hcju/proofwikips huggingface dataset:
Epimorphism Preserves Associativity
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures. Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism. Let $\circ$ be an associative operation. Then $*$ is also an associative operation.
Epimorphism Preserves Semigroups
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures. Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism. Let $\left({S, \circ}\right)$ be a semigroup. Then $\left({T, *}\right)$ is also a semigroup.
Homomorphism Preserves Subsemigroups
Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups. Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism. Let $S'$ be a subsemigroup of $S$. Then $\phi \paren {S'}$ is a subsemigroup of $T$.
Epimorphism Preserves Commutativity
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures. Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism. Let $\circ$ be a commutative operation. Then $*$ is also a commutative operation.
Epimorphism Preserves Identity
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures. Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism. Let $\struct {S, \circ}$ have an identity element $e_S$. Then $\struct {T, *}$ has the identity element $\map \phi {e_S}$.
Epimorphism Preserves Inverses
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures. Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism. Let $\left({S, \circ}\right)$ have an identity $e_S$. Let $x^{-1}$ be an inverse element of $x$ for $\circ$. Then $\phi \left({x^{-1}}\right)$ is an inverse element of $\phi \left({x}\right)$ for $*$. That is: : $\phi \left({x^{-1}}\right) = \left({\phi \left({x}\right)}\right)^{-1}$
Homomorphism with Cancellable Codomain Preserves Identity
Let $\struct{S, \circ}$ and $\struct{T, *}$ be algebraic structures. Let $\phi: \struct{S, \circ} \to \struct{T, *}$ be a homomorphism. Let $\struct{S, \circ}$ have an identity $e_S$. Let $\struct{T, *}$ have an identity $e_T$. Let every element of $\struct{T, *}$ be cancellable. Then $\map \phi {e_S}$ is the identity $e_T$.
Homomorphism with Identity Preserves Inverses
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures. Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a homomorphism. Let $\left({S, \circ}\right)$ have an identity $e_S$. Let $\left({T, *}\right)$ also have an identity $e_T = \phi \left({e_S}\right)$. If $x^{-1}$ is an inverse of $x$ for $\circ$, then $\phi \left({x^{-1}}\right)$ is an inverse of $\phi \left({x}\right)$ for $*$. That is: : $\phi \left({x^{-1}}\right) = \left({\phi \left({x}\right)}\right)^{-1}$
Homomorphism to Group Preserves Identity
Let $\struct {S, \circ}$ be an algebraic structure. Let $\struct {T, *}$ be a group. Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism. Let $\struct {S, \circ}$ have an identity $e_S$. Then: :$\map \phi {e_S} = e_T$
Homomorphism of External Direct Products
Let: : $\left({S_1 \times S_2, \circ}\right)$ be the external direct product of two algebraic structures $\left({S_1, \circ_1}\right)$ and $\left({S_2, \circ_2}\right)$ : $\left({T_1 \times T_2, *}\right)$ be the external direct product of two algebraic structures $\left({T_1, *_1}\right)$ and $\left({T_2, *_2}\right)$ : $\phi_1$ be a homomorphism from $\left({S_1, \circ_1}\right)$ onto $\left({T_1, *_1}\right)$ : $\phi_2$ be a homomorphism from $\left({S_2, \circ_2}\right)$ onto $\left({T_2, *_2}\right)$. Then the mapping $\phi_1 \times \phi_2: \left({S_1 \times S_2, \circ}\right) \to \left({T_1 \times T_2, *}\right)$ defined as: :$\left({\phi_1 \times \phi_2}\right) \left({\left({x, y}\right)}\right) = \left({\phi_1 \left({x}\right), \phi_2 \left({y}\right)}\right)$ is a homomorphism from $\left({S_1 \times S_2, \circ}\right)$ to $\left({T_1 \times T_2, *}\right)$.
Monomorphism Image is Isomorphic to Domain
The image of a monomorphism is isomorphic to its domain. That is, if $\phi: S_1 \to S_2$ is a monomorphism, then: :$\phi: S_1 \to \Img \phi$ is an isomorphism.
