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I'm reading Exner's: Logic in Elementary Mathematics and trying to understand the modus ponens. When we have:

$$\cfrac{P, P\to Q}{Q}$$

And we can prove $P$ and $P\to Q$ is true, then $Q$ is true. I am trying to apply it to an exercise in Lang's Linear Algebra:

For $c\neq0$, if $cV=0$ then $V=0$. ($c$ is a number and $V$ is a vector).

Now I guess that I have:

$$\stackrel{\neq 0}{c}V=0 \to V=0$$

I believe I quite understand the meaning of proving $P\to Q$ in this case. But what is the meaning of proving that $P$ is true? That is: I have to prove that $\stackrel{\neq 0}{c}V=0$ but after all, it can be zero if $V=0$ and not zero if $V\neq 0$. Perhaps I have articulated it wrong? Or proving $P$ means finding a $V$ such that the proposition is true?

Notice that I am asking how to apply the modus ponens to this situation. I know how to prove the proposition "without" the modus ponens.

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    I assume you intend $P = \{cV=0\wedge c\ne0\}$ and $Q= \{V=0\}$? – 2017-02-11
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    @spaceisdarkgreen Yes. Perhaps this is where I messed up? – 2017-02-11
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    If you've proven $P\to Q$ and $P$ then Modus Ponens allows you to conclude $Q.$ So you're proving $Q$ not $P\to Q$ – 2017-02-11
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    @spaceisdarkgreen I know. But I *still didn't prove those two.* I am trying to do it now, but I'm a little bit confused as to the meaning of proving $P$. – 2017-02-11
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    In this linear algebra problem, you don't want to use modus ponens. You want to prove $P\to Q$. Modus ponens is used to prove $Q$, given $P$ and $P\to Q$. – 2017-02-11
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    @OppaHilbertStyle Proving $P$ means proving that $c\ne 0$ and $cV=0$ Proving $P\to Q$ means proving the highlighted theorem you are talking about. Do those however you want and then modus ponens allows you to conclude that $V=0.$ – 2017-02-11
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    @GitGud It's not quite clear why modus ponens is inadequate here. Can you expand a little bit? – 2017-02-11
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    @OppaHilbertStyle Of course we might have $c=0$ or $cV\ne 0$ in which case you wouldn't be able to prove $P$ and thus wouldn't be able to conclude that $V=0$ – 2017-02-11
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    @OppaHilbertStyle I think what GitGud is saying is "You use modus ponens to conclude $Q$ from $P$ and $P\to Q.$ Not to conclude that $P\to Q$" – 2017-02-11
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    Oh. I get it now! – 2017-02-11
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    @spaceisdarkgreen is correct. In mathematical practice one proves $P\to Q$ all the time and one uses such statements to prove $Q$ in another context, where $P$ happens to hold. – 2017-02-11
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    You are not applying *modus ponens*; you are asked to prove the conditional : "if $c \ne 0$, then (if $cV=0$, then $V=0)$, i.e. $p \to (q \to r)$ that is equivalent to: $(p \land q) \to r$. – 2017-02-12
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    The "logical form" of the proof is: assume $c \ne 0$ and assume $cV=0$ and derive $V=0$. The result follows by $\to$-introduction (i.e. by Conditional Proof). In terms of inference rules, *modus ponens* is $\to$-elimination. – 2017-02-12

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