Elements reduction in a general case If $G$ is a group, then each element has an inverse and $\forall x, y, z \in G, xy = xz \Rightarrow x^{-1} \cdot xy = x^{-1} \cdot xz \Rightarrow 1 \cdot y = 1 \cdot z \Rightarrow y = z$
However, we also use this idea, when we say $$(f * g)(x) = \sum\limits_{y + z = x} f(y)g(z) \color{#c00} = \sum\limits_{0 \leq z \leq x} f(x - z)g(z)$$
where $y, z, x \in X = \mathbb{N}_0$ with associative $+$ and $f, g \in R^X$.
We mean that $y + z = x \Rightarrow y = x + (-z)$, so we only sum for the appropriate $z$. We use reduction and even inverse element of $z$ here, although $(\mathbb{N}_0, +)$ is not a group.
Why is it true in this example and what are some general thoughts on whether reduction is possible or not?
 A: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\N}{\mathbb{N}}$$\newcommand{\Z}{\mathbb{Z}}$Consider the integers. 
On $\Z$ you have the unary operation of taking the opposite, which takes $x \in \Z$ to $-x \in \Z$.
Then you have the binary operation of subtraction, which takes a pair $(x, y)$ of integers to their difference $x - y$.
You can define the second operation in terms of the first one, say 
$$
x - y = x + (-y),
$$
where the first "$-$" is binary, and the second one unary.
But you can  also give an independent definition of binary subtraction, for instance by saying that $x - y$ is the (unique) integer $z$ for which $x = y + z$.
When you take $\N_{0}$, only $0$ will have an opposite, and in general the subtraction of a natural number from another will not give a natural number. But subtraction can be  still defined as a function
\begin{align*}
&\Set{ (x, y) : x, y \in \N_{0}, x \ge y}
\to
\N_{0}\\
&(x, y) \mapsto x - y.
\end{align*}
In other words, for $x, y \in \N_{0}$, the equation
$$
x = y + z
$$
in the variable $z$ will have a solution $z \in \N_{0}$ iff $x \ge y$.
