From Lexicographic Order on Totally Ordered Sets is Total Ordering we have that $\struct {S_1 \times S_2, \preccurlyeq}$ is a ordered set. === Necessary Condition === Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be well-ordered sets. By definition of well-ordering we have that $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are totally ordered sets. From Lexicographic Order on Totally Ordered Sets is Total Ordering it follows that $\struct {S_1 \times S_2, \preccurlyeq}$ is a totally ordered set. It remains to be shown that the Cartesian product of arbitrary non-empty subsets of $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ has a smallest element under $\preccurlyeq$. Let $T_1 \subseteq S_1$ and $T_2 \subseteq S_2$. Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in T_1 \times T_2$. As both $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are well-ordered sets it follows that both $\struct {T_1, \preceq_1}$ and $\struct {T_2, \preceq_2}$ have smallest elements. Let those smallest elements be $m_1$ and $m_2$ respectively. Consider the element $\tuple {m_1, m_2} \in T_1 \times T_2$. Let $\tuple {x_1, x_2} \in T_1 \times T_2$. We have that: :$m_1 \preceq_1 x_1$ and: :$m_2 \preceq_2 x_2$ The following cases are to be examined: $(1): \quad m_1 \ne x_1$ By defiinition of $m_1$ as the smallest element of $T_1$: :$m_1 \preceq_1 x_1$ Thus by definition of the lexicographic order on $S_1 \times S_2$: :$\tuple {m_1, m_2} \preccurlyeq \tuple {x_1, x_2}$ $(2): \quad m_1 = x_1$ By definition of $m_2$ as the smallest element of $T_2$: :$m_2 \preceq_2 x_2$ Thus by definition of the lexicographic order on $S_1 \times S_2$: :$\tuple {m_1, m_2} \preccurlyeq \tuple {x_1, x_2}$ Thus it follows that $T_1 \times T_2$ has a smallest element $\tuple {m_1, m_2}$. As $T_1 \times T_2$ is the Cartesian product of arbitrary non-empty subsets of of $S_1$ and $S_2$, it follows that $\struct {S_1 \times S_2, \preccurlyeq}$ is a well-ordered set. {{qed|lemma}} === Sufficient Condition === Let $\struct {S_1 \times S_2, \preccurlyeq}$ be a well-ordered set. By definition of well-ordering we have that $\struct {S_1 \times S_2, \preccurlyeq}$ is a totally ordered set. From Lexicographic Order on Totally Ordered Sets is Total Ordering $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are totally ordered sets. It remains to be shown that arbitrary non-empty subsets of $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ have a smallest elements. Let $T_1 \subseteq S_1$ and $T_2 \subseteq S_2$. Let $\tuple {m_1, m_2} \in T_1 \times T_2$ be the smallest element of $T_1 \times T_2$. Thus: :$\forall \tuple {x_1, x_2} \in T_1 \times T_2: \tuple {m_1, m_2} \preccurlyeq \tuple {x_1, x_2}$ There are two cases to address: $(1): \quad$ Let $m_1 = x_1$. Thus by definition of the lexicographic order on $S_1 \times S_2$: :$m_2 \preceq_2 x_2$ This is the case for all $\tuple {x_1, x_2} \in T_1 \times T_2$ such that $m_1 = x_1$. Therefore: :$\forall x_2 \in T_2: m_2 \preceq_2 x_2$ Thus by definition, $m_2$ is the smallest element of $T_2$. We have that $T_2$ is an arbitrary non-empty subset $\struct {S_2, \preceq_2}$. Therefore by definition $T_2$ is a well-ordered set. $(2): \quad$ Let $m_1 \ne x_1$. Thus by definition of the lexicographic order on $S_1 \times S_2$: :$m_1 \prec_1 x_1$ This is the case for all $\tuple {x_1, x_2} \in T_1 \times T_2$ such that $m_1 \ne x_1$. Therefore: :$\forall x_1 \in T_1: m_1 \preceq_1 x_1$ Thus by definition, $m_1$ is the smallest element of $T_1$. We have that $T_1$ is an arbitrary non-empty subset $\struct {S_1, \preceq_1}$. Therefore by definition $T_1$ is a well-ordered set. {{qed}}