Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Lemma 2

Theorem
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.

Let $\preccurlyeq_l$ denote the lexicographic order on $S_1 \times S_2$''':
 * $\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tuple {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$

Let $x_1$ and $y_1$ be non-comparable elements in $S_1$:
 * $\lnot \paren {x_1 \preccurlyeq_1 y_1}$ and $\lnot \paren {y_1 \preccurlyeq_1 x_1}$

Then $\tuple {x_1, x_2}$ and $\tuple {y_1, y_2}$ are non-comparable elements in $S_1 \times S_2$:
 * $\lnot \paren {\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} }$ and $\lnot \paren {\tuple {y_1, y_2} \preccurlyeq_l \tuple {x_1, x_2} }$

Proof
Suppose $x_1$ and $y_1$ are non-comparable elements in $S_1$:
 * $\lnot \paren {x_1 \preccurlyeq_1 y_1}$ and $\lnot \paren {y_1 \preccurlyeq_1 x_1}$

that:
 * $\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2}$ or $\tuple {y_1, y_2} \preccurlyeq_l \tuple {x_1, x_2}$

Then from Lemma $1$:
 * $x_1 \preccurlyeq_1 y_1$ or $y_1 \preccurlyeq_1 x_1$

This contradicts our assumption.

Hence the result by Proof by Contradiction.