Element of Leibniz Harmonic Triangle as Sum of Elements on Diagonal from Below/Lemma 2

Theorem
Consider the Leibniz harmonic triangle:

Let $\tuple {n, m}$ be the element in the $n$th row and $m$th column.

Then:
 * $\ds \forall r \in \N_{>0}: \tuple {n, m} = \tuple {n + r, m + r} + \sum_{k \mathop = 1}^r \tuple {n + k, m + k - 1}$

That is, each number in the Leibniz harmonic triangle is equal to the sum of the number below it, $\paren {r - 1}$ numbers diagonally below that number, and the number to the right of the last number. This is similar to Sum of r+k Choose k up to n (Hockey-stick Identity) for Pascal's triangle.

Proof
Proof by induction:

For all $r \in \N_{>0}$, let $\map P r$ be the proposition:
 * $\ds \tuple {n, m} = \tuple {n + r, m + r} + \sum_{k \mathop = 1}^r \tuple {n + k, m + k - 1}$

Basis for the Induction
$\map P 1$ is:
 * $\tuple {n, m} = \tuple {n + 1, m + 1} + \tuple {n + 1, m}$

which is shown in Element of Leibniz Harmonic Triangle as Sum of Elements on Diagonal from Below/Lemma 1.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P s$ is true, where $s \ge 1$, then it logically follows that $\map P {s + 1}$ is true.

So this is our induction hypothesis:
 * $\ds \tuple {n, m} = \tuple {n + s, m + s} + \sum_{k \mathop = 1}^s \tuple {n + k, m + k - 1}$

Then we need to show:
 * $\ds \tuple {n, m} = \tuple {n + s + 1, m + s + 1} + \sum_{k \mathop = 1}^{s + 1} \tuple {n + k, m + k - 1}$

Induction Step
This is our induction step:

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

Therefore:
 * $\ds \forall r \in \N_{>0}: \tuple {n, m} = \tuple {n + r, m + r} + \sum_{k \mathop = 1}^r \tuple {n + k, m + k - 1}$