avewright/proofwiki-math huggingface dataset:
Theorem: Kernel of Ring Homomorphism is Ideal

The kernel of $\phi$ is an ideal of $R_1$.

Proof:
By Kernel of Ring Homomorphism is Subring, $\map \ker \phi$ is a subring of $R_1$.
Let $s \in \map \ker \phi$, so $\map \phi s = 0_{R_2}$.
Suppose $x \in R_1$. Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x \circ_1 s}
      | r = \map \phi x \circ_2 \map \phi s
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map \phi x \circ_2 0_{R_2}
      | c = as $s \in \map \ker \phi$
}}
{{eqn | r = 0_{R_2}
      | c = Properties of $0_{R_2}$
}}
{{end-eqn}}
and similarly for $\map \phi {s \circ_1 x}$.
The result follows.
{{Qed}}
Theorem: Ideals of Division Ring

Let $\struct {R, +, \circ}$ be a division ring whose zero is $0_R$.
The only ideals of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.
That is, $\struct {R, +, \circ}$ has no non-null proper ideals.

Proof:
From Null Ring is Ideal, $\set {0_R}$ is an ideal of $\struct {R, +, \circ}$, as $\struct {R, +, \circ}$, being a division ring, is also a ring.
By definition, every non-zero element of a division ring is a unit.
So $S \ne \set {0_R} \implies \exists x \in S: x \ne 0_R$ such that $r$ is a unit of $R$.
The result follows from Ideal of Unit is Whole Ring.
{{qed}}
Theorem: Quotient Ring of Kernel of Ring Epimorphism

There exists a unique ring isomorphism $g: R_1 / K \to R_2$ such that:
:$g \circ q_K = \phi$

Proof:
From the Quotient Theorem for Epimorphisms, there is one and only one isomorphism that satisfies the conditions for each of the operations on $R_1$.
Hence the result.
{{qed}}
Theorem: Ring Epimorphism Preserves Ideals

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $J$ be an ideal of $R_1$.
Then $\phi \sqbrk J$ is an ideal of $R_2$.

Proof:
$J$ is an ideal of $R_1$, so it is also a subring of $R_1$.
From Ring Homomorphism Preserves Subrings, it follows that $\phi \sqbrk J$ is a subring of $R_2$.
Now suppose $u \in \phi \sqbrk J$.
Let $v \in R_2$.
Then $\exists x \in J, y \in R_1$ such that $\map \phi x = u, \map \phi y = v$.
Thus, by the morphism property:
:$u \circ_2 v = \map \phi x \circ_2 \map \phi y = \map \phi {x \circ_1 y}$
So $u \circ_2 v \in \phi \sqbrk J$ because $x \circ_1 y \in J$.
Similarly $v \circ_2 u \in \phi \sqbrk J$ also.
The result follows.
{{qed}}
Theorem: Preimage of Image of Subring under Ring Homomorphism

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $K = \map \ker \phi$ be the kernel of $\phi$. 
Let $J$ be a subring of $R_1$.
Then:
:$\phi^{-1} \sqbrk {\phi \sqbrk J} = J + K$

