Consider the set of all finite directed binary trees with vertices $V$ and edges $E$. If $T_i=(V_i, E_i)$, $i=1,2$ are trees in this set, then we say $T_1 \sqsubseteq T_2$ if and only if $V_1 \subseteq V_2$ and $E_1$ is the set of all edges in $E_2$ with tail in $V_1$.
As a small exercise in the lecture notes, the reader is asked to prove that $\sqsubseteq$ is a well-founded order. Well-foundedness would follow directly from the fact that the single vertex is a minimal element of any FINITE directed tree. Is there more I should say about any tails connected to it?
Secondly, I wish to prove it is an order relation. We need not think hard about the vertices since $V_1 \subseteq V_2$ is a well-defined order relation called set inclusion and this has all the properties of reflexivity, antisymmetry and transitivity. Now the hard part would be to prove that the second statement "$E_1$ is the set of all edges in $E_2$ with tail in $V_1$" also satisfies the transitive, antisymmetric and reflexive definitions. I would say that for any tree, all of its edges are contained in itself and its tail is in itself, so reflexivity is trivial.
But what about the other two properties? This section is building up to induction so I am not sure if we need to use that yet or if it is required. But it feels like overkill.