Inverse of Algebraic Structure Isomorphism is Isomorphism
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures. Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a mapping. Then $\phi$ is an isomorphism iff $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is also an isomorphism.
Isomorphism is Equivalence Relation
Isomorphism is an equivalence on a set of magmas. This result applies to all magmas: rings, groups, R-algebraic structures etc.
Projection is Epimorphism
Let $\struct {S, \circ}$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$. Then: :$\pr_1$ is an epimorphism from $\struct {S, \circ}$ to $\struct {S_1, \circ_1}$ :$\pr_2$ is an epimorphism from $\struct {S, \circ}$ to $\struct {S_2, \circ_2}$ where $\pr_1$ and $\pr_2$ are the first and second projection respectively of $\struct {S, \circ}$.
Quotient Theorem for Epimorphisms
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures. Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism. Let $\mathcal R_\phi$ be the equivalence induced by $\phi$. Let $S / \mathcal R_\phi$ be the quotient of $S$ by $\mathcal R_\phi$. Let $q_{\mathcal R_\phi}: S \to S / \mathcal R_\phi$ be the quotient mapping induced by $\mathcal R_\phi$. Let $\struct {S / \mathcal R_\phi}, {\circ_{\mathcal R_\phi} }$ be the quotient structure defined by $\mathcal R_\phi$. Then: :The induced equivalence $\mathcal R_\phi$ is a congruence relation for $\circ$ :There is one and only one isomorphism $\psi: \struct {S / \mathcal R_\phi}, {\circ_{\mathcal R_\phi} } \to \struct {T, *}$ which satisfies $\psi \bullet q_{\mathcal R_\phi} = \phi$. where, in order not to cause notational confusion, $\bullet$ is used as the symbol to denote composition of mappings.
Homomorphism on Induced Structure
Let $\struct {S, \circ}$ be an algebraic structure. Let $\struct {T, \oplus}$ be a commutative semigroup. Let $T^S$ be the set of all mappings from $S$ to $T$. Let $f$ and $g$ be homomorphisms from $S$ into $T$. Let $f \oplus' g$ be the pointwise operation on $T^S$ induced by $\oplus$. Then $f \oplus' g$ is a homomorphism from $\struct {S, \circ}$ into $\struct {T, \oplus}$.
Inverse Mapping in Induced Structure
Let $\left({S, \circ}\right)$ be an algebraic structure. Let $\left({T, \oplus}\right)$ be an abelian group. Let $f$ be a homomorphism from $S$ into $T$. Let $f^*$ be the induced structure inverse of $f$. Then $f^*$ is a homomorphism from $\left({S, \circ}\right)$ into $\left({T, \oplus}\right)$.
Set of Homomorphisms is Subgroup of All Mappings
Let $\struct {S, \circ}$ be an algebraic structure. Let $\struct {T, \oplus}$ be an abelian group. Let $\struct {T^S, \oplus}$ be the algebraic structure on $T^S$ induced by $\oplus$. Then the set of all homomorphisms from $\struct {S, \circ}$ into $\struct {T, \oplus}$ is a subgroup of $\struct {T^S, \oplus}$.
Transplanting Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure. Let $f: S \to T$ be a bijection. Then there exists one and only one operation $\oplus$ such that $f: \left({S, \circ}\right) \to \left({T, \oplus}\right)$ is an isomorphism. The operation $\oplus$ is defined by: :$\forall x, y \in T: x \oplus y = f \left({f^{-1} \left({x}\right) \circ f^{-1} \left({y}\right)}\right)$ The operation $\oplus$ is called the '''transplant of $\circ$ under $f$'''.
Exists Bijection to a Disjoint Set
Let $S$ and $T$ be sets. Then there exists a bijection from $T$ onto a set $T'$ disjoint from $S$.
