Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is: :$\displaystyle T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$


Sum of Sequence of Squares

:$\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$


Union is Associative

Set union is associative: :$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$


Pythagoras's Theorem

Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse. Then: :$a^2 + b^2 = c^2$


Euclid's Theorem

For any finite set of prime numbers, there exists a prime number not in that set. {{:Euclid:Proposition/IX/20}}


Limsup Squeeze Theorem

Let $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ be sequences in $\R$. Let: : $(1): \quad \forall n \ge n_0: \left|{x_n}\right| \le y_n$ : $(2): \quad \displaystyle \limsup_{n \mathop \to \infty} \left({y_n}\right) = 0$, where $\limsup$ denotes the limit superior. Then: : $\displaystyle \lim_{n \mathop \to \infty} x_n = 0$


Square Root of Prime is Irrational

The square root of any prime number is irrational.


Derivative of Exponential Function

Let $\exp$ be the exponential function. Then: :$\map {\dfrac \d {\d x} } {\exp x} = \exp x$


Derivative of Sine Function

:$\map {\dfrac \d {\d x} } {\sin x} = \cos x$


Cauchy-Bunyakovsky-Schwarz Inequality

=== Semi-Inner Product Spaces === {{:Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces}}


0.999...=1

:$0.999 \ldots = 1$


Sum of Sequence of Cubes

:$\displaystyle \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$


Law of Cosines

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$. Then: :$c^2 = a^2 + b^2 - 2 a b \cos C$


Euler's Formula

Let $z \in \C$ be a complex number. Then: :$e^{i z} = \cos z + i \sin z$


Lagrange's Theorem (Group Theory)

Let $G$ be a finite group. Let $H$ be a subgroup of $G$. Then: : $\order H$ divides $\order G$ where $\order G$ and $\order H$ are the order of $G$ and $H$ respectively. In fact: :$\index G H = \dfrac {\order G} {\order H}$ where $\index G H$ is the index of $H$ in $G$. When $G$ is an infinite group, we can still interpret this theorem sensibly: :A subgroup of finite index in an infinite group is itself an infinite group. :A finite subgroup of an infinite group has infinite index.


ProofWiki:Sandbox

$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^{n} \binom {2n} {2k} E_{2n - 2k } = 0$ where $E_k$ denotes the $k$th Euler number.


Carathéodory's Theorem (Analysis)

Let $I \subseteq \R$. Let $c \in I$ be an interior point of $I$. {{Disambiguate|Definition:Interior Point}} {{explain|In this case, there appears not to be a definition for "interior point" which appropriately captures the gist of this. It is clearly a point inside a real interval, but the concept has not yet been defined.}} Let $f : I \to \R$ be a real function. Then $f$ is differentiable at $c$ {{iff}}: :There exists a real function $\varphi : I \to \R$ that is continuous at $c$ and satisfies: :$(1): \quad \forall x \in I: f \left({x}\right) - f \left({c}\right) = \varphi \left({x}\right) \left({x - c}\right)$ :$(2): \quad \varphi \left({c}\right) = f' \left({c}\right)$


One-Step Subgroup Test

Let $\struct {G, \circ}$ be a group. Let $H$ be a subset of $G$. Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ {{iff}}: : $(1): \quad H \ne \O$, that is, $H$ is non-empty : $(2): \quad \forall a, b \in H: a \circ b^{-1} \in H$.


Two-Step Subgroup Test

Let $\struct {G, \circ}$ be a group. Let $H$ be a subset of $G$. Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ {{iff}}: :$(1): \quad H \ne \O$, that is, $H$ is non-empty :$(2): \quad a, b \in H \implies a \circ b \in H$ :$(3): \quad a \in H \implies a^{-1} \in H$. That is, $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ {{iff}} $\struct {H, \circ}$ is a $H$ be a nonempty subset of $G$ which is: :closed under its operation and: :closed under inversion.


Fundamental Theorem of Arithmetic

For every integer $n$ such that $n > 1$, $n$ can be expressed as the product of one or more primes, uniquely up to the order in which they appear.


Euclid's Lemma

Let $a, b, c \in \Z$. Let $a \divides b c$, where $\divides$ denotes divisibility. Let $a \perp b$, where $\perp$ denotes relative primeness. Then $a \divides c$.


