avewright/proofwiki-math huggingface dataset:
Theorem: Composite of Increasing Mappings is Increasing

Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ and $\struct {S_3, \preceq_3}$ be ordered sets.
Let:
:$\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$
and:
:$\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$
be increasing mappings.
Then $\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is also an increasing mapping.

Proof:
By definition of composition of mappings:
:$\map {\paren {\psi \circ \phi} } x = \map \psi {\map \phi x}$
As $\phi$ is an increasing mapping, we have:
:$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \map \phi {x_1} \preceq_2 \map \phi {y_1}$
As $\psi$ is an increasing mapping, we have:
:$\forall x_2, y_2 \in S_2: x_2 \preceq_2 y_2 \implies \map \psi {x_2} \preceq_3 \map \psi {y_2}$
By setting $x_2 = \map \phi {x_1}, y_2 = \map \phi {y_1}$, it follows that:
:$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \map \psi {\map \phi {x_1} } \preceq_3 \map \psi {\map \phi {y_1} }$
That is:
:$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \map {\paren {\psi \circ \phi} } {x_1} \preceq_3 \map {\paren {\psi \circ \phi} } {y_1}$
Hence the result.
{{qed}}
Category:Increasing Mappings
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Theorem: Order-Preserving Identity Mapping between Ordered Structures not necessarily Isomorphism

Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$ such that:
:$\forall a, b \in A: a \mathrel \RR b \implies a \mathrel \SS b$
Let $I_A$ denote the identity mapping on $A$.
Then it is not necessarily the case that $I_A$ is an order isomorphism from the ordered structures $\struct {A, \RR}$ and $\struct {A, \SS}$.

Proof:
Proof by Counterexample
Let $\RR: \N \to \N$ be the ordering on the natural numbers $\N$ defined as:
:$\forall a, b \in \N: a \mathrel \RR b \iff a \le b \text { and } \map P a = \map P b$
where $\map P x$ is the parity of $x$.
That is:
:$0 \mathrel \RR 2 \mathrel \RR 4 \mathrel \RR \cdots$
and:
:$1 \mathrel \RR 3 \mathrel \RR 5 \mathrel \RR \cdots$
and so on, but (for example):
:$\lnot \paren {0 \mathrel \RR 1}$
Let $\SS: \N \to \N$ be the ordering on the natural numbers $\N$ defined as:
:$\forall a, b \in \N: a \mathrel \SS b \iff a \le b$
We have that:
{{begin-eqn}}
{{eqn | q = \forall a, b \in \N
      | l = a
      | o = \RR
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a
      | o = \le
      | r = b
      | c = Definition of $\RR$
}}
{{eqn | ll= \leadsto
      | l = a
      | o = \SS
      | r = b
      | c = Definition of $\SS$
}}
{{end-eqn}}
But note that:
{{begin-eqn}}
{{eqn | l = 0
      | o = \le
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 0
      | o = \SS
      | r = 1
      | c = Definition of $\SS$
}}
{{end-eqn}}
while it is not the case that $0 \mathrel \RR 1$.
Hence while it is true that:
:$\forall a, b \in \N: a \mathrel \RR b \implies a \mathrel \SS b$
it is not the case that:
:$\forall a, b \in \N: a \mathrel \SS b \implies a \mathrel \RR b$
and so the identity mapping on $\NN$ is not an order isomorphism.
{{qed}}
Theorem: Relation Isomorphism preserves Ordering

Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be an ordered set.
Then $\struct {B, \SS}$ is also an ordered set.

Proof:
Let $\struct {A, \RR}$ be an ordered set.
Recall the definition:
{{:Definition:Ordering/Definition 1}}
From Relation Isomorphism Preserves Reflexivity:
:$\SS$ is reflexive.
From Relation Isomorphism Preserves Antisymmetry:
:$\SS$ is antisymmetric.
From Relation Isomorphism Preserves Transitivity:
:$\SS$ is transitive.
So by definition $\struct {B, \SS}$ is an ordered set.
{{qed}}
Theorem: Relation Isomorphism preserves Total Ordering

Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be a totally ordered set.
Then $\struct {B, \SS}$ is also a totally ordered set.

