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I don't understand why the following argument is invalid:

        $5$ is not an even number.
        If $5$ is an even number, then $7$ is an even number.
$\therefore\quad 7$ is not an even number.

I was told that to show an argument is invalid, the hypotheses must be true while the conclusion is false.

In the above argument, we have

            True
            False $\rightarrow$ False
    $\therefore\quad$ True,

so the hypotheses and conclusion are all true, so isn't the argument actually valid?

ryang
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Not Friedrich gauss
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  • You are probably misinterpreting "$P\implies Q$ is false when $P$ is true and $Q$ is false". An "argument", i.e. a proof attempt, may be invalid (and may even not prove anything) even when the conclusion may be proved true by another (valid) argument. – Anne Bauval Jan 15 '23 at 17:59
  • Here "agument" means scheme of a proof. Then, there can be an "argument" that gives a true conclusion, in a specific case, but it does not in general cases. It would be wrong, Please, notice that the second sentence, "if 5 is...", has no meaning. You cannot state an implication between boolean values, you have to make an implication between assertions. E.g. "if x is even, then x+2 is even". Then you use the "modus ponens", e.g. giving a specific value of x. – Doriano Brogioli Jan 15 '23 at 18:01
  • Each statement is true, but the conclusion does not follow from the previous statements. In more fancy language, there is a model which includes the first two statements, but which proves that 7 is not an even number. Nothing about the first two statements entails the conclusion. When we say an argument is valid, we mean more strictly that any model including all of the statements in the argument must entail the conclusion. – Charles Hudgins Jan 15 '23 at 18:09
  • If you don't make this more technical requirement, you end up saying "grass is green, so 2 + 2 = 4" is a valid argument, which is absurd. – Charles Hudgins Jan 15 '23 at 18:11
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    You have to consider the form of the argument, and not a single example: the argument-form: "not-P, if P, then Q; therefore Q" is invalid because we can easily find counter-examples. – Mauro ALLEGRANZA Jan 16 '23 at 15:36
  • I think more context is needed here. This argument is perfectly valid (although rather pointless) if carried out in any reasonable theory of arithmetic and don't just take "$\therefore$" to mean "therefore by the theory of arithmetic". This argument is not valid if you interpret "$\therefore$" as meaning "therefore by modus ponens". – Rob Arthan Jan 16 '23 at 22:16
  • Please ignore the spurious words "don't just" in my comment above. – Rob Arthan Jan 16 '23 at 22:26

5 Answers5

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Generally speaking for statements $A,B$ from $A\Rightarrow B$ you may not conclude, that $\neg A \Rightarrow \neg B$, which is what is being done in the argument.

StiftungWarentest
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If $A$ means "5 is an even number" and $B$ means "7 is an even number", then formally the argument is:

$1.\; \lnot A$

$2.\; A \implies B$

$3.\; \lnot B$

However, notice that (2) is equivalent to:

$2a. \lnot B \implies \lnot A$

This is called the contrapositive of (2). Assuming that $\lnot A$ is true, means that $\lnot B$ could be either true or false, because $\lnot A$ being true means it doesn't matter what truth value $\lnot B$ takes, (2a) will be true regardless (the only way for it to be false would be if $\lnot B$ was true AND $\lnot A$ was false). Therefore it doesn't follow that $\lnot B$ is necessarily true.

656475
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bianchiniah
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... i was told that to show an argument is invalid the hypotheses must be true while the conclusion is false.

Please know that an argument is invalid if and only if it is possible for the premises to be true and the conclusion to be false. Whether it is actually the case that the premises are true and conclusion is false is not relevant when it comes to validity.

Of course, one way to demonstrate that an argument is invalid, i.e. to demonstrate that it is possible for the premises to be true and the conclusion to be false, is to create a possible world or scenario in which the premises are true and the conclusion is false.

So, while in the 'normal' world of arithmetic 5 and 7 are both not even, try to imagine a world where 5 is not even, but 7 is. Now, I understand that may be difficult to do, but one way to think about this is that ultimately logic is only about symbol structures ... logic has no 'understanding' of what '5', '7', or 'even' mean ... as far as logic is concerned, to say that '7 is even' is not any different from a claim like 'snow is white': it just says that there is some object that happens to be labeled '7', has some property that happens to be labeled 'even'. And if we have some further object labeled '5', and if we assume that that object does not have the property of 'even', then in that imagined scenario or world, both premises are true, but the conclusion is false, meaning that the argument is invalid.

So again, the premises don't have to be true and the conclusion false in our world for an argument to be invalid. You just need to have a logically possible world (which for classical logic means: claims cannot both be true and false at the same time) in which that is so ... so that need not be our world (or, in this case, the classical world of arithmetic).

Bram28
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As others have mentioned, the general form of the argument is invalid, except on a technicality. That technicality is that most proof systems allow you to conclude anything that’s already been proven, so assuming “7 is not even” has already been proven, then you can use any assumption to re-prove it. This is the gist behind the common axiom schema $P \to (Q \to P)$. This same general idea can be used as an inference rule which allows you to introduce anything that already has a proof into another proof or derivation.

PW_246
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        $5$ is not an even number.
        If $5$ is an even number, then $7$ is an even number.
$\therefore\quad 7$ is not an even number.

the hypotheses and conclusion are all true, so isn't the argument actually valid?

An argument's validity is assessed by considering its logical form: a valid argument's corresponding conditional is logically true—not merely true in your specific interpretation.

bianchiniah's answer points out that the above argument corresponds to the implication $$¬A\quad\text{and}\quad (A→B)\quad\quad\to\quad\quad¬B;$$ since this conditional is invalid (i.e., not logically true: at least one interpretation, $(A,B)=(\text{False, True}),$ falsifies it), the argument is invalid.

But I was told that to show an argument is invalid, the hypotheses must be true while the conclusion is false.

No, you are characterising a false conditional rather than an invalid argument: observe that the invalid argument this is the second planet from the Sun; therefore, this planet takes 12 Earth months to revolve around the Sun has a false premise and a true conclusion (though on Venus the conditional becomes false).

ryang
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