Proof:
Let $x \in \phi^{-1} \sqbrk {\phi \sqbrk J}$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \phi^{-1} \sqbrk {\phi \sqbrk J}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi x
      | o = \in
      | r = \phi \sqbrk J
      | c = {{Defof|Preimage of Element under Mapping}}
}}
{{eqn | ll= \leadsto
      | q = \exists b \in J
      | l = \map \phi x
      | r = \map \phi b
      | c = {{Defof|Image of Element under Mapping}}
}}
{{eqn | ll= \leadsto
      | l = \map \phi x + \paren {-\map \phi b}
      | r = 0_{R_2}
      | c = {{Defof|Ring Negative}} on $R_2$
}}
{{eqn | ll= \leadsto
      | l = \map \phi {x + \paren {-b} }
      | r = 0_{R_2}
      | c = as $\phi$ is a homomorphism
}}
{{eqn | ll= \leadsto
      | l = x + \paren {-b}
      | o = \in
      | r = K
      | c = {{Defof|Kernel of Ring Homomorphism}}
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = J + K
      | c = as $x = b + \paren {x + \paren {-b} }$
}}
{{end-eqn}}
So we have shown that:
:$\phi^{-1} \sqbrk {\phi \sqbrk J} \subseteq J + K$
Now suppose that $x \in J + K$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = J + K
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists b \in J, a \in K
      | l = x
      | r = b + a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi x
      | r = \map \phi b + \map \phi a
      | c = 
}}
{{eqn | r = \map \phi b + 0_{R_2}
      | c = as $a \in K$
}}
{{eqn | r = \map \phi b
      | c =
}}
{{eqn | ll= \leadsto
      | l = \map \phi x
      | o = \in
      | r = \phi \sqbrk J
      | c = as $b \in J$
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = \phi^{-1} \sqbrk {\phi \sqbrk J}
      | c = {{Defof|Preimage of Subset under Mapping}}
}}
{{end-eqn}}
So we have shown that:
:$J + K \subseteq \phi^{-1} \sqbrk {\phi \sqbrk J}$
Hence the result by definition of set equality.
{{Qed}}
Theorem: Preimage of Subring under Ring Homomorphism is Subring

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S_2$ be a subring of $R_2$.
Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is a subring of $R_1$ such that $\map \ker \phi \subseteq S_1$.

Proof:
Let $K = \map \ker \phi$ be the kernel of $R_1$.
We have that $0_{R_2} \in S_2$ and so $\set {0_{R_2} } \subseteq S_2$.
From Subset Maps to Subset:
:$\phi^{-1} \sqbrk {\set {0_{R_2} } } \subseteq \phi^{-1} \sqbrk {S_2} = S_1$
But by definition, $K = \phi^{-1} \sqbrk {\set {0_{R_2} } }$
and so $S_1$ is a subset of $R_1$ containing $K$, that is:
:$K \subseteq S_1 \subseteq R_1$
Now we need to show that $S_1$ is a subring of $R_1$.
Let $r, r' \in S_1$.
Then $\map \phi r, \map \phi {r'} \in S_2$.
Hence:
{{begin-eqn}}
{{eqn | l = \map \phi {r +_1 r'}
      | r = \map \phi r +_2 \map \phi {r'}
      | c = as $\phi$ is a ring homomorphism
}}
{{eqn | o = \in
      | r = S_2
      | c = because $S_2$ is a subring
}}
{{end-eqn}}
So:
:$r + r' \in S_1$
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {-r}
      | r = -\map \phi r
      | c = Group Homomorphism Preserves Inverses
}}
{{eqn | o = \in
      | r = S_2
      | c = because $S_2$ is a subring
}}
{{end-eqn}}
So:
:$-r \in \phi^{-1} S_1$
Finally:
{{begin-eqn}}
{{eqn | l = \map \phi {r \circ_1 r'}
      | r = \map \phi r \circ_2 \map \phi {r'}
      | c = as $\phi$ is a ring homomorphism
}}
{{eqn | o = \in
      | r = S_2
      | c = because $S_2$ is a subring
}}
{{end-eqn}}
So:
:$r \circ_1 r' \in S_1$
So from the Subring Test we have that $\phi^{-1} \sqbrk {S_2}$ is a subring of $R$ containing $K$.
{{qed}}
Theorem: Preimage of Ideal under Ring Homomorphism is Ideal

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S_2$ be an ideal of $R_2$.
Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is an ideal of $R_1$ such that $\map \ker \phi \subseteq S_1$.