Proof:
Let $\struct {A, \RR}$ be a totally ordered set.
Recall the definition:
{{:Definition:Total Ordering/Definition 1}}
From Relation Isomorphism preserves Ordering:
:$\SS$ is an ordering.
Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.
Let $x, y \in A$.
Then either $x \mathrel \RR y$ or $y \mathrel \RR x$.
From the definition of relation isomorphism, either:
:$\map \phi x \mathrel \SS \map \phi y$
or:
:$\map \phi y \mathrel \SS \map \phi x$
and so by definition $\SS$ is also a total ordering.
So by definition $\struct {B, \SS}$ is also a totally ordered set.
{{qed}}
Theorem: Relation Isomorphism preserves Well-Ordering

Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be a well-ordered set.
Then $\struct {B, \SS}$ is also a well-ordered set.

Proof:
Let $\struct {A, \RR}$ be a well-ordered set.
Recall the definition:
{{:Definition:Well-Ordering/Definition 1}}
By Well-Ordering is Total Ordering, $\RR$ is a total ordering.
From Relation Isomorphism preserves Total Ordering:
:$\SS$ is a total ordering on $B$.
Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.
Let $A' \subseteq A$.
As $\struct {A, \RR}$ is a well-ordered set:
:$\exists m_1 \in A': \forall a \in A': a = m_1 \lor \neg \paren {a \mathrel \RR m_1}$
Now consider the image $m_2$ of $m_1$ in $B$:
:$m_2 = \map \phi {m_1}$
{{AimForCont}} $\struct {B, \SS}$ is not a well-ordered set.
Let there exist $B' \subseteq B$ such that:
:$\exists x \in B': x \ne m_2, x \mathrel \SS m_2$
By Inverse of Relation Isomorphism is Relation Isomorphism, $\phi^{-1}$ is a relation isomorphism.
Hence:
:$\map {\phi^{-1} } x \ne \map {\phi^{-1} } {m_2}, \map {\phi^{-1} } x \mathrel \SS \map {\phi^{-1} } {m_2}$
But this contradicts the fact that $\struct {A, \RR}$ is well-ordered.
Hence by Proof by Contradiction $\struct {B, \SS}$ must be a well-ordered set.
{{qed}}
Theorem: Relation Isomorphism preserves Lattice Structure

Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be a lattice.
Then $\struct {B, \SS}$ is also a lattice.

Proof:
Let $\struct {A, \RR}$ be a lattice.
Recall the definition:
{{:Definition:Lattice (Ordered Set)}}
From Relation Isomorphism preserves Ordering:
:$\SS$ is an ordering on $B$.
Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.
We need to show that for all $x, y \in A$, the ordered set $\struct {\set {\map \phi x, \map \phi y}, \SS}$ admits both a supremum and an infimum. 
Let $x, y \in A$.
Then $\struct {\set {x, y}, \RR}$ admits both a supremum and an infimum.
Let $c = \map \sup {\set {x, y}, \RR}$.
Then by definition of supremum:
:$\forall a \in \set {x, y}: a \mathrel \RR c$
:$\forall d \in A: c \mathrel \RR d$
where $d$ is an upper bound of $\struct {\set {x, y}, \RR} \subseteq A$.
Now consider the image of $\set {x, y}$ under $\phi$.
By definition of relation isomorphism:
:$\forall \map \phi s \in \set {\map \phi x, \map \phi y}: \map \phi c \mathrel \SS \map \phi s$
:$\forall \map \phi d \in B: \map \phi c \mathrel \SS \map \phi d$
where $\map \phi d$ is an upper bound of $\struct {\set {\map \phi x, \map \phi y}, \SS} \subseteq B$.
So by definition of supremum:
:$\map \phi c = \map \sup {\set {\map \phi x, \map \phi y}, \SS}$
That is, $\struct {\set {\map \phi x, \map \phi y}, \SS}$ admits a supremum.
Using a similar technique it can be shown that:
:If $c = \map \inf {\set {x, y}, \SS}$, then:
:$\map \phi c = \map \inf {\set {\map \phi x, \map \phi y}, \SS}$
Hence $\struct {\set {\map \phi x, \map \phi y}, \SS}$ admits both a supremum and an infimum.
That is, $\SS$ is a lattice ordering and so $\struct {B, \SS}$ is a lattice.
{{qed}}
Theorem: Ordered Set with Greatest Element whose Subsets have Infimum is Complete Lattice

Let $\struct {S, \preccurlyeq}$ be an ordered set such that:
:$\struct {S, \preccurlyeq}$ has a greatest element $u$
:every non-empty subset of $S$ admits an infimum.
Then $\struct {S, \preccurlyeq}$ is a complete lattice.

