@Bram28
In your last paragraph of your answer http://math.stackexchange.com/q/2171958, you would benefit from changing the line: "All we assume is that P(n), so it will take some work to go from there to P(n+1)."
To
All we assume is that $P(n)$ implies $P(n+1)$, so it will take some work to go from there to $\forall n P(n)$.
It appears that the misinterpretation some people have made is reading that last line as containing $\forall n P(n)$ and $\forall n P(n+1)$ as opposed to $\forall n [P(n) \implies P(n+1)]$ .
In your last paragraph of your answer http://math.stackexchange.com/q/2171958, you would benefit from changing the line: "All we assume is that P(n), so it will take some work to go from there to P(n+1)."
To
All we assume is that $P(n)$ implies $P(n+1)$, so it will take some work to go from there to $\forall n P(n)$.
It appears that the misinterpretation some people have made is reading that last line as containing $\forall n P(n)$ and $\forall n P(n+1)$ as opposed to $\forall n [P(n) \implies P(n+1)]$ .