Proof:
From Preimage of Subring under Ring Homomorphism is Subring we have that $S_1 = \phi^{-1} \sqbrk {S_2}$ is a subring of $R_1$ such that $\map \ker \phi \subseteq S_1$.
We now need to show that $S_1$ is an ideal of $R_1$.
Let $s_1 \in S_1, r_1 \in R_1$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {r_1 \circ_1 s_1}
      | r = \map \phi {r_1} \circ_2 \map \phi {s_1}
      | c = as $\phi$ is a homomorphism
}}
{{eqn | r = S_2
      | c = as $S_2$ is an ideal of $R_2$
}}
{{end-eqn}}
Thus:
:$r_1 \circ_1 s_1 \in \phi^{-1} \sqbrk {S_2} = S_1$
Similarly for $s_1 \circ_1 r_1$.
So $S_1$ is an ideal of $R_1$.
{{qed}}
Theorem: Image of Preimage of Subring under Ring Epimorphism

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $S_2$ be a subring of $R_2$.
Then:
:$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$

Proof:
As $\phi$ is an epimorphism, it is a surjection, and so:
:$\Img \phi = R_2$
So:
:$S_2 \subseteq \Img {R_1}$
The result then follows from Image of Preimage under Mapping.
{{qed}}
Theorem: Diagonal Relation is Universally Congruent

The diagonal relation $\Delta_S$ on a set $S$ is universally congruent on $S$.

Proof:
We have that the diagonal relation is an equivalence relation.
Let $\struct {S, \circ}$ be any algebraic structure.
{{begin-eqn}}
{{eqn | o = 
      | r = x_1 \mathrel {\Delta_S} x_2 \land y_1 \mathrel {\Delta_S} y_2
      | c = 
}}
{{eqn | o = \leadsto
      | r = x_1 = x_2 \land y_1 = y_2
      | c = {{Defof|Diagonal Relation}}
}}
{{eqn | o = \leadsto
      | r = x_1 \circ y_1 = x_2 \circ y_2
      | c = as a consequence of equality
}}
{{eqn | o = \leadsto
      | r = \paren {x_1 \circ y_1} \mathrel {\Delta_S} \paren {x_2 \circ y_2}
      | c = {{Defof|Diagonal Relation}}
}}
{{end-eqn}}
$\Delta_S$ can therefore be described as universally congruent.
{{Qed}}
Theorem: Positive Elements of Ordered Ring

Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $P$ be the set of positive elements of $R$ , that is, $P = R_{\ge 0}$.
Then:
: $(1): \quad P + P \subseteq P$
: $(2): \quad P \cap \paren {-P} = \set {0_R}$
: $(3): \quad P \circ P \subseteq P$
If $\le$ is a total ordering, that is, if $\struct {R, +, \circ, \le}$ is a totally ordered ring, then:
: $(4): \quad P \cup \paren {-P} = R$
The converse is also true:
Let $\struct {R, +, \circ}$ be a ring.
Let $P \subseteq R$ such that $(1)$, $(2)$ and $(3)$ are satisfied.
Then there is one and only one ordering $\le$ compatible with the ring structure of $R$ such that $P = R_{\ge 0}$.
Also, if $(4)$ is also satisfied, then $\le$ is a total ordering.

Proof:
=== Necessary Condition ===
First, suppose that $\le$ is compatible with the ring structure of $R$.
Hence:
:$(OR1): \quad \le$ is compatible with $+$
:$(OR2): \quad \forall x, y \in R: 0_R \le x, 0_R \le y \implies 0_R \le x \circ y$.
$(1)$: Let $x, y \in R: 0_R \le x, 0_R \le y$.
Then $0_R + 0_R \le x + y$ by the fact that $\le$ is compatible with $+$.
Thus $0_R \le x + y$ and thus $x + y \in P$.
$(2)$: By Properties of Ordered Ring item 4, $-P = \set {x \in R: x \le 0_R}$.
Let $x \in P \cap \paren {-P}$.
Then $x \le 0_R$ and $0_R \le x$.
So from the antisymmetric nature of $\le$, $x = 0_R$.
$(3)$: This is equivalent to $0_R \le x, 0_R \le y \implies 0_R \le x \circ y$ which is one of the properties of being compatible with the ring structure of $R$.
$(4)$: Now if $\le$ is a total ordering, then $\forall x \in R$, either $x \le 0_R$ or $0_R \le x$, and the result follows.
Theorem: Symmetric Difference with Intersection forms Ring

Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cap$ denote the set intersection operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity, in which the unity is $S$.
This ring is not an integral domain.

