Strictly Increasing Sequence on Ordered Set

Lemma
Let $\left({S, \preceq}\right)$ be a totally ordered set.

Let $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ be a finite sequence of elements of $\left({S, \preceq}\right)$.

Then $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is strictly increasing :


 * $\forall k \in \left[{p + 1 \,.\,.\, q}\right]: r_{k - 1} \prec r_k$

Proof
Let $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ be strictly increasing.

Because $\forall k \in \N_{>0}: k - 1 < k$, it follows directly that:
 * $\forall k \in \left[{p + 1 \,.\,.\, q}\right]: r_{k - 1} \prec r_k$

Aiming for a contradiction, suppose $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is not strictly increasing.

Let $K$ be the set of all $k \in \left[{p \,.\,.\, q}\right]$ such that:
 * $\exists j \in \left[{p \,.\,.\, q}\right]$ such that $j < k$ and $r_k \preceq r_j$

The set $K$ is not empty because $\left \langle {r_k} \right \rangle_{p \mathop \le k \mathop \le q}$ is not strictly increasing.

As $K \subset \N$ and the latter is well-ordered, then so is $K$.

Thus $K$ has a minimal element $m$.

Thus there exists $j \in \left[{p \,.\,.\, q}\right]$ such that:
 * $j < m$

and:
 * $r_m \preceq r_j$

Because $m - 1 < m$:
 * $j \le m - 1$

and so:
 * $m - 1 \notin K$

So:
 * $r_j \preceq r_{m-1} \prec r_m \preceq r_j$

which is a contradiction.

Therefore by Proof by Contradiction, there can be no such element $m$

Therefore $K$ is empty.

The result follows.