Can I infer $P$ given only $P\to (Q\wedge R)$ and $R=true$ In an introductory class, the professor showed us the following example:
Anyone who loves nature($P$) certainly also loves animals($Q$) and plants($R$). Which was translated to the expression $P\to (Q\wedge R)$, which, also sounds strange, but I'll use it as an example.
Then, given the information that Celina loves plants ($R=true$), we were told that it's possible to infer that Celinas does not love nature.
My initial thought was that I could not infer that because I don't have any information regarding Celina's love to animals.
I figure it as:
$P\to (Q\wedge R)$
$P\to (?\wedge true)$
$P\to ?$
So, I can not infer any value to $P$ for this situation.
Am I missing something?
 A: Somewhere along the way from the professor to us a "not" must have gone missing. Given that Celina doesn't love plants, you can infer that she doesn't love nature. You're right that the inference as you stated it isn't valid.
A: I think that you are correct. By your understanding of your professor, he asserts $(P \to Q \wedge R) \to (R \to \lnot P)$, but since $(P \to Q \wedge R) \to (P \to R)$, if that understanding is correct we have $(P \to Q \wedge R) \to (R \to \lnot P) \wedge (P \to R)$, from which we can deduce $\lnot P$ (since $P \to R \to \lnot P$). Since $\lnot P$ is obviously not deducible from $P \to Q \wedge R$, neither is $R \to \lnot P$.
I think that he meant to add a not ($\lnot R \to \lnot P$); we can obtain $P \to R$ from $P \to Q \wedge R$, and that proposition follows by contraposition. Nothing is deducible from $R$ (in general, affirming the truth of part of the consequent gets you nowhere. It is actually not even relevant that you do not know whether $Q$; $P \wedge Q$ also proves nothing. 