Proof:
From Symmetric Difference on Power Set forms Abelian Group, $\struct {\powerset S, \symdif}$ is an abelian group, where $\O$ is the identity and each element is self-inverse.
From Power Set with Intersection is Monoid, $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$.
Also Intersection Distributes over Symmetric Difference.
Thus $\struct {\powerset S, \cap}$ is a commutative ring with a unity which is $S$.
From Intersection with Empty Set:
:$\forall A \in \powerset S: A \cap \O = \O = \O \cap A$
Thus $\O$ is indeed the zero.
However, from Set Intersection Not Cancellable, it follows that $\struct {\powerset S, \symdif, \cap}$ is not an integral domain.
{{qed}}
Theorem: Symmetric Difference with Intersection forms Ring

Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cap$ denote the set intersection operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity, in which the unity is $S$.
This ring is not an integral domain.

Proof:
From Power Set is Closed under Symmetric Difference and Power Set is Closed under Intersection, we have that both $\struct {\powerset S, \symdif}$ and $\struct {\powerset S, \cap}$ are closed.
Hence $\powerset S$ is a ring of sets, and hence a commutative ring.
From Intersection with Subset is Subset, we have $A \subseteq S \iff A \cap S = A$.
Thus we see that $S$ is the unity.
Also during the proof of Power Set with Intersection is Monoid, it was established that $S$ is the identity of $\struct {\powerset S, \cap}$.
We also note that set intersection is not cancellable, so $\struct {\powerset S, \symdif, \cap}$ is not an integral domain.
The result follows.
{{qed}}
Theorem: Field of Quotients of Subdomain

Let $\struct {F, +, \circ}$ be a field whose unity is $1_F$.
Let $\struct {D, +, \circ}$ be a subdomain of $\struct {F, +, \circ}$ whose unity is $1_D$.
Let:
:$K = \set {\dfrac x y: x \in D, y \in D^*}$
where $\dfrac x y$ is the division product of $x$ by $y$.
Then $\struct {K, +, \circ}$ is a field of quotients of $\struct {D, +, \circ}$.

Proof:
$1_D = 1_F$ by Subdomain Test.
The sum and product of two elements of $K$ are also in $K$ by Addition of Division Products and Product of Division Products.
The additive and product inverses of $K$ are also in $K$ by Negative of Division Product and Inverse of Division Product.
Thus by Subfield Test, $\struct {K, +, \circ}$ is a subfield of $\struct {F, +, \circ}$ which clearly contains $\struct {D, +, \circ}$.
Hence the result.
{{qed}}
Theorem: Existence of Field of Quotients

Let $\struct {D, +, \circ}$ be an integral domain.
Then there exists a field of quotients of $\struct {D, +, \circ}$.

Proof:
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Theorem: Quotient Theorem for Monomorphisms

Let $K, L$ be fields of quotients of integral domains $\struct {R, +_R, \circ_R}, \struct {S, +_S, \circ_S}$ respectively.
Let $\phi: R \to S$ be a monomorphism.
Then there is one and only one monomorphism $\psi: K \to L$ extending $\phi$, and:
:$\forall x \in R, y \in R^*: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$
Also, if $\phi$ is a ring isomorphism, then so is $\psi$.

