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Mathematics LibreTexts

2.7: Validity of Arguments and Common Errors

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    An argument is said to be valid or to have a valid form if each deduction in it can be justified with one of the rules of inference listed in the previous section. The form of an argument might be valid, but still the conclusion may be false if some of the premises are false. So to show that an argument is good we have to be able to do two things: show that the argument is valid (i.e. that every step can be justified) and that the argument is sound which means that all the premises are true. If you start off with a false premise, you can prove anything!

    Consider, for example the following “proof” that \(2 = 1\).

    Suppose that \(a\) and \(b\) are two real numbers such that \(a = b\).

    \[a^2 = ab \tag{by hypothesis, \(a\) and \(b\) are equal, so} \]

    \[a^2 − b^2 = ab − b^2 \tag{subtracting \(b^2\) from both sides} \]

    \[(a + b)(a − b) = b(a − b) \tag{factoring both sides} \]

    \[ a + b = b \tag{canceling \((a − b)\) from both sides} \]

    Now let \(a\) and \(b\) both have a particular value, \(a = b = 1\), and we see that \(1 + 1 = 1\), i.e. \(2 = 1\).

    This argument is not sound (thank goodness!) because one of the premises – actually the bad premise appears as one of the justifications of a step – is false. You can argue with perfect logic to achieve complete nonsense if you include false premises.


    It is not true that you can always cancel the same thing from both sides of an equation. Under what circumstances is such cancellation disallowed?

    So, how can you tell if an argument has a valid form? Use a truth table. As an example, we’ll verify that the rule of inference known as “destructive dilemma” is valid using a truth table. This argument form contains \(4\) predicate variables so the truth table will have \(16\) rows. There is a column for each of the variables, the premises of the argument and its conclusion.

    \(A\;B\;C\;D\) \(A \implies B\) \(C \implies D\) \(¬B ∨ ¬D\) \(¬A ∨ ¬C\)
    \(\text{T} \;\text{T} \;\text{T} \;\text{T}\) \(\text{T}\) \(\text{T}\) \(\phi\) \(\phi\)
    \(\text{T} \;\text{T} \;\text{T} \; \phi\) \(\text{T}\) \(\phi\) \(\text{T}\) \(\phi\)
    \(\text{T} \;\text{T} \;\phi \;\text{T}\) \(\text{T}\) \(\text{T}\) \(\phi\) \(\text{T}\)
    \(\text{T} \;\text{T} \;\phi \;\phi\) \(\text{T}\) \(\text{T}\) \(\text{T}\) \(\text{T}\)
    \(\text{T} \;\phi \;\text{T} \;\text{T}\) \(\phi\) \(\text{T}\) \(\text{T}\) \(\phi\)
    \(\text{T} \;\phi \;\text{T} \;\phi\) \(\phi\) \(\phi\) \(\text{T}\) \(\phi\)
    \(\text{T} \;\phi \;\phi \;\text{T}\) \(\phi\) \(\text{T}\) \(\text{T}\) \(\text{T}\)
    \(\text{T} \;\phi \;\phi \;\phi\) \(\phi\) \(\text{T}\) \(\text{T}\) \(\text{T}\)
    \(\phi \;\text{T} \;\text{T} \;\text{T}\) \(\text{T}\) \(\text{T}\) \(\phi\) \(\text{T}\)
    \(\phi \;\text{T} \;\text{T} \;\phi\) \(\text{T}\) \(\phi\) \(\text{T}\) \(\text{T}\)
    \(\phi \;\text{T} \;\phi \;\text{T}\) \(\text{T}\) \(\text{T}\) \(\phi\) \(\text{T}\)
    \(\phi \;\text{T} \;\phi \;\phi\) \(\text{T}\) \(\text{T}\) \(\text{T}\) \(\text{T}\)
    \(\phi \;\phi \;\text{T} \;\text{T}\) \(\text{T}\) \(\text{T}\) \(\text{T}\) \(\text{T}\)
    \(\phi \;\phi \;\text{T} \;\phi\) \(\text{T}\) \(\phi\) \(\text{T}\) \(\text{T}\)
    \(\phi \;\phi \;\phi \;\text{T}\) \(\text{T}\) \(\text{T}\) \(\text{T}\) \(\text{T}\)
    \(\phi \;\phi \;\phi \;\phi\) \(\text{T}\) \(\text{T}\) \(\text{T}\) \(\text{T}\)

