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The Modus Ponens inference rule is generally expressed as:

$$ \begin{array}{rl} & P\rightarrow Q \\ & P \\ \hline \therefore & Q\end{array} $$

Is the below rule also considered to be Modus Ponens?

$$ \begin{array}{rl} & P \lor \lnot Q \\ & Q \\ \hline \therefore & P\end{array} $$

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    $\begingroup$ Translate $P \lor \lnot Q$ into $\lnot Q \lor P$ (by commutativity of $\lor$) and then to an implication $\endgroup$
    – Bernard Masse
    Mar 18, 2019 at 20:16
  • $\begingroup$ Ah! I don't know why I didn't think of that! Thanks! $\endgroup$
    – Abhilash Kishore
    Mar 18, 2019 at 20:27
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    $\begingroup$ You didn't think of it because you are new to the game. By practice and by asking questions when you are stuck, you'll become an expert or at least a good player in the game of logic. $\endgroup$
    – Bernard Masse
    Mar 18, 2019 at 20:34
  • $\begingroup$ Yes; it is a particular case of the Resolution rule. $\endgroup$
    – Mauro ALLEGRANZA
    Mar 19, 2019 at 8:10

2 Answers 2

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Yes.

Thanks to @Bernard Massé for pointing me in the right direction.

Here's the proof:

  1. $(P \lor \lnot Q)$ can be written as $ (\lnot Q \lor P) $ - Commutative Property

  2. $ (\lnot Q \lor P) $ can be written as $ (Q \rightarrow P) $ - Material Implication

  3. By Modus Ponens :

$$ \begin{array}{rl} & Q\rightarrow P \\ & Q \\ \hline \therefore & P\end{array} $$ This is equivalent to

$$ \begin{array}{rl} & P \lor \lnot Q \\ & Q \\ \hline \therefore & P\end{array} $$

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    $\begingroup$ This is an answer to a slightly different question than what you asked which is whether the latter rule would be considered Modus Ponens. I, personally, would consider it (a special case of) resolution. For me to call it Modus Ponens, I would have had to define $P\to Q$ as a syntactic abbrevation for $\neg P\lor Q$. Part of the reason I have this view is that the "Material Implication" equivalence is not derivable in constructive logic while both of the rules you list are, so these rules are talking about different things. $\endgroup$
    – Derek Elkins left SE
    Mar 19, 2019 at 0:03
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I would consider it an application of Disjunctive Syllogism, which is typically stated as:

$P \lor Q$

$\neg P$

$\therefore Q$

Of course, that is not exactly the same pattern, but the basic idea of Disjunctive Syllogism is that you have two options ... but it isn't one of them, and therefore you are left with the other one. Your argument is like that too: it is either P, or $\neg Q$, but given $Q$ it is not $\neg Q$, and so you are left with $P$

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