Strictly Increasing Sequence on Ordered Set

Lemma
Let $\struct {S, \preceq}$ be a totally ordered set.

Let $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ be a finite sequence of elements of $\struct {S, \preceq}$.

Then $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ is strictly increasing :


 * $\forall k \in \closedint {p + 1} q: r_{k - 1} \prec r_k$

Proof
Let $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ be strictly increasing.

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

For the other direction, we use a Proof by Contraposition.

To that end, suppose $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ is not strictly increasing.

Let $K$ be the set of all $k \in \closedint p q$ such that:
 * $\exists j \in \closedint p q$ such that $j < k$ and $r_k \preceq r_j$

The set $K$ is not empty because $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ 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 $j \in \closedint p q$ 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:
 * $r_m \preceq r_j\prec r_{m - 1}$

Since orderings are transitive, it follows
 * $r_m \preceq r_{m - 1}$

From Rule of Transposition it follows that
 * $\forall k \in \closedint {p + 1} q: r_{k - 1} \prec r_k \implies \sequence {r_k}_{p \mathop \le k \mathop \le q}$ is strictly increasing.

The result follows.