    Now, mark the lines in which all of the premises of this argument form are true. You should note that in every single situation in which all the premises are true the conclusion is also true. That’s what makes “destructive dilemma” – and all of its friends – a rule of inference. Whenever all the premises are true so is the conclusion. You should also notice that there are several rows in which the conclusion is true but some one of the premises isn’t. That’s okay too, isn’t it reasonable that the conclusion of an argument can be true, but at the same time the particulars of the argument are unconvincing?

    As we’ve noted earlier, an argument by deductive reasoning can go wrong in only certain well-understood ways. Basically, either the form of the argument is invalid, or at least one of the premises is false. Avoiding false premises in your arguments can be trickier than it sounds – many statements that sound appealing or intuitively clear are actually counter-factual. The other side of the coin, being sure that the form of your argument is valid, seems easy enough – just be sure to only use the rules of inference as found in Table 2.6.1. Unfortunately, most arguments that you either read or write will be in prose, rather than appearing as a formal list of deductions. When dealing with that setting – using natural rather than formalized language – making errors in form is quite common.

    Two invalid forms are usually singled out for criticism, the converse error and the inverse error. In some sense, these two apparently different ways to screw up are really the same thing. Just as a conditional statement and its contrapositive are known to be equivalent, so too are the other related statements – the converse and the inverse – equivalent. The converse error consists of mistaking the implication in a modus ponens form for its converse.

    The converse error:

    \[\begin{array} &&B \\ &\underline{A \implies B} \\ ∴ &A \end{array}\]

    Consider, for a moment the following argument.

    If a rhinoceros sees something on fire, it will stomp on it.

    A rhinoceros stomped on my duck.

    Therefore, the rhino must have thought that my duck was on fire.

    It is true that rhinoceroses have an instinctive desire to extinguish fires. Also, we can well imagine that if someone made this ridiculous argument that their duck must actually have been crushed by a rhino. But, is the conclusion that the duck was on fire justified? Not really, what the first part of the argument asserts is that “(on fire) implies (rhino stomping)” but couldn’t a rhino stomp on something for other reasons? Perhaps the rhino was just ill-tempered. Perhaps the duck was just horrifically unlucky.

    The closer the conditional is to being a biconditional, the more reasonable sounding is an argument exhibiting the converse error. Indeed, if the argument actually contains a biconditional, the “converse error” is not an error at all.

    The following is a perfectly valid argument, that (sadly) has a false premise.

    You will get an A in your Foundations class if and only if you read Dr. Fields’ book.

    You read Dr. Fields’ book.

    Therefore, you will get an A in Foundations.

    Suppose that we try changing the major premise of that last argument to something more believable.

    If you read Dr. Fields’ book, you will pass your Foundations class.

    You did not read Dr. Fields’ book.

    Therefore, you will not pass Foundations.

    This last argument exhibits the so-called inverse error. It is by no means meant as a guarantee, but nevertheless, it seems reasonable that if someone reads this book they will pass a course on this material. The second premise is also easy to envision as true, although the “you” that it refers to obviously isn’t you, because you are reading this book! But even if we accept the premises as true, the conclusion doesn’t follow. A person might have read some other book that addressed the requisite material in an exemplary way.

    Notice that the names for these two errors are derived from the change that would have to be made to convert them to modus ponens. For example, the inverse error is depicted formally by:

    \[\begin{array} &&¬A \\ &\underline{A \implies B} \\ ∴ &¬B \end{array}\]

    If we replaced the conditional in this argument form by its inverse \((¬A \implies ¬B)\) then the revised argument would be modus ponens. Similarly, if we replace the conditional in an argument that suffers from the converse error by its converse, we’ll have modus ponens.


