hcju/proofwikips huggingface dataset:
Idempotent Elements of Ring with No Proper Zero Divisors
Let $\left({R, +, \circ}\right)$ be a non-null ring with no (proper) zero divisors. Let $x \in R$. Then: :$x \circ x = x \iff x \in \left\{{0_R, 1_R}\right\}$ That is, the only elements of $R$ that are idempotent are zero and unity.
Non-Zero Elements of Division Ring form Group
Let $\struct {R, +, \circ}$ be a division ring. Then $\struct {R^*, \circ}$ is a group.
Null Ring and Ring Itself Subrings
In any ring $R$, the null ring and $R$ itself are subrings of $R$.
Subring Test
Let $S$ be a subset of a ring $\struct {R, +, \circ}$. Then $\struct {S, +, \circ}$ is a subring of $\struct {R, +, \circ}$  these all hold: :$(1): \quad S \ne \O$ :$(2): \quad \forall x, y \in S: x + \paren {-y} \in S$ :$(3): \quad \forall x, y \in S: x \circ y \in S$
Subdomain Test
Let $S$ be a subset of an integral domain $\struct {R, +, \circ}$. Then $\struct {S, +\restriction_S, \circ \restriction_S}$ is a subdomain of $\struct {R, +, \circ}$  these conditions hold: :$(1): \quad$ $\struct {S, + \restriction_S, \circ \restriction_S}$ is a subring of $\struct {R, +, \circ}$ :$(2): \quad$ The unity of $R$ is also in $S$, that is $1_R = 1_S$.
Centralizer of Ring Subset is Subring
Let $S$ be a subset of a ring $\struct {R, +, \circ}$ Then $\map {C_R} S$, the centralizer of $S$ in $R$, is a subring of $R$.
Center of Ring is Commutative Subring
The center $\map Z R$ of a ring $R$ is a commutative subring of $R$.
Ideal is Subring
Let $\left({R, +, \circ}\right)$ be a ring, and let $J$ be an ideal of $R$. Then $J$ is a subring of $R$.
Ring is Ideal of Itself
Let $\struct {R, +, \circ}$ be a ring. Then $R$ is an ideal of $R$.
Ideal of Unit is Whole Ring
Let $\struct {R, +, \circ}$ be a ring with unity. Let $J$ be an ideal of $R$. If $J$ contains a unit of $R$, then $J = R$.
Test for Ideal
Let $J$ be a subset of a ring $\struct {R, +, \circ}$. Then $J$ is an ideal of $\struct {R, +, \circ}$  these all hold: : $(1): \quad J \ne \O$ : $(2): \quad \forall x, y \in J: x + \paren {-y} \in J$ : $(3): \quad \forall j \in J, r \in R: r \circ j \in J, j \circ r \in J$
Epimorphism Preserves Rings
Let $\struct {R_1, +_1, \circ_1}$ be a ring, and $\struct {R_2, +_2, \circ_2}$ be a closed algebraic structure. Let $\phi: R_1 \to R_2$ be an epimorphism. Then $\struct {R_2, +_2, \circ_2}$ is a ring.
Ring Homomorphism of Addition is Group Homomorphism
Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Then $\phi: \left({R_1, +_1}\right) \to \left({R_2, +_2}\right)$ is a group homomorphism.
Element of Integral Domain Divides Zero
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$. Then every element of $D$ is a divisor of zero: :$\forall x \in D: x \divides 0_D$
Unity Divides All Elements
Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$. Then unity is a divisor of every element of $D$: :$\forall x \in D: 1_D \divides x$ Also: :$\forall x \in D: -1_D \divides x$
Element of Integral Domain is Divisor of Itself
Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$. Then every element of $D$ is a divisor of itself: :$\forall x \in D: x \divides x$
Unit of Integral Domain divides all Elements
Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$. Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$. Then: :$\forall x \in D: \forall u \in U_D: u \divides x$ That is, every unit of $D$ is a divisor of every element of $D$.
Divisor Relation in Integral Domain is Transitive
Let $\struct {D, +, \circ}$ be an integral domain. Let $x, y, z \in D$. Then: :$x \divides y \land y \divides z \implies x \divides z$
Integers form Unique Factorization Domain
The integers $\struct {\Z, +, \times}$ form a unique factorization domain.