Proof:
For $\struct {S, \preccurlyeq}$ to be a complete lattice, it has to be such that:
:every non-empty subset of $S$ admits both a supremum and an infimum.
Let $T \subseteq S$ be an arbitrary subset of $S$.
We already have {{hypothesis}} that $T$ admits an infimum.
We are to show that $T$ admits a supremum.
By definition of greatest element:
:$\forall x \in T: x \preccurlyeq u$
Thus by definition $u$ is an upper bound of $T$.
Let $H \subseteq S$ be the set of all upper bounds of $T$.
Because $u$ is an upper bound of $T$, $u \in H$.
Hence $H$ is not empty.
We have {{hypothesis}} that $H$ has an infimum.
That is:
:$\exists v \in H: \forall x \in H: v \preccurlyeq x$
Hence $v$ is by definition a supremum of $T$.
Thus $T$ admits a supremum.
Hence {{apriori}} $T$ admits both a supremum and an infimum.
Hence the result.
{{qed}}
Theorem: Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice

Let $A$ be a set.
Let $\SS$ be a set of subsets of $A$ such that:
:$A \in \SS$
:for every non-empty subset $\TT$ of $\SS$, $\ds \bigcap \TT \in \SS$, where $\ds \bigcap \TT$ denotes the intersection of $\TT$.
Then:
:the ordered set $\struct {\SS, \subseteq}$ is a complete lattice
where:
:$\ds \bigcap \TT$ is the infimum necessarily admitted by $\TT$.

Proof:
From Subset Relation is Ordering, $\struct {\SS, \subseteq}$ is indeed an ordered set.
We have by definition of $\SS$ that:
:$\forall H \in \SS: H \subseteq A$
Hence $A$ is the greatest element of $\SS$ with respect to the ordered set $\struct {A, \subseteq}$.
Then from Intersection is Largest Subset:
:$\forall K \in \TT: \ds \bigcap \TT \subseteq K$
and so $\TT$ admits an infimum, which is $\ds \bigcap \TT$.
Hence the conditions of Ordered Set with Greatest Element whose Subsets have Infimum is Complete Lattice.
The result follows.
{{qed}}
Theorem: Intersection of Submagmas is Largest Submagma

Let $\HH$ be a set of submagmas of $\struct {S, \odot}$, where $\HH \ne \O$.
Then the intersection $\bigcap \HH$ of the elements of $\HH$ is the largest submagma of $\struct {S, \odot}$ contained in each element of $\HH$.

Proof:
Let $K = \bigcap \HH$.
Let $K_i$ be an arbitrary element of $\HH$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
      | o = \in
      | r = K
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall i
      | l = a, b
      | o = \in
      | r = K_i
      | c = {{Defof|Intersection of Set of Sets}}
}}
{{eqn | ll= \leadsto
      | q = \forall i
      | l = a \odot b
      | o = \in
      | r = K_i
      | c = as $K_i$ is itself a magma, therefore $\odot$ is closed
}}
{{eqn | ll= \leadsto
      | l = a \odot b
      | o = \in
      | r = K
      | c = {{Defof|Intersection of Set of Sets}}
}}
{{end-eqn}}
That is, $\odot$ is closed in $K$.
Hence by definition $\struct {K, \odot}$ is a submagma of $\struct {S, \odot}$.
{{qed|lemma}}
Now to show that $\struct {K, \odot}$ is the largest submagma of $\struct {S, \odot}$ contained in each element of $\HH$.
Let $N$ be a submagma of $\struct {S, \odot}$ such that:
:$\forall H \in \HH: N \subseteq H$
Then by definition $N \subseteq K$.
Let $x, y \in N$.
Then:
:$x \odot y \in N \implies x \odot y \in K$
Thus any submagma of all elements of $\HH$ is also a  submagma of $K$ and so no larger than $K$.
Thus $K = \bigcap \HH$ is the largest submagma of $H$ contained in each element of $\HH$.
{{qed}}
Category:Submagmas
Category:Set Intersection
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Theorem: Set of Submagmas of Magma under Subset Relation forms Complete Lattice