Embedding Theorem
Let: :$(1): \quad \left({T_2, \oplus_2}\right)$ be a submagma of $\left({S_2, *_2}\right)$ :$(2): \quad f: \left({T_1, \oplus_1}\right) \to \left({T_2, \oplus_2}\right)$ be an isomorphism then there exists: :$(1): \quad$ a magma $\left({S_1, *_1}\right)$ which algebraically contains $\left({T_1, \oplus_1}\right)$ :$(2): \quad g: \left({S_1, *_1}\right) \to \left({S_2, *_2}\right)$ where $g$ is an isomorphism which extends $f$. {{Expand|Add the corollary that extends this theorem to structures with two operations.}}
Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup
Let $\left({S, \circ}\right)$ be a commutative semigroup. Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$. Let $\left({T, \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$. Then: :$S \circ' C^{-1}$ is a commutative semigroup where $S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$ under $\circ'$ in $T$.
Inverse Completion of Commutative Semigroup is Inverse Completion of Itself
Let $\left({S, \circ}\right)$ be a commutative semigroup. Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$. Let $\left({T, \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$. Then $\left({T, \circ'}\right)$ is its own inverse completion.
Identity of Inverse Completion of Commutative Monoid
Let $\left({S, \circ}\right)$ be a commutative monoid whose identity is $e$. Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$. Let $\left({T, \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$. Then $e \in T$ is the identity for $\circ'$.
Inverse Completion of Commutative Semigroup is Abelian Group
Let $\struct {S, \circ}$ be a commutative semigroup Let all the elements of $\struct {S, \circ}$ be cancellable. Then an inverse completion of $\struct {S, \circ}$ is an abelian group.
Taylor's Theorem
Every infinitely differentiable function can be approximated by a series of polynomials.
Inverse Completion Theorem
Every commutative semigroup containing cancellable elements admits an inverse completion.
Extension Theorem for Homomorphisms
Let $\struct {S, \circ}$ be a commutative semigroup with cancellable elements Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of all cancellable elements of $S$ Let $\struct {S', \circ'}$ be an inverse completion of $\struct {S, \circ}$ Let $\phi$ be a (semigroup) homomorphism from $\struct {S, \circ}$ into a semigroup $\struct {T, *}$ such that $\map \phi y$ is invertible for all $y \in C$. Then: :$(1): \quad$ There is one and only one homomorphism $\psi$ from $\struct {S', \circ'}$ into $\struct {T, *}$ extending $\phi$ :$(2): \quad \forall x \in S, y \in C: \map \psi {x \circ' y^{-1} } = \map \phi * \paren {\map \phi y}^{-1}$ :$(3): \quad$ If $\phi$ is a monomorphism, then so is $\psi$.
Extension Theorem for Isomorphisms
Let the following conditions be fulfilled: : Let $\left({S, \circ}\right)$ be a commutative semigroup with cancellable elements : Let $\phi$ be an isomorphism from $\left({S, \circ}\right)$ into a semigroup $\left({T, *}\right)$ : Let $\left({S', \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$ : Let $\left({T', \circ'}\right)$ be an inverse completion of $\left({T, \circ}\right)$. Then there is a unique isomorphism $\phi': S' \to T'$ extending $\phi$.
Inverse Completion is Unique
An inverse completion of a commutative semigroup is unique up to isomorphism.
Extension Theorem for Distributive Operations
Let $\struct {R, *}$ be a commutative semigroup, all of whose elements are cancellable. Let $\struct {T, *}$ be an inverse completion of $\struct {R, *}$. Let $\circ$ be an operation on $R$ which distributes over $*$. Then: :$(1): \quad$ There is a unique operation $\circ'$ on $T$ which distributes over $*$ in $T$ and induces on $R$ the operation $\circ$ :$(2): \quad$ If $\circ$ is associative, then so is $\circ'$ :$(3): \quad$ If $\circ$ is commutative, then so is $\circ'$ :$(4): \quad$ If $e$ is an identity for $\circ$, then $e$ is also an identity for $\circ'$ :$(5): \quad$ Every element cancellable for $\circ$ is also cancellable for $\circ'$.
Monoid is not Empty
A monoid cannot be empty.
Group is not Empty
A group cannot be empty.
Identity is only Idempotent Element in Group
Every group has exactly one idempotent element: the identity.
