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From what I understand, this is how to prove that modus ponens is a valid argument:

An argument form is valid iff its corresponding conditional is a Tautology
[(P→Q)∧P]→Q is a Tautology
∴ modus ponens is a valid argument form

But how do we know that the above argument is valid? It seems to me that modus ponens is valid because modus ponens itself is valid.

ncc291203
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  • Ever since humans first noticed that if it is raining, then it is cloudy, modus ponens has been pretty much self-evident "common sense." The perfect candidate for "axiomhood" thousands of years later. IMHO – Dan Christensen Mar 22 '21 at 14:43
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    Consider a $P,Q$ truth table that examines the 4 possibilities, re $P$ is either true or its false, $Q$ is either true or its false. The premise $(P \implies Q)$ eliminates one of the $4$ possibilities, namely $(P \wedge [\neg Q]).$ Then, the 2nd premise, $(P)$ eliminates two of the $4$ possibilities, namely $([\neg P] \wedge Q)$ and $([\neg P] \wedge [\neg Q]).$ What you are left with is only 1 entry left in the truth table: $(P \wedge Q).$ – user2661923 Mar 22 '21 at 14:50
  • See Modus ponens – Mauro ALLEGRANZA Mar 22 '21 at 15:29
  • See this post – Mauro ALLEGRANZA Mar 22 '21 at 16:06
  • Actually, your argument isn't quite Modus Ponens, since it uses a biconditional 'iff': "An argument form is valid iff its corresponding conditional is a Tautology" But, this is easily remedied, since you can simply use "An argument form is valid if its corresponding conditional is a Tautology", and using that as a premise instead you indeed have Modus Ponens. The problem that you describe here is well-known, and is called "The Problem of Deduction". There is also a Problem of Induction which is very similar. Here is a nice video: https://www.youtube.com/watch?v=xO3Gw0zlmWM – Bram28 Mar 22 '21 at 17:11
  • @user2661923 So what you are saying is: "If you work out a truth-table for an argument and find that [blah], then the argument is valid. Well, we work out the truth-table for the Modus Ponens argument and find that [blah]. Therefore, Modus Ponens is valid." But stated as such, the OP will have the same objection: Your argument has the Modus Ponens pattern, and thus we are relying on the validity of Modus Ponens to justify the validity of Modus Ponens! – Bram28 Mar 22 '21 at 18:04
  • @Bram28 No, my argument will use a truth table + that $(P \implies Q)$ is equivalent to $\neg(P \wedge [\neg Q])$ to prove Modus Ponens. This means that I am not using Modus Ponens itself, so I am not using circular reasoning. I am however, using the truth table equivalent of $(P \implies Q)$. – user2661923 Mar 22 '21 at 18:18
  • @Bram28 By the way, in the youtube video that you cited, the use of truth tables to justify Modus Morons is in error. If you use the truth table approach, with the premises of $B$ and $A \implies B$, the truth table does not allow you to conclude $A$, because, within the truth table, the case of $A$ false, $B$ true has not been eliminated. – user2661923 Mar 22 '21 at 18:28
  • @user2661923 But in order to show that (P⟹Q) is equivalent to ¬(P∧[¬Q]) you have to use Modus Ponens so it still results in circular reasoning. So according to the video, I should just see Modus Ponens as a fundamental rule of inference that is simply valid because it intuitively makes sense that way and it works? – ncc291203 Mar 23 '21 at 14:35
  • @ncc291203 "But in order to show that $(P\implies Q)$ is equivalent to $\neg(P \wedge [\neg Q])$ you have to use Modus Ponens". No, you don't. What you have to do is present the truth table to the student and ask the student which of the 4 rows violates the premise that $(P \implies Q)$? Thus, the student does have to be able to relate the statement $(P \implies Q)$ to something. Note that asking the student to comprehend $(P \implies Q)$ is a simpler task than asking the student to comprehend Modus Ponens. – user2661923 Mar 23 '21 at 14:41

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