Divisor of Integer/Examples/6 divides n (n+1) (n+2)/Proof 2

Proof
Proof by induction:

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $6 \divides n \paren {n + 1} \paren {n + 2}$

$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $6 \divides k \paren {k + 1} \paren {k + 2}$

from which it is to be shown that:
 * $6 \divides \paren {k + 1} \paren {k + 2} \paren {k + 3}$

Induction Step
This is the induction step:

From the induction hypothesis we have that:
 * $6 \divides k \paren {k + 1} \paren {k + 2}$

Hence by definition of divisibility, we have:
 * $(1): \quad \exists r \in \Z: k \paren {k + 1} \paren {k + 2} = 6 r$

From $2$ divides $n \paren {n + 1}$ we have:
 * $(2): \quad \exists s \in \Z: \paren {k + 1} \paren {k + 2} = 2 s$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: 6 \divides n \paren {n + 1} \paren {n + 2}$