Group Product Identity therefore Inverses
Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$. Then if either: :$g h = e$ or: :$h g = e$ it follows that: :$g = h^{-1}$ and: :$h = g^{-1}$
Self-Inverse Elements Commute iff Product is Self-Inverse
Let $\struct {G, \circ}$ be a group. Let $x, y \in \struct {G, \circ}$, such that $x$ and $y$ are self-inverse. Then $x$ and $y$ commute  $x \circ y$ is also self-inverse.
Power Set of Group under Induced Operation is Semigroup
Let $\struct {G, \circ}$ be a group. Let $\struct {\powerset G, \circ_\mathcal P}$ be the algebraic structure consisting of the power set of $G$ and the operation induced on $\powerset G$ by $\circ$. Then $\struct {\powerset G, \circ_\mathcal P}$ is a semigroup.
Inverse of Product of Subsets of Group
Let $\struct {G, \circ}$ be a group. Let $X, Y \subseteq G$. Then: :$\paren {X \circ Y}^{-1} = Y^{-1} \circ X^{-1}$ where $X^{-1}$ is the inverse of $X$.
Regular Representations in Group are Permutations
Let $\struct {G, \circ}$ be a group. Let $a \in G$ be any element of $G$. Then the left regular representation $\lambda_a$ and the right regular representation $\rho_a$ are permutations of $G$.
Set Equivalence of Regular Representations
If $S$ is a finite subset of a group $G$, then: :$\card {a \circ S} = \card S = \left|{S \circ a}\right|$ That is, $a \circ S$, $S$ and $S \circ a$ are equivalent: $a \circ S \sim S \sim S \circ a$.
Composition of Regular Representations
Let $\left({S, *}\right)$ be a semigroup. Let $\lambda_x, \rho_x$ be the left and right regular representations of $\left({S, *}\right)$ with respect to $x$. Let $\lambda_x \circ \lambda_y$, $\rho_x \circ \rho_y$ etc. be defined as the composition of the mappings $\lambda_x$ and $\lambda_y$ etc. Then $\forall x, y \in S$, the following results hold:
All Elements Self-Inverse then Abelian
Let $\struct {G, \circ}$ be a group. Suppose that every element of $G$ is self-inverse. Then $G$ is abelian.
Commutation Property in Group
Let $\struct {G, \circ}$ be a group. Then $x$ and $y$ commute  $x \circ y \circ x^{-1} = y$.
Identity Mapping is Automorphism
The identity mapping $I_S: \left({S, \circ}\right) \to \left({S, \circ}\right)$ on the algebraic structure $\left({S, \circ}\right)$ is an automorphism. Its image is $S$.
Group Homomorphism of Product with Inverse
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism. Then: :$(1): \quad \forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$ :$(2): \quad \forall x, y \in G: \map \phi {y^{-1} \circ x} = \paren {\map \phi y}^{-1} * \map \phi x$
Mapping to Square is Endomorphism iff Abelian
Let $\struct {G, \circ}$ be a group. Let $\phi: G \to G$ be defined as: :$\forall g \in G: \map \phi g = g \circ g$ Then $\struct {G, \circ}$ is abelian  $\phi$ is a (group) endomorphism.
Induced Group Product is Homomorphism iff Commutative
Let $\struct {G, \circ}$ be a group. Let $H_1, H_2$ be subgroups of $G$. Let $\phi: H_1 \times H_2 \to G$ be defined such that: :$\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map \phi {h_1, h_2} = h_1 \circ h_2$ Then $\phi$ is a homomorphism  every element of $H_1$ commutes with every element of $H_2$.
Isomorphism of Abelian Groups
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group isomorphism. Then $\struct {G, \circ}$ is abelian  $\struct {H, *}$ is abelian.
Group Example: x inv c y
Let $\struct {G, \circ}$ be a group. Let $c \in G$. We define a new operation $*$ on $G$ as: :$\forall x, y \in G: x * y = x \circ c^{-1} \circ y$ Then $\struct {G, *}$ is a group.
