# 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?

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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.

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That is perhaps the case, a poorly written example. Thank you, joriki. – Asakeron Dec 3 '12 at 14:24

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.

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By "Nothing is deducible from $R$" you mean not even $¬P$ given $¬R$? – Asakeron Dec 3 '12 at 14:38
I mean that $R$ being true proves nothing. – isturdy Dec 3 '12 at 15:34
I see. Thank you. – Asakeron Dec 3 '12 at 15:53