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