 Solution

Read ${\mathrm{prop}}_3(u,v)$ as the existence of a directed successor path from $u$ to $v$. Then the property says every reachable ordered pair of even-valued nodes admits a connecting route through an odd-valued node. More exactly, for all even-valued $u,v$ with $u$ reaching $v$, there is an odd-valued $w$ such that $u$ reaches $w$ and $w$ reaches $v$.

The two reachability witnesses concatenate to a directed walk through $w$. In an acyclic graph this yields an intervening node on a path. In a graph with directed cycles, the concatenation may revisit vertices. The existential witness does not say that every connecting path contains an odd-valued node, or even that an even-to-even edge is forbidden. For example edges $u\to v$, $u\to w$, $w\to v$, with even endpoints and odd $w$, satisfy the property for this pair despite the direct even-to-even edge.

There are two source-notation qualifications. The displayed outer universal quantifier binds both symbols appearing as formal arguments to prop, so the whole formula is a global property of $T$, rather than a genuinely free two-node predicate. Also the indexed ellipsis defining prop3 does not formalize path length zero. With positive-length reachability, even nodes on a cycle impose additional requirements; with reflexive reachability, choosing $u=v$ also imposes a requirement, and an isolated even-valued node would fail. The usual intended reading here is positive-length successor reachability. No uniqueness, acyclicity or specific path-length convention is proved by the displayed formula alone.