Let $A$ be a noetherian regular local ring, $x_1,\dots,x_n$ a regular $A$-sequence and $B = A / (x_1,\dots,x_n)$. Then $B$ is a complete intersection ring by definition.
If $(x_1,\dots,x_n)$ is a maximal $A$-sequence, does it follow that $B$ is Artinian?
Found the answer myself:
Yes it is. Because $A$ is regular, every maximal $A$-sequence has $\text{dim}(A)$ elements, i.e. $\text{dim} A = n$. By Krulls Hauptidealsatz we know, that every minimal prime over $x_{i+1}$ in $A/(x_1,\dots,x_i)$ has height 0 or 1, and because $x_{i+1}$ is not a zero divisor, every minimal prime actually has height 1. Thus modding out $x_{i+1}$ reduces the dimension by at least $1$, so $B=A/(x_1,\dots,x_n)$ has dimension 0. Noetherian, zero dimensional rings are artinian.