[proofplan]
We compute $\mathbb{P}(X + Y = n)$ by partitioning the event $\{X + Y = n\}$ according to the value of $X$, then use independence to factor each joint probability $\mathbb{P}(X = k, Y = n - k)$.
[/proofplan]
[step:Partition the event $\{X + Y = n\}$ by the value of $X$]
The event $\{X + Y = n\}$ can be decomposed as a disjoint union over all possible values $k$ of $X$:
\begin{align*}
\{X + Y = n\} = \bigcup_k \{X = k, Y = n - k\},
\end{align*}
where the union runs over all $k$ in the range of $X$. These events are pairwise disjoint (since $X$ takes a single value). By [countable](/page/Countable%20Set) additivity,
\begin{align*}
\mathbb{P}(X + Y = n) = \sum_k \mathbb{P}(X = k, Y = n - k).
\end{align*}
[/step]
[step:Factor using independence to obtain the convolution formula]
Since $X$ and $Y$ are independent, the joint probability factors:
\begin{align*}
\mathbb{P}(X = k, Y = n - k) = \mathbb{P}(X = k)\,\mathbb{P}(Y = n - k).
\end{align*}
Substituting:
\begin{align*}
\mathbb{P}(X + Y = n) = \sum_k \mathbb{P}(X = k)\,\mathbb{P}(Y = n - k).
\end{align*}
[/step]