Trivial Ordering Compatibility in Boolean Ring
Let $\struct {S, +, \circ}$ be a Boolean ring. Then the trivial ordering is the only ordering on $S$ compatible with both its operations.
Field is Integral Domain
Every field is an integral domain.
Finite Integral Domain is Galois Field
A finite integral domain is a Galois field.
Center of Division Ring is Subfield
Let $\struct {K, +, \circ}$ be an division ring. Then $\map Z K$, the center of $K$, is a subfield of $K$.
Ideals of Field
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$. Then $\struct {R, +, \circ}$ is a field  the only ideals of $\struct {R, +, \circ}$ are $\struct {R, +, \circ}$ and $\set {0_R}$.
Epimorphism from Division Ring to Ring
Let $\left({K, +, \circ}\right)$ be a division ring whose zero is $0_K$. Let $\left({R, +, \circ}\right)$ be a ring whose zero is $0_R$. Let $\phi: K \to R$ be a ring epimorphism. Then one of the following applies: :$(1): \quad R$ is a null ring :$(2): \quad R$ is a division ring and $\phi$ is a ring isomorphism.
Congruence Class Modulo Subgroup is Coset
Let $G$ be a group. Let $H$ be a subgroup of $G$.
Cosets are Equivalent
All left cosets of a group $G$ with respect to a subgroup $H$ are equivalent. That is, any two left cosets are in one-to-one correspondence. The same applies to right cosets. As a special case of this: :$\forall x \in G: \order {x H} = \order H = \order {H x}$ where $H$ is a subgroup of $G$.
Left and Right Coset Spaces are Equivalent
Let $\struct {G, \circ}$ be a group. Let $H$ be a subgroup of $G$. Let: :$x H$ denote the left coset of $H$ by $x$ :$H y$ denote the right coset of $H$ by $y$. Then: :$\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$
Cosets in Abelian Group
Let $G$ be an abelian group. Then every right coset modulo $H$ is a left coset modulo $H$. That is: :$\forall x \in G: x H = H x$ In an abelian group, therefore, we can talk about '''congruence modulo $H$''' and not worry about whether it is left or right.
Conjugate of Set by Identity
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $S \subseteq G$. Then the conjugate of $S$ by $e$ is $S$: :$S^e = S$
Conjugate of Set by Group Product
Let $\struct {G, \circ}$ be a group. Let $S \subseteq G$. Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as: :$S^a := \set {y \in G: \exists x \in S: y = a \circ x \circ a^{-1} } = a \circ S \circ a^{-1}$ Then: :$\paren {S^a}^b = S^{b \circ a}$
Conjugate of Subgroup is Subgroup
Let $G$ be a group. Let $H \le G$ be a subgroup of $G$. Then the conjugate of $H$ by $a$ is a subgroup of $G$: :$\forall H \le G, a \in G: H^a \le G$
Inner Automorphisms form Normal Subgroup of Automorphism Group
Let $G$ be a group. Then the set $\Inn G$ of all inner automorphisms of $G$ is a normal subgroup of the automorphism group $\Aut G$ of $G$: :$\Inn G \lhd \Aut G$
Subgroup of Abelian Group is Normal
Every subgroup of an abelian group is normal.
Intersection with Normal Subgroup is Normal
Let $G$ be a group. Let $H$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$. Then $H \cap N$ is a normal subgroup of $H$.
Kernel is Normal Subgroup of Domain
Let $\phi$ be a group homomorphism. Then the kernel of $\phi$ is a normal subgroup of the domain of $\phi$: :$\map \ker \phi \lhd \Dom \phi$
Coset Product is Well-Defined
Let $\struct {G, \circ}$ be a group. Let $N$ be a normal subgroup of $G$. Let $a, b \in G$. Then the coset product: :$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$ is well-defined.
Quotient Group of Abelian Group is Abelian
Let $G$ be an abelian group. Let $N \le G$. Then the quotient group $G / N$ is abelian.
Quotient Theorem for Group Epimorphisms
Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be groups. Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a group epimorphism. Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively. Let $K = \map \ker \phi$ be the kernel of $\phi$. There is one and only one group isomorphism $\psi: G / K \to H$ satisfying: :$\psi \circ q_K = \phi$ where $q_K$ is the quotient epimorphism from $G$ to $G / K$.
