"Modus moron" rule of inference?

Question

This is an exercise I got from the book "First Order Mathematical Logic" by Angelo Margaris (1967). I have never heard of this rule before, the question is whether what Margaris calls the modus moron rule of inference is correct or not and to explain why I think so.

$$\frac{P\Rightarrow Q, Q}{\therefore P}\qquad \text{(modus moron)}$$

It seems correct to me, my reasoning is that if $P\Rightarrow Q$ and $Q$ it does not matter whether $P$ or $\neg P$ since a false antecedent makes a true conditional, which I would show by the rows of the truth table of $(P\Rightarrow Q)$ where $Q$ is true.

Is this a valid argument?

Answer 1

It's true that if $P$ is false then $P\Rightarrow Q$ is true. But the question is not asking if $P\Rightarrow Q$ is true, it's asking you if you can infer $P$ from $P\Rightarrow Q$ and $Q$.

Let's be concrete. Suppose "if Mark is drunk, then Mark is happy" is a true statement, because Mark is a happy drunk. Given that Mark is presently happy, may we infer that Mark is drunk? No; there may be other circumstances in which Mark is happy besides being drunk.

(This example is drawn from a real life discussion between friends about putative alcoholism which devolved into a debate about logical implication.)

Answer 2

This pattern is a logical fallacy called Affirming the Consequent, though I often call it Modus Bogus.

To show it is not a valid inference, here is a simple Refutation by Logical Analogy:

If I have blond hair, I have hair

I have hair

Therefore, I have blond hair

Here is my favorite logical fallacy:

$$\frac{}{\therefore P}\qquad \text{(hokus ponens)}$$

Answer 3

Even if affirming the consequent is not valid, other logical rules still work. Other logical rules still work. Therefore, affirming the consequent is not valid?

Answer 4

Not stated in the other answers so far is that you misunderstood the meaning of logical validity, which means that, in every situation where the premises hold, the conclusion also holds. Now you may have observed that if $Q$ is true then $(P \to Q)$ is also true, by looking at the truth table of implication. But that is not what you should be looking for. Rather you should check whether $P$ is true for every possible choice of truth values for $P,Q$ such that $(P \to Q)$ and $Q$ are both true. If so, then the argument is valid. If not, then the argument is invalid and you can identify precisely the situations where the argument will fail (meaning that it gives a false conclusion despite true premises). Two such failure situations have already been mentioned by Bram28 and arctic tern.

Answer 5

As others have said, the question is asking whether it is necessarily the case that whenever $P \Rightarrow Q$ and $Q$ are true, $P$ must be true.

My personal favoured instantiations of $P$ and $Q$ are 'it is Saturday' and 'it is the weekend' respectively; modus morons would enable us to infer:

(1) If it is Saturday, then it is the weekend. [premise]
(2) It is the weekend. [premise]
Therefore, it is Saturday. [MM 1,2]

Modus morons has wider interest as an illustration of how justifications of deductive logic are nearly always circular. In 'The Justification of Deduction' Susan Haack (1976) compares a proposed justification of modus ponens (which we do consider valid) to a parallel justification of modus morons (which we do not).

Answer 6

if P⇒Q and Q it does not matter whether P or ¬P

That's the point. It doesn't matter whether $P$ or $\neg P$: The statement becomes true for $P$ and $\neg P$ either.
Therefore, you can not deduce that it must be $P$. It could be $P$, but it might just as well be $\neg P$, because both would entail the truth of the statement.

By writing

$$\frac{A}{\therefore B}$$

you say that $B$ necessarily follows from $A$, i.e., "In all cases where $A$ is true, $B$ must be true as well", so, given that the premise is true, there would be no other logical possiblity than $B$ being true.

But this is isn't the case in our example: $P$ might as well be false, thereby still pertaining the truth of $P \to Q, Q$. It isn't a logical necessity for $P$ to be true; because it doesn't matter whether $P$ or $\neg P$, it could just as well be the case that $P$ is false and $\neg P$ is true instead.

Therfore, $P$ doesn't logically follow from $P \to Q, Q$: It isn't a necessity for $P$ to hold given the premises; there would be no contradiction if $P$ didn't hold.
This is why making a logical inference from $P \to Q, Q$ to $P$, i.e. a statement of the form "If $P \to Q, Q$ holds, then $P$ must hold as well" is invalid.

Answer 7

The other answers have perfectly settled this question, however I thought I would share another funny example. It is not exactly the same logical structure, but very similar. This example originates from the satirical play Erasmus Montanus written in 1722 by Danish-Norwegian writer Ludvig Holberg. In Danish it goes like this

En sten kan ikke flyve.
Morlille kan ikke flyve.
Ergo er morlille en sten!

In English it translates to

A rock cannot fly.
Little mother cannot fly.
Ergo [Therefore] little mother is a rock!

Answer 8

I strongly disagree with demeaning this rule of inference. It is a very common problem solving technique to use the inference:

$$\begin{array} {rl} & P \to Q \\ & Q \\ & Q \text{ is really rare} \\ \hline \therefore & P \text{ is almost true} \end{array}$$

For example, your task might be to prove $P$ as "the maximum of $f$ occurs at $x$ (for the sake of discussion ignoring endpoints)". And you know that at the maximum of the function, $f' = 0$, which is the $Q$. So $P \to Q$. But you also know that $f' = 0$ is an infrequent occurrence on an arbitrary graph. So now you have the 3 ingredients you need to almost establish $P$, all that is left is finding that little bit of missing information, $f'' = 0$ and all other (finite number of usually) maximal points are lesser.

If this inference were useless we wouldn't even have a definition for the concept of stationary points. I've also used this technique used in algorithm design, "what do we know about the best possible result" and use that to find the algorithm, then use further investigation to prove what is found is optimal.

So rather than "modus moron", I would suggest calling it "modus hopeful".

Answer 9

This will, in all likelihood, be drowned out by the other answers. But, a mathematical version of this, is as follows:

$$p>3,p\in \mathbb P\implies p\in\{x:\text{Mod}(x,6)\in\{1,5\}\},25\in\{x:\text{Mod}(x,6)\in\{1,5\}\}\over\therefore 25\in\mathbb P$$