t choose the speech, so she must have chosen the paper.
Keep in mind that, when you are determining the validity of an argument, you must assume that the premises are true. If you don’t agree with one of the premises, you need to keep your personal opinion out of it. Your job is to pretend that the premises are true and then determine whether they force you to accept the conclusion. You may attack the premises in a court of law or a political discussion, of course, but here we are focusing on the structure of the arguments, not the truth of what they actually say.
We have just looked at four forms of valid arguments; there are two common forms that represent invalid arguments, which are also called fallacies.
The fallacy of the converse is:
\(\begin{array} {ll} \text{Premise:} & p \rightarrow q \\ \text{Premise:} & q \\ \text{Conclusion:} & p \end{array}\)
Notice that the second premise and the conclusion look like the converse of the first premise, \(q \rightarrow p\), but they have been detached. The fallacy of the converse incorrectly tries to assert that the converse of a statement is equivalent to that statement.
\(\begin{array} {ll} \text{Premise:} & \text{If I drink coffee after noon, then I have a hard time falling asleep that night.} \\ \text{Premise:} & \text{I had a hard time falling asleep last night.} \\ \text{Conclusion:} & \text{I drank coffee after noon yesterday.} \end{array}\)
If we let \(c=\mathrm{I}\) drink coffee after noon and \(h=\mathrm{I}\) have a hard time falling asleep, then our argument looks like this:
\(\begin{array} {ll} \text{Premise:} & c \rightarrow h \\ \text{Premise:} & h \\ \text{Conclusion:} & c \end{array}\)
This argument uses converse reasoning, so it is an invalid argument. There could be plenty of other reasons why I couldn’t fall asleep: I could be worried about money, my neighbors might have been setting off fireworks, …
Is this argument valid?
\(\begin{array} {ll} \text{Premise:} & \text{If you pull that fire alarm, you will get in big trouble.} \\ \text{Premise:} & \text{You got in big trouble.} \\ \text{Conclusion:} & \text{You must have pulled the fire alarm.} \end{array}\)
- Answer
-
Let \(f=\) pulled fire alarm and \(t=\) got in big trouble.
\(\begin{array} {ll} \text{Premise:} & f \rightarrow t \\ \text{Premise:} & t \\ \text{Conclusion:} & f \end{array}\)
The fallacy of the inverse is:
\(\begin{array} {ll} \text{Premise:} & p \rightarrow q \\ \text{Premise:} & \sim p \\ \text{Conclusion:} & \sim q \end{array}\)
Again, notice that the second premise and the conclusion look like the inverse of the first premise, \(\sim p \rightarrow \sim q\), but they have been detached. The fallacy of the inverse incorrectly tries to assert that the inverse of a statement is equivalent to that statement.
\(\begin{array} {ll} \text{Premise:} & \text{If you listen to the Grateful Dead, then you are a hippie.} \\ \text{Premise:} & \text{Sky doesn’t listen to the Grateful Dead.} \\ \text{Conclusion:} & \text{Sky is not a hippie.} \end{array}\)
If we let \(g=\) listen to the Grateful Dead and \(h=\) is a hippie, then this is the argument:
\(\begin{array} {ll} \text{Premise:} & g \rightarrow h \\ \text{Premise:} & \sim g \\ \text{Conclusion:} & \sim h \end{array}\)
This argument is invalid because it uses inverse reasoning. The first premise does not imply that all hippies listen to the Grateful Dead; there could be some hippies who listen to Phish instead.
Is this argument valid?
\(\begin{array} {ll} \text{Premise:} & \text{If a hockey player trips an opponent, he will be assessed a 2-minute penalty.} \\ \text{Premise:} & \text{Alexei did not trip an opponent.} \\ \text{Conclusion:} & \text{Alexei will not be assessed a 2-minute penalty.} \end{array}\)
- Answer
-
Let \(t=\) tripped and \(p=\) got a penalty.
\(\begin{array} {ll} \text{Premise:} & t \rightarrow p \\ \text{Premise:} & \sim t \\ \text{Conclusion:} & \sim p \end{array}\)
This argument is invalid because it has the form of the Fallacy of the Inverse. Alexei may have gotten a penalty for an infraction other than tripping.
Of course, arguments are not limited to these six basic forms; some arguments have more premises, or premises that need to be rearranged before you can see what is really happening. There are plenty of other forms of arguments that are valid or invalid. If an argument doesn’t seem to fit the pattern of any of these common forms, though, you may want to use a truth table instead.