Principle of Induction applied to Interval of Naturally Ordered Semigroup

Theorem
Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup.

Let $\left[{p \,. \, . \, q}\right]$ be a closed interval of $\left({S, \circ, \preceq}\right)$.

Let $T \subseteq \left[{p \,. \, . \, q}\right]$ such that the minimal element of $\left[{p \,. \, . \, q}\right]$ is in $T$.

Let $x \in T: x \prec q \implies x \circ 1 \in T$.

Then $T = \left[{p \,. \, . \, q}\right]$.

Proof
Let $T' = T \cup \left\{{x \in S: q \prec x}\right\}$.

Then $T'$ satisfies the conditions of the Principle of Finite Induction.

Therefore $T' = \left\{{x \in S: p \preceq x}\right\}$.

Therefore $T = \left[{p \,. \, . \, q}\right]$.