Symmetric Difference on Power Set forms Abelian Group
Let $S$ be a set such that $\O \subset S$ (that is, $S$ is non-empty). Let $A * B$ be defined as the symmetric difference between $A$ and $B$. Let $\powerset S$ denote the power set of $S$. Then the algebraic structure $\struct {\powerset S, *}$ is an abelian group.
Group is Subgroup of Itself
Let $\struct {G, \circ}$ be a group. Then: :$\struct {G, \circ} \le \struct {G, \circ}$ That is, a group is always a subgroup of itself.
Identity of Subgroup
Let $G$ be a group whose identity is $e$. Let $H$ be a subgroup of group $G$. Then the identity of $H$ is also $e$.
Inverses in Subgroup
Let $G$ be a group. Let $H$ be a subgroup of $G$. Then for each $h \in H$, the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.
Subgroup of Abelian Group is Abelian
A subgroup of an abelian group is itself abelian.
Intersection of Subgroups is Subgroup
The intersection of two subgroups of a group is itself a subgroup of that group: :$\forall H_1, H_2 \le \struct {G, \circ}: H_1 \cap H_2 \le G$ It also follows that $H_1 \cap H_2 \le H_1$ and $H_1 \cap H_2 \le H_2$.
Union of Subgroups
Let $\struct {G, \circ}$ be a group. Let $H, K \le G$ be subgroups of $G$. Let neither $H \subseteq K$ nor $K \subseteq H$. Then $H \cup K$ is ''not'' a subgroup of $G$.
Elements of Group with Equal Images under Homomorphisms form Subgroup
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups. Let $f: G \to H$ and $g: G \to H$ be group homomorphisms. Then the set: :$S = \set {x \in G: \map f x = \map g x}$ is a subgroup of $G$.
Product of Subgroup with Itself
Let $\struct {G, \circ}$ be a group. Then: : $\forall H \le \struct {G, \circ}: H \circ H = H$
Inverse of Subgroup
Let $\struct {G, \circ}$ be a group. Let $H$ be a subgroup of $G$. Then: :$H^{-1} = H$ where $H^{-1}$ is the inverse of $H$.
Subset Product of Subgroups
Let $\struct {G, \circ}$ be a group. Let $H, K$ be subgroups of $G$. Then $H \circ K$ is a subgroup of $G$  $H$ and $K$ are permutable. That is: $H \circ K$ is a subgroup of $G$ : :$H \circ K = K \circ H$ where $H \circ K$ denotes subset product.
Group Homomorphism Preserves Subgroups
Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups. Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group homomorphism.  Then: :$H \le G_1 \implies \phi \sqbrk H \le G_2$ where: :$\phi \sqbrk H$ denotes the image of $H$ under $\phi$ :$\le$ denotes subgroup. That is, group homomorphism preserves subgroups.
Image of Group Homomorphism is Subgroup
Let $\phi: G_1 \to G_2$ be a group homomorphism. Then: :$\Img \phi \le G_2$ where $\le$ denotes the relation of being a subgroup.
Conjugacy is Equivalence Relation
Conjugacy of group elements is an equivalence relation.
Kernel of Group Homomorphism is Subgroup
The kernel of a group homomorphism is a subgroup of its domain: :$\map \ker \phi \le \Dom \phi$
Commutative Semigroup is Entropic Structure
A commutative semigroup is an entropic structure.
Abelian Group Induces Entropic Structure
Let $\struct {G, \circ}$ be an abelian group. Let the operation $*$ be defined on $G$ such that: :$\forall x, y \in G: x * y = x \circ y^{-1}$ Then $\struct {G, *}$ is an entropic structure.
Cancellable Semiring with Unity is Additive Semiring
Let $\struct {S, *, \circ}$ be a cancellable semiring with unity $1_S$. Then the distributand $*$ is commutative. That is to say, $\struct {S, *, \circ}$ is also an additive semiring.
Ring is not Empty
A ring cannot be empty.
Ring Product with Zero
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$. Then: :$\forall x \in R: 0_R \circ x = 0_R = x \circ 0_R$ That is, the zero is a zero element for the ring product, thereby justifying its name.
