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I am taking a course in discrete structures and have just started using rules of inference. One of my homework problems contains the following step:

$$P \vee Q \to S$$

I also know from the problem that P is true, so I can write:

$$P \vee Q \to S$$

$$P$$

Therefore S.

Could I call this Modus Ponens? Or is it something else?

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    $\begingroup$ No, it's not quite Modus Ponens. The first step would be deriving $P\lor Q$ from $P$; now $S$ follows by Modus Ponens from $P\lor Q$ and $P\lor Q\to S$. $\endgroup$ – David C. Ullrich Sep 27 '20 at 14:10
  • $\begingroup$ You have to use Addition: $P \to (P\lor Q)$ $\endgroup$ – Mauro ALLEGRANZA Sep 27 '20 at 14:53
  • $\begingroup$ Okay, I see now. Thank you both $\endgroup$ – InsularPortrayal Sep 27 '20 at 15:20
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Well you skipped Addition. ie. If you are given a true statement P then you can infer P or Q where Q is some other statement. And then you can use Modus Ponens.

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  • $\begingroup$ "or" is used in Inclusive sense. $\endgroup$ – user824627 Sep 27 '20 at 15:43
  • $\begingroup$ "or" refers to Disjunction $\endgroup$ – user824627 Sep 27 '20 at 15:44

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