Let $A = kQ/I$ be a bound quiver algebra for some algebraically closed field $k$, $Q$ a finite connected quiver without oriented cycles, and $I$ an admissible ideal. Say that $I'$ is also an admissible ideal such that $I \subseteq I'$.
My question is this: if $A$ is representation-finite, is $A' = kQ/I'$ also representation-finite?
It seems like this should be true, and it seems like there should be an "easy" way to prove this (maybe if I knew more algebra or representation theory!), but I can't seem to find this in the literature as a proposition/remark/exercise. Specifically, I've been looking at "Elements of the Representation Theory of Associative Algebras, Vol I" by Assem, Simson, and Skowronski, and at "Quiver Representations" by Schiffler.
Does anyone know if this is a known result? Or is there an easy way to see if this is true/false?