Congruence Relation induces Normal Subgroup
Let $\left({G, \circ}\right)$ be a group whose identity is $e$. Let $\mathcal R$ be a congruence relation for $\circ$. Let $H = \left[\!\left[{e}\right]\!\right]_\mathcal R$, where $\left[\!\left[{e}\right]\!\right]_\mathcal R$ is the equivalence class of $e$ under $\mathcal R$. Then: : $(1): \quad \left({H, \circ \restriction_H}\right)$ is a normal subgroup of $G$ : $(2): \quad \mathcal R$ is the equivalence relation $\mathcal R_H$ defined by $H$ : $(3): \quad \left({G / \mathcal R, \circ_\mathcal R}\right)$ is the subgroup $\left({G / H, \circ_H}\right)$ of the semigroup $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$.
Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal
Let $G$ be a group. Let $H \lhd G$ where $\lhd$ denotes that $H$ is a normal subgroup of $G$. Let $K \lhd G/H$ and $L = q_H^{-1} \left[{K}\right]$, where: :$q_H: G \to G/H$ is the quotient epimorphism from $G$ to the quotient group $G/H$ :$q_H^{-1} \left[{K}\right]$ is the preimage of $K$ under $q_H$. Then: :$L \lhd G$
Trivial Quotient Group is Quotient Group
Let $G$ be a group. Then the trivial quotient group: :$G / \set {e_G} \cong G$ where: :$\cong$ denotes group isomorphism :$e_G$ denotes the identity element of $G$ is a quotient group.
Correspondence Theorem (Group Theory)
Let $G$ be a group. Let $N \lhd G$ be a normal subgroup of $G$. Then every subgroup of the quotient group $G / N$ is of the form $H / N = \set {h N: h \in H}$, where $N \le H \le G$. Conversely, if $N \le H \le G$ then $H / N \le G / N$. The correspondence between subgroups of $G / N$ and subgroups of $G$ containing $N$ is a bijection. This bijection maps normal subgroups of $G / N$ onto normal subgroups of $G$ which contain $N$.
Centralizer of Group Element is Subgroup
Let $\struct {G, \circ}$ be a group and let $a \in G$. Then $\map {C_G} a$, the centralizer of $a$ in $G$, is a subgroup of $G$.
Centralizer in Subgroup is Intersection
Let $G$ be a group. Let $H$ be a subgroup of $G$. Then: :$\forall x \in G: \map {C_H} x = \map {C_G} x \cap H$ That is, the centralizer of an element in a subgroup is the intersection of that subgroup with the centralizer of the element in the group.
Kernel of Inner Automorphism Group is Center
Let the mapping $\kappa: G \to \Inn G$ from a group $G$ to its inner automorphism group $\Inn G$ be defined as: :$\map \kappa a = \kappa_a$ where $\kappa_a$ is the inner automorphism of $G$ given by $a$. Then $\kappa$ is a group epimorphism, and its kernel is the center of $G$: :$\map \ker \kappa = \map Z G$
Group equals Center iff Abelian
Let $G$ be a group. Then $G$ is abelian  $\map Z G = G$, that is,  $G$ equals its center.
Normalizer is Subgroup
Let $G$ be a group. The normalizer of a subset $S \subseteq G$ is a subgroup of $G$. :$S \subseteq G \implies \map {N_G} S \le G$
Subgroup is Normal Subgroup of Normalizer
Let $G$ be a group. A subgroup $H \le G$ is a normal subgroup of its normalizer: :$H \le G \implies H \lhd \map {N_G} H$
Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup
Let $G$ be a group. Let $H$ be a subgroup of $G$. Then $\map {N_G} H$, the normalizer of $H$ in $G$, is the largest subgroup of $G$ containing $H$ as a normal subgroup.
Normal Subgroup iff Normalizer is Group
Let $G$ be a group. Let $H$ be a subgroup of $G$. Then $H$ is normal in $G$  the normalizer of $H$ is equal to $G$: :$H \lhd G \iff \map {N_G} H = G$
Normalizer of Conjugate is Conjugate of Normalizer
The normalizer of a conjugate is the conjugate of the normalizer: :$S \subseteq G \implies \map {N_G} {S^a} = \paren {\map {N_G} S}^a$
Quotient Ring is Ring/Quotient Ring Product is Well-Defined
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$. Then $\circ$ is well-defined on $R / J$, that is: :$x_1 + J = x_2 + J, y_1 + J = y_2 + J \implies x_1 \circ y_1 + J = x_2 \circ y_2 + J$
Quotient Ring of Commutative Ring is Commutative
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$. If $\struct {R, +, \circ}$ is a commutative ring, then so is $\struct {R / J, +, \circ}$.
