Is this valid Modus Ponens?

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?

• 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 $P\lor Q\to S$. – David C. Ullrich Sep 27 '20 at 14:10
• You have to use Addition: $P \to (P\lor Q)$ – Mauro ALLEGRANZA Sep 27 '20 at 14:53
• Okay, I see now. Thank you both – InsularPortrayal Sep 27 '20 at 15:20