  A finite tree $T$ with $|V(T)| \geq 2$ is called {\it automorphism-free} if
there is no non-trivial automorphism of $T$. Let $\mathcal{AFT}$ be the poset
with the element set of all finite automorphism-free trees (up to graph
isomorphism) ordered by $T_1 \preceq T_2$ if $T_1$ can be obtained from $T_2$
by successively deleting one leaf at a time in such a way that each
intermediate tree is also automorphism-free. In this paper, we prove that
$\mathcal{AFT}$ has a unique minimal element. This result gives an affirmative
answer to the question asked by Rupinski.