    Exercise \(\PageIndex{1}\)

    Determine the logical form of the following arguments. Use symbols to express that form and determine whether the form is valid or invalid. If the form is invalid, determine the type of error made. Comment on the soundness of the argument as well, in particular, determine whether any of the premises are questionable.

    1. All who are guilty are in prison. George is not in prison. Therefore, George is not guilty.
    2. If one eats oranges one will have high levels of vitamin C. You do have high levels of vitamin C. Therefore, you must eat oranges.
    3. All fish live in water. The mackerel is a fish. Therefore, the mackerel lives in water.
    4. If you’re lazy, don’t take math courses. Everyone is lazy. Therefore, no one should take math courses.
    5. All fish live in water. The octopus lives in water. Therefore, the octopus is a fish.
    6. If a person goes into politics, they are a scoundrel. Harold has gone into politics. Therefore, Harold is a scoundrel.
    Exercise \(\PageIndex{2}\)

    Below is a rule of inference that we call extended elimination.

    \[\begin{array} &&(A ∨ B) ∨ C \\ &¬A \\ &\underline{¬B\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ ∴ &C ​\end{array}\]

    Use a truth table to verify that this rule is valid.

    Exercise \(\PageIndex{3}\)

    If we allow quantifiers and open sentences in an argument form we get a couple of new argument forms. Arguments involving existentially quantified premises are rare – the new forms we are speaking of are called “universal modus ponens” and “universal modus tollens.” The minor premises may also be quantified or they may involve particular elements of the universe of discourse – this leads us to distinguish argument subtypes that are termed “universal” and “particular.”

    For example,

    \(\begin{array} &&∀x, A(x) \implies B(x) \\ &\underline{A(p)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ ∴ &B(p) ​​\end{array}\)

    is the particular form of universal modus ponens (here, \(p\) is not a variable – it stands for some particular element of the universe of discourse) and

    \(\begin{array} &&∀x, A(x) \implies B(x) \\ &\underline{∀x, ¬B(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ ∴ &∀x, ¬A(x) ​​\end{array}\)

    universal form of (universal) modus tollens.

    Reexamine the arguments from problem (1), determine their forms (including quantifiers) and whether they are universal or particular.

    Exercise \(\PageIndex{4}\)

    Identify the rule of inference being used.

    1. The Buley Library is very tall. Therefore, either the Buley Library is very tall or it has many levels underground.
    2. The grass is green. The sky is blue. Therefore, the grass is green and the sky is blue.
    3. \(g\) has order \(3\) or it has order \(4\). If \(g\) has order \(3\), then \(g\) has an inverse. If \(g\) has order \(4\), then \(g\) has an inverse. Therefore, \(g\) has an inverse.
    4. \(x\) is greater than \(5\) and \(x\) is less than \(53\). Therefore, \(x\) is less than \(53\).
    5. If \(a|b\), then \(a\) is a perfect square. If \(a|b\), then \(b\) is a perfect square. Therefore, if \(a|b\), then \(a\) is a perfect square and \(b\) is a perfect square.
    Exercise \(\PageIndex{5}\)

    Read the following proof that the sum of two odd numbers is even. Discuss the rules of inference used.

    Proof: Let \(x\) and \(y\) be odd numbers. Then \(x = 2k + 1\) and \(y = 2j + 1\) for some integers \(j\) and \(k\). By algebra,

    \(x + y = 2k + 1 + 2j + 1 = 2(k + j + 1).\)

    Note that \(k +j +1\) is an integer because \(k\) and \(j\) are integers. Hence \(x + y\) is even.


    Exercise \(\PageIndex{6}\)

    Sometimes in constructing a proof we find it necessary to “weaken” an inequality. For example, we might have already deduced that \(x < y\) but what we need in our argument is that \(x ≤ y\). It is okay to deduce \(x ≤ y\) from \(x < y\) because the former is just shorthand for \(x < y ∨ x = y\). What rule of inference are we using in order to deduce that \(x ≤ y\) is true in this situation?