Proof:
By definition, $\struct {K, \circ_R}$ and $\struct {L, \circ_S}$ are inverse completions of $\struct {R, \circ_R}$ and $\struct {S, \circ_S}$ respectively.
{{Questionable|this is not the definition}}
So by the Extension Theorem for Homomorphisms, there is one and only one monomorphism $\psi: \struct {K, \circ_R} \to \struct {L, \circ_S$ extending $\phi$.
Thus:
:$\forall x \in R, y \in R^*: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$
By the Extension Theorem for Isomorphisms, $\psi$ is an isomorphism if $\phi$ is.
Thus, $\forall x, y \in R, z, w \in R^*$:
{{begin-eqn}}
{{eqn | l = \map \psi {\frac x z +_R \frac y w}
      | r = \map \psi {\frac {\paren {x \circ_R w} +_R \paren {y \circ_R z} } {z \circ_R w} }
      | c = Addition of Division Products
}}
{{eqn | r = \frac {\map \phi {\paren {x \circ_R w} +_R \paren {y \circ_R z} } } {\map \phi {z \circ_R w} }
      | c = Definition of $\psi$
}}
{{eqn | r = \frac {\paren {\map \phi x \circ_S \map \phi w} +_S \paren {\map \phi y \circ_S \map \phi z} } {\map \phi z \circ_S \map \phi w}
      | c = Morphism Property
}}
{{eqn | r = \frac {\map \phi x} {\map \phi z} +_S \frac {\map \phi y} {\map \phi w}
      | c = Addition of Division Products
}}
{{eqn | r = \map \psi {\frac x z} +_S \map \psi {\frac y w}
      | c = Definition of $\psi$
}}
{{end-eqn}}
Thus $\psi: K \to L$ is a monomorphism.
{{qed}}
Theorem: Field of Quotients is Unique

Let $\struct {D, +, \circ}$ be an integral domain.
Let $K, L$ be field of quotients of $\struct {D, +, \circ}$.
Then there is one and only one (field) isomorphism $\phi: K \to L$ satisfying:
:$\forall x \in D: \map \phi x = x$

Proof:
Follows directly from the Quotient Theorem for Monomorphisms.
{{qed}}
Theorem: Divided by Positive Element of Field of Quotients

Let $\struct {K, +, \circ}$ be the field of quotients of a totally ordered integral domain $\struct {D, +, \circ, \le}$.
Then:
:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{>0}$

Proof:
By definition of field of quotients:
:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{\ne 0}$
Suppose $z = x' / y'$ such that $y' \notin D_{>0}$.
Then $y' < 0$ as $D$ is totally ordered.
Then:
{{begin-eqn}}
{{eqn | l = x' / y'
      | r = x' \circ \paren {y'}^{-1}
      | c = {{Defof|Division over Field}}
}}
{{eqn | r = \paren {-x'} \circ \paren {-\paren {y'}^{-1} }
      | c = Product of Ring Negatives
}}
{{eqn | r = \paren {-x'} \circ \paren {-y'}^{-1}
      | c = Negative of Product Inverse
}}
{{eqn | r = \paren {-x'} / \paren {-y'}
      | c = {{Defof|Division over Field}}
}}
{{end-eqn}}
If $y' < 0$, then $\paren {-y'} > 0$ from Properties of Ordered Ring $(4)$.
So all we need to do is set $x = -x', y = -y'$ and the result follows.
{{qed}}
Category:Fields of Quotients
Category:Integral Domains
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Theorem: Order Embedding between Quotient Fields is Unique

Let $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ be totally ordered integral domains.
Let $K, L$ be totally ordered fields of quotients of $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ respectively.
Let $\phi: R \to S$ be a order embedding.
Then there exists exactly one order embedding $\psi: K \to L$ extending $\phi$.
Also:
:$\forall x \in R, y \in R_{\ne 0}: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$
If $\phi: R \to S$ is an order isomorphism, then so is $\psi$.