Let $\struct {A, \odot}$ be a magma.
Let $\SS$ be the set of submagmas of $A$.
Then:
:the ordered set $\struct {\SS, \subseteq}$ is a complete lattice
where for every subset $\TT$ of $\SS$:
:the infimum of $\TT$ necessarily admitted by $\TT$ is $\bigcap \TT$.

Proof:
From Magma is Submagma of Itself:
:$\struct {A, \odot} \in \SS$
Let $\TT$ be a non-empty subset of $\SS$.
From Intersection of Submagmas is Largest Submagma:
:$\bigcap \TT \in \SS$
Hence, from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$\struct {\SS, \subseteq}$ is a complete lattice
where $\bigcap \TT$ is the infimum of $\TT$.
Theorem: Intersection of Subsemigroups/General Result/Proof 1

Let $\struct {S, \circ}$ be a semigroup.
{{:Intersection of Subsemigroups/General Result}}

Proof:
Let $T = \bigcap \mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
      | o = \in
      | r = T
      | c =
}}
{{eqn | ll= \leadsto
      | q = \forall K \in \mathbb S
      | l = a, b
      | o = \in
      | r = K
      | c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
      | q = \forall K \in \mathbb S
      | l = a \circ b
      | o = \in
      | r = K
      | c = Subsemigroups are closed
}}
{{eqn | ll= \leadsto
      | l = a \circ b
      | o = \in
      | r = T
      | c = {{Defof|Set Intersection}}
}}
{{end-eqn}}
So by Subsemigroup Closure Test, $\struct {T, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Now to show that $\struct {T, \circ}$ is the largest such subsemigroup.
Let $x, y \in T$.
Then $\forall K \subseteq T: x \circ y \in K \implies x \circ y \in T$.
Thus $\forall K \in \mathbb S: K \subseteq T$.
Thus $T$ is the largest subsemigroup of $S$ contained in each member of $\mathbb S$.
{{qed}}
Theorem: Intersection of Subsemigroups/General Result/Proof 2

Let $\struct {S, \circ}$ be a semigroup.
{{:Intersection of Subsemigroups/General Result}}

Proof:
From Set of Subsemigroups forms Complete Lattice:
:$\struct {\mathbb S, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subsemigroups of $S$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.
Hence the result, by definition of infimum.
{{qed}}
Theorem: Elements of Semigroup with Equal Images under Homomorphisms form Subsemigroup

Let $\struct {A, \circ}$ and $\struct {B, *}$ be semigroups.
Let $f: A \to B$ and $g: A \to B$ be semigroup homomorphisms.
Then the set:
:$S = \set {x \in A: \map f x = \map g x}$
is a subsemigroup of $A$.

Proof:
Let $x, y \in A$. Then:
{{begin-eqn}}
{{eqn | l = \map f {x \circ y}
      | r = \map f x * \map f y
      | c = Morphism Property
}}
{{eqn | r = \map g x * \map g y
      | c = Definition of $A$
}}
{{eqn | r = \map g {x \circ y}
      | c = Morphism Property
}}
{{end-eqn}}
Thus $x \circ y \in A$.
So, by the Subsemigroup Closure Test:
:$S$ is a subsemigroup of $A$.
{{Qed}}
Theorem: Set of Congruence Relations on Algebraic Structure forms Complete Lattice

Let $\struct {S, \odot}$ be an algebraic structure.
Let $\map \RR \odot$ be the set of all congruence relations on $\struct {S, \odot}$.
Then $\struct {\map \RR \odot, \subseteq}$ is a complete lattice.