Quotient Ring of Ring with Unity is Ring with Unity
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$. Then $\struct {R / J, +, \circ}$ is a ring with unity, and its unity is $1_R + J$.
Ring Epimorphism with Trivial Kernel is Isomorphism
Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring epimorphism. Then $\phi$ is a ring isomorphism iff $\ker \left({\phi}\right) = \left\{ {0_{R_1} }\right\}$.
Quotient Epimorphism is Epimorphism/Ring
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$. Let $\phi: R \to R / J$ be the quotient (ring) epimorphism from $R$ to $R / J$: :$x \in R: \map \phi x = x + J$ Then $\phi$ is a ring epimorphism whose kernel is $J$.
Ring Homomorphism Preserves Subrings
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism. If $S$ is a subring of $R_1$, then $\phi \sqbrk S$ is a subring of $R_2$.
Kernel of Ring Homomorphism is Subring
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism. Then the kernel of $\phi$ is a subring of $R_1$.
Kernel of Ring Homomorphism is Ideal
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism. Then the kernel of $\phi$ is an ideal of $R_1$.
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.
Quotient Ring of Kernel of Ring Epimorphism
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism. Let $K = \map \ker \phi$. Then there is a unique ring isomorphism $g: R_1 / K \to R_2$ such that: :$g \circ q_K = \phi$ $\phi$ is an isomorphism  $K = \set {0_{R_1} }$.
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 $R_2$.
Preimage of Image of Subring under Ring Homomorphism
Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Let $K = \ker \left({\phi}\right)$, where $\ker \left({\phi}\right)$ is the kernel of $\phi$.  Let $J$ be a subring of $R_1$. Then: :$\phi^{-1} \left[{\phi \left[{J}\right]}\right] = J + K$
Preimage of Subring under Ring Homomorphism is Subring
Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Let $S_2$ be a subring of $R_2$. Then $S_1 = \phi^{-1} \left[{S_2}\right]$ is a subring of $R_1$ such that $\ker \left({\phi}\right) \subseteq S_1$.
Preimage of Ideal under Ring Homomorphism is Ideal
Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Let $S_2$ be an ideal of $R_2$. Then $S_1 = \phi^{-1} \left[{S_2}\right]$ is an ideal of $R_1$ such that $\ker \left({\phi}\right) \subseteq S_1$.
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: :$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$
Properties of Ordered Ring
Let $\struct {R, +, \circ, \le}$ be an ordered ring whose zero is $0_R$ and whose unity is $1_R$. Let $U_R$ be the group of units of $R$. Let $x, y, z \in \struct {R, +, \circ, \le}$. Then the following properties hold: : $(1): \quad x < y \iff x + z < y + z$. Hence $x \le y \iff x + z \le y + z$ (because $\struct {R, +, \le}$ is an ordered group). : $(2): \quad x < y \iff 0 < y + \paren {-x}$. Hence $x \le y \iff 0 \le y + \paren {-x}$ : $(3): \quad 0 < x \iff \paren {-x} < 0$. Hence $0 \le x \iff \paren {-x} \le 0$ : $(4): \quad x < 0 \iff 0 < \paren {-x}$. Hence $x \le 0 \iff 0 \le \paren {-x}$ : $(5): \quad \forall n \in \Z_{>0}: x > 0 \implies n \cdot x > 0$ : $(6): \quad x \le y, 0 \le z: x \circ z \le y \circ z, z \circ x \le z \circ y$ : $(7): \quad x \le y, z \le 0: y \circ z \le x \circ z, z \circ y \le z \circ x$
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.
Symmetric Difference with Intersection forms Ring
Let $S$ be a set. Then $\struct {\powerset S, *, \cap}$ is a commutative ring with unity, in which the unity is $S$. This ring is not an integral domain.