Proof:
By Field of Quotients is Unique, all we need to show is:
:$\forall x_1, x_2 \in R, y_1, y_2 \in R_{> 0}: \dfrac {x_1} {y_1} \le \dfrac {x_2} {y_2} \iff \dfrac {\map \phi {x_1}} {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} }$
Let $x_1 / y_1 \le x_2 / y_2$, where $y_1, y_2 \in R_{> 0}$.
As $y_1, y_2 \in R_{> 0}$, it follows that $0 < y_1 \circ_1 y_2$ and $0 < 1 / \paren {y_1 \circ_1 y_2}$.
We also have:
:$0 < \map \phi {y_1} \circ_2 \map \phi {y_2} = \map \phi {y_1 \circ_1 y_2}$
Therefore:
:$x_1 \circ_1 y_2 = \frac {x_1} {y_1} \circ_1 \paren {y_1 \circ y_2} \le \dfrac {x_2} {y_2} \circ_1 \paren {y_1 \circ_1 y_2} = x_2 \circ_1 y_1$
Conversely, let $x_1 \circ_1 y_2 \le x_2 \circ_1 y_1$.
Then:
:$\dfrac {x_1} {y_1} = x_1 \circ_1 y_2 \circ_1 \paren {\dfrac 1 {y_1 \circ_1 y_2} } \le x_2 \circ_1 y_1 \circ_1 \paren {\dfrac 1 {y_1 \circ_1 y_2} } = \dfrac {x_2} {y_2}$
That is, we have:
:$\dfrac {x_1} {y_1} \le \dfrac {x_2} {y_2} \iff x_1 \circ_1 y_2 \le x_2 \circ_1 y_1$
Similarly:
:$\dfrac {\map \phi {x_1} } {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} } \iff \map \phi {x_1} \circ_2 \map \phi {y_2} \le \map \phi {x_2} \circ_2 \map \phi {y_1}$
Now $\phi: R \to S$ is an order embedding.
Therefore:
:$x_1 \circ_1 y_2 \le x_2 \circ_1 y_1 \iff \map \phi {x_1 \circ_1 y_2} \le \map \phi {x_2 \circ_1 y_1}$
The result follows.
{{qed}}
Theorem: Strict Ordering Preserved under Product with Cancellable Element

Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $x, y, z \in S$ be such that:
:$(1): \quad z$ is cancellable for $\circ$
:$(2): \quad x \prec y$
Then:
:$x \circ z \prec y \circ z$
:$z \circ x \prec z \circ y$

Proof:
Let $z$ be cancellable and $x \prec y$.
Then by the definition of ordered semigroup:
:$x \circ z \preceq y \circ z$
From the fact that $z$ is cancellable:
:$x \circ z = y \circ z \iff x = y$
Thus as $x \circ z \ne y \circ z$ it follows from Strictly Precedes is Strict Ordering that:
:$x \circ z \prec y \circ z$
Similarly, $z \circ x \prec z \circ y$ follows from $z \circ x \preceq z \circ y$.
{{qed}}
Theorem: Ordering of Inverses in Ordered Monoid

Let $\struct {S, \circ, \preceq}$ be an ordered monoid whose identity is $e$.
Let $x, y \in S$ be invertible.
Then:
:$x \prec y \iff y^{-1} \prec x^{-1}$

Proof:
=== Necessary Condition ===
{{begin-eqn}}
{{eqn | l = x
      | o = \prec
      | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e
      | r = x^{-1} \circ x \prec x^{-1} \circ y
      | c = Strict Ordering Preserved under Product with Cancellable Element
}}
{{eqn | ll= \leadsto
      | l = y^{-1}
      | r = e \circ y^{-1} \prec x^{-1} \circ y \circ y^{-1} = x^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y^{-1}
      | o = \prec
      | r = x^{-1}
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Monomorphism from Total Ordering

Let the following conditions hold:
:$(1): \quad$ Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
:$(2): \quad$ Let $\phi: S \to T$ be a mapping.
:$(3): \quad$ Let $\preceq$ be a total ordering on $S$.
Then $\phi \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ is a (structure) monomorphism {{iff}}:
:$(1): \quad \phi$ is strictly increasing from $\struct {S, \preceq}$ into $\struct {T, \preccurlyeq}$
:$(2): \quad \phi$ is a homomorphism from $\struct {S, \circ}$ into $\struct {T, *}$.