Proof:
Elements of $\map \RR \odot$ are subsets of $S \times S$.
First we consider the trivial relation $S \times S$ itself.
From Trivial Relation is Universally Congruent, $S \times S$ is a congruence relation on $\struct {S, \odot}$.
Let $\TT$ be a subset of $\map \RR \odot$.
Consider the intersection $\HH = \ds \bigcap \TT$.
Let $x_1, y_1, x_2, y_2 \in S$ be arbitrary, such that:
:$x_1 \mathrel \HH y_1$ and $x_2 \mathrel \HH y_2$
We have:
{{begin-eqn}}
{{eqn | l = x_1
      | o = \HH
      | r = y_1
      | c = 
}}
{{eqn | lo= \land
      | l = x_2
      | o = \HH
      | r = y_2
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall \KK \in \TT
      | l = x_1
      | o = \KK
      | r = y_1
      | c = {{Defof|Intersection of Set of Sets}}
}}
{{eqn | lo= \land
      | l = x_2
      | o = \KK
      | r = y_2
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall \KK \in \TT
      | l = x_1 \odot x_2
      | o = \KK
      | r = y_1 \odot y_2
      | c = {{Defof|Congruence Relation}}
}}
{{eqn | ll= \leadsto
      | l = x_1 \odot x_2
      | o = \HH
      | r = y_1 \odot y_2
      | c = {{Defof|Intersection of Set of Sets}}
}}
{{end-eqn}}
Hence $\HH = \ds \bigcap \TT$ is a congruence relation on $\struct {S, \odot}$.
That is:
:$\ds \bigcap \TT \in \map \RR \odot$
The appropriate conditions are fulfilled, and from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$\struct {\map \RR \odot, \subseteq}$ is a complete lattice.
{{qed}}
Theorem: Group Epimorphism Preserves Generator

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: G \to H$ be a group epimorphism.
Let $A$ be a generator for $\struct {G, \circ}$.
Then $\phi \sqbrk A$ is a generator for $\struct {H, *}$.

Proof:
By definition of generator:
:$A$ is the intersection of all subgroups of $G$ containing $A$.
{{ProofWanted}}
Theorem: Lattice of Subgroups forming Totally Ordered Set is Indecomposable

Let $\struct {G, \circ}$ be a Group.
Let $\mathbb G$ be the set of subgroups of $G$.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by the subset ordering on $\mathbb G$.
Let $\struct {\mathbb G, \subseteq}$ be totally ordered.
Then $\struct {G, \circ}$ is an indecomposable group.

Proof:
First we note that from Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ is indeed a complete lattice.
As postulated, let $\struct {\mathbb G, \subseteq}$ be totally ordered.
{{AimForCont}} there exists a decomposition of $\struct {G, \circ}$.
Then:
:$\exists H, K \in \mathbb G: H \cap K = \set e$
where $\struct {H, \circ}$ and $\struct {K, \circ}$ are non-trivial normal subgroups of $\struct {G, \circ}$.
But then neither $H \subseteq K$ nor $K \subseteq H$.
That is, $\struct {\mathbb G, \subseteq}$ is not totally ordered.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Indecomposable Lattice of Subgroups does not necessarily form Totally Ordered Set

Let $\struct {G, \circ}$ be a Group.
Let $\mathbb G$ be the set of subgroups of $G$.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by the subset ordering on $\mathbb G$.
Let $\struct {G, \circ}$ be an indecomposable group.
Then it is not necessarily the case that $\struct {\mathbb G, \subseteq}$ is totally ordered.

Proof:
First we note that from Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ is indeed a complete lattice.
;Proof by Counterexample
Let $D_4$ denote the dihedral group of order $8$, also known as the symmetry group of the square.
From Internal Group Direct Product Examples: $D_4$, it is seen that $D_4$ is indecomposable group.
But from the Hasse diagram, we see:
:540px
it is seen that $\struct {\mathbb G, \subseteq}$, where $\mathbb G$ denotes the set of subsets of $D_4$, is not totally ordered.
Hence the result.
{{qed}}
Theorem: Conditions for Relation to be Well-Ordering

Let $\struct {S, \RR}$ be a relational structure.
$\RR$ is a well-ordering {{iff}}:
:$(1): \quad$ For all $x, y \in S$, either $x \mathrel \RR y$ or $y \mathrel \RR x$
:$(2): \quad$ For every non-empty subset $T$ of $S$, there exists $z \in T$ such that:
::::$\forall x \in T: \paren {x \mathrel \RR z \iff x = z}$

