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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?

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    $\begingroup$ (Why do you think they called it modus moron?) $\endgroup$ – arctic tern Mar 7 '17 at 1:09
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    $\begingroup$ While the official name for this logical fallacy is Affirming the Consequent, I have always known this one as Modus Bogus! Another one is Hokus Ponens: $$\frac{}{\therefore P} \quad \text{(Hokus Ponens)}$$ $\endgroup$ – Bram28 Mar 7 '17 at 1:13
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    $\begingroup$ @Bram28: How about Modus Pokus? =) $\endgroup$ – user21820 Mar 7 '17 at 4:07
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    $\begingroup$ "It doesn't matter whether not-P or P" -- then surely you can't conclude P, right? $\endgroup$ – BallpointBen Mar 7 '17 at 22:12
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    $\begingroup$ Note that technically, you can define define whatever inferences rules you like in your logic, so it's not inherently incorrect. It just flies in the face of common intuition $\endgroup$ – gardenhead Mar 8 '17 at 16:33
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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.)

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    $\begingroup$ You should add a link to an AA site. $\endgroup$ – Mikhail Katz Mar 8 '17 at 14:19
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    $\begingroup$ You have impressive friends! $\endgroup$ – Nox Mar 8 '17 at 15:58
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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:

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

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    $\begingroup$ I think an example where the premises are both true may be much clearer. The second premise, "I am a coin", is false and obfuscates the counterexample. $\endgroup$ – 6005 Mar 7 '17 at 16:29
  • $\begingroup$ @6005 good point! I'll change it to something more obvious $\endgroup$ – Bram28 Mar 7 '17 at 17:13
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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?

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    $\begingroup$ Lol. Nice joke, but not obvious unless you already know the fallacy of affirming the consequent. =) $\endgroup$ – user21820 Mar 7 '17 at 12:28
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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.

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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).

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  • $\begingroup$ I haven’t read that book, but it’s pretty easy to derive logical contradictions from a putative rule that allows us to flip any arrow from right to left. I made a joke about it in my answer, but a more formal one might be that False → True is vacuously true, since the thing on the right is (literally) true. So this rule trivially proves a logical contradiction. I’ve usually seen modus ponens proved by truth table. $\endgroup$ – Davislor Mar 7 '17 at 16:02
  • $\begingroup$ @Davislor But as Haack explains, when you use a truth table to justify MP you're really arguing something like "Suppose that P > Q and P are true. By the truth table, P > Q and P imply Q. So Q is true." But this is just a use of MP in the metalanguage, and we could write something parallel using MM to justify MM. $\endgroup$ – dbmag9 Mar 7 '17 at 21:17
  • $\begingroup$ One difference being that MM can equally well be used to disprove itself, while MP is consistent. In some reasoning systems, such as Natural Deduction, MP is indeed an axiom. But you could also. e.g. start with some identities of Boolean algebra, transform p∧(p→q) into p∧(¬p∨q), then (p∧¬p)∨(p∧q), false∨(p∧q), p∧q, and finally extract q. Or MP is derivable from Hilbert’s system of axiomatic propositional logic. As you know, the question of which axioms to use can get pretty arbitrary. But we want them to be powerful enough to prove interesting things, consistent, intuitive and elegant. $\endgroup$ – Davislor Mar 7 '17 at 22:25
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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.

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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!

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    $\begingroup$ A funny counterexample, though I think we nowadays know the similar argument: "Witches burn, wood burns, wood floats, ducks float, ergo someone weighing the same as a duck means they're made of wood, hence they're a witch." $\endgroup$ – Kevin Long Mar 7 '17 at 18:43
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    $\begingroup$ If it has a tail it's a monkey; if it doesn't have a tail it's an ape. My desk doesn't have a tail... it's an ape... my dog has a tail ... it's a monkey. A hilarious song from a children's video involving animated vegetables called VeggieTales. If you don't know VeggieTales, you have my sympathies. :) youtube.com/watch?v=--szrOHtR6U $\endgroup$ – Kevin Mar 8 '17 at 19:35
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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".

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

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

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