 Solution

Use the strict-order convention appropriate to the membership relation. A bounded logical formula isThe first clause is irreflexivity, the second transitivity, and the third comparability of distinct elements. Irreflexivity and transitivity exclude $u\in v\in u$, so distinct elements cannot be ordered in both directions. These are exactly the axioms of a strict total order on the set $x$, with $\in$ as its order relation.

Every displayed quantifier is bounded by $x$, so $TO(x)$ is ${\Sigma }_0$. The logical formula does not assert that $x$ is transitive as a set, and no such additional assertion is needed to say that the restriction of membership to $x$ is a total order.