Tree predecessors and compatibility

Let $\alpha =\mathrm{ht}(t)$ and let $e:\alpha \to P_t$ be its order isomorphism. For $\beta <\alpha$, transitivity and order reflection give $P_{e(\beta )}=e[\beta ]$. Thus $\mathrm{ht}(e(\beta ))=\beta$. Conversely every predecessor is $e(\gamma )$ for exactly one $\gamma <\alpha$ and has height $\gamma$, proving existence and uniqueness.

If $s,t{\le }_{T}u$ and either equals $u$, they are comparable. Otherwise both belong to the well-ordered set $P_u$, so its linear order compares them. This includes $s=t$.

If $s{<}_{T}t$, step 1.1 puts $s=e(\beta )$ at some $\beta <\mathrm{ht}(t)$, so $\mathrm{ht}(s)<\mathrm{ht}(t)$. Distinct nodes of one level therefore cannot be comparable; each level is an antichain. For a root the predecessor assertion has no indices, and empty levels satisfy the antichain assertion vacuously.