Subfield Test
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$. Let $K$ be a subset of $F$. Then $\struct {K, +, \circ}$ is a subfield of $\struct {F, +, \circ}$  these all hold: :$(1): \quad K^* \ne \O$ :$(2): \quad \forall x, y \in K: x + \paren {-y} \in K$ :$(3): \quad \forall x, y \in K: x \circ y \in K$ :$(4): \quad x \in K^* \implies x^{-1} \in K^*$ where $K^*$ denotes $K \setminus \set {0_F}$.
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 $\struct {D, +, \circ}$.
Existence of Field of Quotients
Let $\struct {D, +, \circ}$ be an integral domain. Then there exists a field of quotients of $\struct {D, +, \circ}$.
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 $\psi$.
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: :$\forall x \in D: \map \phi x = x$
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: :$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{>0}$
Total Ordering on Field of Quotients is Unique
Let $\struct {K, +, \circ}$ be a field of quotients of an ordered integral domain $\struct {D, +, \circ, \le}$. Then there is one and only one total ordering $\le'$ on $K$ which is compatible with its ring structure and induces on $D$ its given total ordering $\le$. That ordering is the one defined by: :$P = \set {\dfrac x y \in K: x \in D_+, y \in D_+^*}$
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 is one and only 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 $\psi$.
Strict Ordering Preserved under Product with Cancellable Element
Let $\left({S, \circ, \preceq}\right)$ 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: : $x \circ z \prec y \circ z$ : $z \circ x \prec z \circ y$
Ordering of Inverses in Ordered Monoid
Let $\left({S, \circ, \preceq}\right)$ be an ordered monoid whose identity is $e$. Let $x, y \in S$ be invertible. Then: :$x \prec y \iff y^{-1} \prec x^{-1}$
Monomorphism from Total Ordering
Let the following conditions hold: : $(1): \quad$ Let $\left({S, \circ, \preceq}\right)$ and $\left({T, *, \preccurlyeq}\right)$ be ordered semigroups. : $(2): \quad$ Let $\phi: S \to T$ be a mapping. : $(3): \quad$ Let $\preceq$ be a total ordering on $S$. Then $\phi \left({S, \circ, \preceq}\right) \to \left({T, *, \preccurlyeq}\right)$ is a (structure) monomorphism iff: : $(1): \quad \phi$ is strictly increasing from $\left({S, \preceq}\right)$ into $\left({T, \preccurlyeq}\right)$; : $(2): \quad \phi$ is a homomorphism from $\left({S, \circ}\right)$ into $\left({T, *}\right)$.
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 $\forall x_1, x_2, y_1, y_2 \in S: x_1 \circ \paren {y_1}^{-1} \preceq' x_2 \circ \paren {y_2}^{-1} \iff x_1 \circ y_2 \preceq x_2 \circ y_1$ 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 $S$.
Cancellability in Naturally Ordered Semigroup
Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup. Then:
Strict Lower Closure of Sum with One
Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup. Then: :$\forall n \in \left({S, \circ, \preceq}\right): \left({n \circ 1}\right)^\prec = n^\prec \cup \left\{{n}\right\}$ where $n^\prec$ is defined as the strict lower closure of $n$, that is, the set of elements strictly preceding $n$.
Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor
Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup. Then: :$\forall m, n \in \left({S, \circ, \preceq}\right): m \preceq n \implies \left[{m \,.\,.\, n \circ 1}\right] = \left[{m \,.\,.\, n}\right] \cup \left\{{n \circ 1}\right\}$ where $\left[{m \,.\,.\, n}\right]$ is the closed interval between $m$ and $n$.
Homomorphism of Powers
Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be semigroups. Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism.
Naturally Ordered Semigroup is Unique
Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be naturally ordered semigroups. Let: :$0'$ be the smallest element of $S'$ :$1'$ be the smallest element of $S' \setminus \set {0'} = S'^*$. Then the mapping $g: S \to S'$ defined as: :$\forall a \in S: \map g a = \circ'^a 1'$ is an isomorphism from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$. This isomorphism is unique. Thus, up to isomorphism, there is only one naturally ordered semigroup.
Consecutive Subsets of N
Let $\N_k$ denote the initial segment of the natural numbers determined by $k$: :$\N_k = \left\{{0, 1, 2, 3, \ldots, k - 1}\right\}$ Then: :$\N_k = \N_{k + 1} \setminus \left\{{k}\right\}$ In particular: :$\N_{k - 1} = \N_k \setminus \left\{{k - 1}\right\}$