Product with Ring Negative
Let $\struct {R, +, \circ}$ be a ring. Then: :$\forall x, y \in \struct  {R, +, \circ}: \paren {-x} \circ y = -\paren {x \circ y} = x \circ \paren {-y}$ where $\paren {-x}$ denotes the negative of $x$.
Unity of Ring is Unique
A ring can have no more than one unity.
Null Ring iff Zero and Unity Coincide
The null ring is the only ring in which the unity and zero coincide.
Unity is Unit
The unity in a ring is a unit.
Unity and Negative form Subgroup of Units
Let $\struct {R, +, \circ}$ be a ring with unity. Then: :$\struct {\set {1_R, -1_R}, \circ} \le U_R$ That is, the set consisting of the unity and its negative forms a subgroup of the group of units.
Negative of Product Inverse
Let $\left({R, +, \circ}\right)$ be a ring with unity. Let $z \in U_R$, where $U_R$ is the set of units. Then: :$\left({- z}\right)^{-1} = - \left({z^{-1}}\right)$. where $z^{-1}$ is the ring product inverse of $z$.
Product of Negative with Product Inverse
Let $\left({R, +, \circ}\right)$ be a ring with unity. Let $z \in U_R$, where $U_R$ is the set of units. Then: : $(1): \quad \forall x \in R: -\left({x \circ z^{-1}}\right) = \left({- x}\right) \circ z^{-1} = x \circ \left({\left({- z}\right)^{-1}}\right)$ : $(2): \quad \forall x \in R: -\left({z^{-1} \circ x}\right) = z^{-1} \circ \left({- x}\right) = \left({\left({- z}\right)^{-1}}\right) \circ x$
Negative of Division Product
Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$. Then: :$\displaystyle \forall x \in R: -\frac x z = \frac {-x} z = \frac x {-z}$ where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is the division product of $x$ by $z$.
Addition of Division Products
Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$. Let $a, c \in R, b, d \in U_R$. Then: :$\dfrac a b + \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$ where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$. The operation $+$ is well-defined. That is: :$\dfrac a b = \dfrac {a'} {b'}, \dfrac c d = \dfrac {c'} {d'} \implies \dfrac a b + \dfrac c d = \dfrac {a'} {b'} + \dfrac {c'} {d'}$ {{questionable|This is an existing operation and thus already well-defined. See talk.}}
Equality of Division Products
Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$. Let $a, b \in R, c, d \in U_R$. Then: :$\dfrac a c = \dfrac b d \iff a \circ d = b \circ c$ where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.
Product of Division Products
Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$. Let $a, b \in R, c, d \in U_R$. Then: :$\dfrac a c \circ \dfrac b d = \dfrac {a \circ b} {c \circ d}$ where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.
Inverse of Division Product
Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$. Let $a, b \in U_R$. Then: :$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac b a$ where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.
Zero Product with Proper Zero Divisor is with Zero Divisor
Let $\struct {R, +, \circ}$ be a ring. Let $x \in R$ be a proper zero divisor of $R$. Then: :$\paren {x \divides 0_R} \land \paren {x \circ y = 0_R} \land \paren {y \ne 0_R} \implies y \divides 0_R$ That is, if $x$ is a proper zero divisor, then whatever non-zero element you form the product with it by to get zero must itself be a zero divisor.
Unit Not Zero Divisor
A unit of a ring is not a zero divisor.
Zero Divisor Product is Zero Divisor
The ring product of a zero divisor with any ring element is a zero divisor.
Product is Zero Divisor means Zero Divisor
If the ring product of two elements of a ring is a zero divisor, then one of the two elements must be a zero divisor.
Ring Element is Zero Divisor iff not Cancellable
Let $\struct {R, +, \circ}$ be a ring which is not null. Let $z \in R^*$. Then $z$ is a zero divisor  $z$ is not cancellable for $\circ$.
Ring Less Zero is Semigroup for Product iff No Proper Zero Divisors
Let $\struct {R, +, \circ}$ be a non-null ring. Then $R$ has no zero divisors  $\struct {R^*, \circ}$ is a semigroup.