Proof:
=== Sufficient Condition ===
Let $\RR$ be a well-ordering.
Then $\RR$ is {{afortiori}} a total ordering, and so by definition:
:$\forall x, y \in S: x \mathrel \RR y \text { or } y \mathrel \RR x$
Thus condition $(1)$ is fulfilled.
Let $T$ be a non-empty subset of $S$.
Then by definition of well-ordering:
:$\exists z \in T: \forall x \in T: z \mathrel \RR x$
As $\RR$ is a well-ordering, {{afortiori}} it is an ordering and so an asymmetric relation.
Hence:
:$\forall x \in T: x \mathrel \RR z \implies x = z$
But then as $\RR$ is an ordering and so reflexive:
:$x = z \implies x \mathrel \RR z$
Hence:
:$\exists z \in T: \forall x \in T: x \mathrel \RR z \iff x = z$
Thus condition $(2)$ is fulfilled.
{{qed|lemma}}
Theorem: Simple Order Product of Pair of Ordered Sets is Lattice iff Factors are Lattices

Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the '''simple (order) product''' of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is a lattice {{iff}} both $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ are lattices.

Proof:
Let $\struct {T, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
First we note that from Simple Order Product of Pair of Ordered Sets is Ordered Set, $\struct {T, \preccurlyeq_s}$ is an ordered set.
Theorem: Simple Order Product of Pair of Totally Ordered Sets is Total iff One Factor is Singleton

Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the '''simple (order) product''' of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is a totally ordered set {{iff}}:
:either $S_1$ or $S_2$ is a singleton
and:
:both $\preccurlyeq_1$ and $\preccurlyeq_2$ are total orderings.

Proof:
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {T, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
First we note that from Simple Order Product of Pair of Ordered Sets is Ordered Set, $\struct {T, \preccurlyeq_s}$ is an ordered set.
We also note that from Ordering on Singleton is Total Ordering, if either $S_1$ or $S_2$ is a singleton then $\struct {S_1, \preccurlyeq_1}$ or $\struct {S_2, \preccurlyeq_2}$ respectively is a totally ordered set.
Theorem: Ordering on Singleton is Total Ordering

Let $S = \set s$ be a singleton.
Let $\RR$ be an ordering on $S$.
Then $\RR$ is a total ordering on $S$.

Proof:
By definition of ordering, $\RR$ is {{afortiori}} a reflexive relation.
Hence from Reflexive Relation on Singleton is Well-Ordering:
:$\struct {S, \RR}$ is a well-ordered set.
Hence by definition of well-ordered set:
:$\RR$ is a total ordering on $S$.
Hence the result.
{{qed}}
Category:Singletons
Category:Total Orderings
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Theorem: Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice

Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\preccurlyeq_l$ denote the lexicographic order on $S_1 \times S_2$''':
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tuple {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$
Then:
:$\struct {S_1 \times S_2, \preccurlyeq_l}$ is a lattice
{{iff}} all of the following conditions hold:
:$(1): \quad \struct {S_1, \preccurlyeq_1}$ is a lattice
:$(2): \quad$ Either $\preccurlyeq_1$ is a total ordering, or $\struct {S_2, \preccurlyeq_2}$ has a greatest element and a smallest element
:$(3): \quad$ Every doubleton subset of $S_2$ is either unbounded above or admits a supremum, and is also either unbounded below or admits an infimum
:$(4): \quad$ Either every doubleton subset of $S_2$ admits a supremum, or every element of $S_1$ has an immediate successor and $\struct {S_2, \preccurlyeq_2}$ has a smallest element
:$(5): \quad$ Either every doubleton subset of $S_2$ admits an infimum, or every element of $S_1$ has an immediate predecessor and $\struct {S_2, \preccurlyeq_2}$ has a greatest element.

Proof:
Let $\struct {T, \preccurlyeq_l} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
From Lexicographic Order is Ordering we have that $\struct {T, \preccurlyeq_l}$ is an ordered set.
Recall the definition of lexicographic order:
{{:Definition:Lexicographic Order}}