I don't quite get the logical leap from the proof of the proposition above (but with positive integer vectors; I understand the proof of this) to the proof of the following proposition regarding nonnegative integer-valued vectors.
Screenshot of the proof that I don't understand
I mean, I just don't get the transformation into a sum involving y terms. Surely it's the same equality as before (if you subtract r from both sides?)
I would be grateful if someone provided intuition!
The book is 'A first course in probability' by Sheldon Ross (Chapter 1: combinatorial analysis)
If $(x_1,\dots,x_r)$ is a solution for $z_1+\cdots+ z_r=n$ under the condition that the $z_i$ are nonnegative integers then $(x_1+1,\dots, x_r+1)$ is a solution for $z_1+\cdots+ z_r=n+r$.
Substituting $y_i=x_i+1$ we can say that $(y_1,\dots, y_r)$ is a solution for $z_1+\cdots+ z_r=n+r$ under the condition that the $z_i$ are positive.
So there is a one-one relation between solutions of $z_1+\cdots+ z_r=n$ where the $z_i$ are demanded to be nonnegative and solutions of $z_1+\cdots+ z_r=n+r$ where the $z_i$ are demanded to be positive.
Consequently the number of solutions is in both cases the same.