Proof:
This follows:
:$(1): \quad$ As a direct consequence of Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
:$(2): \quad$ From the definition of monomorphism as a homomorphism which is an injection.
{{Qed}}
Theorem: Extension Theorem for Total Orderings

Let the following conditions be fulfilled:
:$(1):\quad$ Let $\struct {S, \circ, \preceq}$ be a totally ordered commutative semigroup
:$(2):\quad$ Let all the elements of $\struct {S, \circ, \preceq}$ be cancellable
:$(3):\quad$ Let $\struct {T, \circ}$ be an inverse completion of $\struct {S, \circ}$.
Then:
:$(1):\quad$ The relation $\preceq'$ on $T$ satisfying $\forall x_1, x_2, y_1, y_2 \in S: x_1 \circ {y_1}^{-1} \preceq' x_2 \circ {y_2}^{-1} \iff x_1 \circ y_2 \preceq x_2 \circ y_1$ is a well-defined relation
:$(2):\quad$ $\preceq'$ is the only total ordering on $T$ compatible with $\circ$
:$(3):\quad$ $\preceq'$ is the only total ordering on $T$ that induces the given ordering $\preceq$ on $S$.

Proof:
By Inverse Completion is Commutative Semigroup:
:every element of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$.
Theorem: Strict Lower Closure of Sum with One

Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Then:
:$\forall n \in \struct {S, \circ, \preceq}: \paren {n \circ 1}^\prec = n^\prec \cup \set n$
where $n^\prec$ is defined as the strict lower closure of $n$, that is, the set of elements strictly preceding $n$.

Proof:
First note that as $\struct {S, \circ, \preceq}$ is well-ordered and hence totally ordered, the Trichotomy Law applies.
Thus:
{{begin-eqn}}
{{eqn | q = \forall m \in S
      | o = 
      | r = m \notin n^\prec
      | c = 
}}
{{eqn | o = \leadstoandfrom
      | r = \neg \ m \prec n
      | c = {{Defof|Strict Lower Closure of Element}}
}}
{{eqn | o = \leadstoandfrom
      | r = m = n \lor n \prec m
      | c = Trichotomy Law
}}
{{eqn | o = \leadstoandfrom
      | r = n \preceq m
      | c = {{Defof|Strictly Precede}}
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | q = \forall p \in S
      | o = 
      | r = p \notin \paren {n \circ 1}^\prec
      | c = 
}}
{{eqn | o = \leadstoandfrom
      | r = n \circ 1 \preceq p
      | c = from the above
}}
{{eqn | o = \leadstoandfrom
      | r = n \prec p
      | c = Sum with One is Immediate Successor in Naturally Ordered Semigroup
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | q = \forall p \in S
      | o = 
      | r = p \notin n^\prec \cup \set n
      | c = 
}}
{{eqn | o = \leadstoandfrom
      | r = n \preceq p \land n \ne p
      | c = from the above
}}
{{eqn | o = \leadstoandfrom
      | r = n \prec p
      | c = {{Defof|Strictly Precede}}
}}
{{end-eqn}}
So:
:$p \notin n^\prec \cup \set n \iff p \notin \paren {n \circ 1}^\prec$
Thus:
:$\relcomp S {\paren {n \circ 1}^\prec} = \relcomp S {n^\prec \cup \set n}$
from the definition of relative complement.
So:
{{begin-eqn}}
{{eqn | l = \paren {n \circ 1}^\prec
      | r = \relcomp S {\relcomp S {\paren {n \circ 1}^\prec} }
      | c = Relative Complement of Relative Complement
}}
{{eqn | r = \relcomp S {\relcomp S {n^\prec \cup \set n} }
      | c = from above
}}
{{eqn | r = n^\prec \cup \set n
      | c = Relative Complement of Relative Complement
}}
{{end-eqn}}
{{qed}}