I want to give a direct proof of a conditionnal $$\forall\, n\in \Bbb Z : \left[ P(n) \Rightarrow \lnot Q(n) \right]$$ such that $$P(n)= n>2$$ and$$ Q(n)= \exists\ m \in \Bbb Z: (\,m+n=mn \wedge n|m\,)$$ hence, I want to prove that $$\forall\, n\in \Bbb Z : \left[\,n > 2 \Rightarrow \lnot \exists\ m \in \Bbb Z: (\,m+n=mn \wedge n|m\,) \,\right] $$ is true.
I think that I proved the statement:
I take the converse, so $Q(n) \Rightarrow \lnot P(n)$ and assume $Q(n)$ hence, $n|m$.
It follows that $$m=nk: k\in\Bbb Z $$ Thus, $$k=\frac 1{n-1} \Rightarrow \left[m=\frac n{n-1}\in \Bbb Z \iff n=2\right] \Rightarrow \lnot P(n)$$
Therefore, $$\left[ Q(n) \Rightarrow \lnot P(n) \right]\Rightarrow \left[P(n) \Rightarrow \lnot Q(n)\right]$$ is true, by contrapositive.
What would be a direct proof of this statement? Is the contrapositive considered a form of direct proof?