Fermat's Little Theorem

Let $p$ be a prime number. Let $n \in \Z_{>0}$ be a positive integer such that $p$ is not a divisor of $n$. Then: :$n^{p - 1} \equiv 1 \pmod p$


De Morgan's Laws (Set Theory)

{{:De Morgan's Laws (Set Theory)/Set Difference}}


De Moivre's Formula

Let $z \in \C$ be a complex number expressed in complex form: :$z = r \paren {\cos x + i \sin x}$ Then: :$\forall \omega \in \C: \paren {r \paren {\cos x + i \sin x} }^\omega = r^\omega \paren {\map \cos {\omega x} + i \, \map \sin {\omega x} }$


First Isomorphism Theorem

=== Groups === {{:First Isomorphism Theorem/Groups}} === Rings === {{:First Isomorphism Theorem/Rings}}


Power Rule for Derivatives

Let $n \in \R$. Let $f: \R \to \R$ be the real function defined as $\map f x = x^n$. Then: :$\map {f'} x = n x^{n - 1}$ everywhere that $\map f x = x^n$ is defined. When $x = 0$ and $n = 0$, $\map {f'} x$ is undefined.


Basel Problem

:$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^\infty {\frac 1 {n^2} } = \frac {\pi^2} 6$ where $\zeta$ denotes the Riemann zeta function.


1+2+...+n+(n-1)+...+1 = n^2

:$\forall n \in \N: 1 + 2 + \cdots + n + \paren {n - 1} + \cdots + 1 = n^2$


Necessary Conditions for Existence of Skolem Sequence

A Skolem sequence of order $n$ can only exist if $n \equiv 0, 1 \pmod 4$.


Divisibility by 9

A number expressed in decimal notation is divisible by $9$ {{iff}} the sum of its digits is divisible by $9$. That is: :$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $9$  {{iff}}: :$a_0 + a_1 + \ldots + a_n$ is divisible by $9$.


Existence of Rational Powers of Irrational Numbers

There exist irrational numbers $a$ and $b$ such that $a^b$ is rational.


Intersection is Associative

Set intersection is associative: :$A \cap \paren {B \cap C} = \paren {A \cap B} \cap C$


Center of Symmetric Group is Trivial

Let $n \in \N$ be a natural number. Let $S_n$ denote the symmetric group of order $n$. Let $n \ge 3$. Then the center $\map Z {S_n}$ of $S_n$ is trivial.


Square Root of 2 is Irrational

:$\sqrt 2$ is irrational.


Empty Set is Unique

The empty set is unique.


Principle of Non-Contradiction

The '''Principle of Non-Contradiction''' is a valid deduction sequent in propositional logic.


Rule of Substitution

Let $S$ be a sequent that has been proved. Then a proof can be found for any substitution instance of $S$.


Rule of Sequent Introduction

Let the statements $P_1, P_2, \ldots, P_n$ be conclusions in a proof, on various assumptions. Let $P_1, P_2, \ldots, P_n \vdash Q$ be a substitution instance of a sequent for which we already have a proof. {{explain|Question the use of "substitution instance": can we not "just" allow for $P_1, \ldots, P_n$ to be "just" statements?}} Then we may introduce, at any stage of a proof (citing '''SI'''), either: :The conclusion $Q$ of the sequent already proved or: :A substitution instance of such a conclusion, together with a reference to the sequent that is being cited. This conclusion depend upon the pool of assumptions upon which $P_1, P_2, \ldots, P_n \vdash Q$ rests.


Law of Identity

Every proposition entails itself:


Rule of Idempotence

The '''rule of idempotence''' is two-fold:


Hypothetical Syllogism

The '''(rule of the) hypothetical syllogism''' is a valid deduction sequent in propositional logic: :If we can conclude that $p$ implies $q$, and if we can also conclude that $q$ implies $r$, then we may infer that $p$ implies $r$.


Extended Rule of Implication

Any sequent can be expressed as a theorem. That is: :$P_1, P_2, P_3, \ldots, P_n \vdash Q$ means the same thing as: :$\vdash P_1 \implies \paren {P_2 \implies \paren {P_3 \implies \paren {\ldots \implies \paren {P_n \implies Q} \ldots} } }$ The latter expression is known as the corresponding conditional of the former. Thus every sequent containing the symbol $\vdash$ can, if so desired, be expressed in the form of a theorem which has $\implies$.


Equivalences are Interderivable

If two propositional formulas are interderivable, they are equivalent: :$\paren {p \dashv \vdash q} \dashv \vdash \paren {p \iff q}$


Union is Commutative

Set union is commutative: :$S \cup T = T \cup S$


Intersection is Commutative

Set intersection is commutative: :$S \cap T = T \cap S$


Russell's Paradox

The comprehension principle leads to a contradiction.


De Morgan's Laws (Logic)

{{:De Morgan's Laws (Logic)/Disjunction of Negations}}


Solution to Quadratic Equation

The quadratic equation of the form $a x^2 + b x + c = 0$ has solutions: :$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}$


Normal Subgroup Test

Let $G$ be a group and $H \le G$. Then $H$ is a normal subgroup of $G$ {{iff}}: :$\forall x \in G: x H x^{-1} \subseteq H$.


Equality of Ordered Pairs

Two ordered pairs are equal {{iff}} corresponding coordinates are equal: :$\tuple {a, b} = \tuple {c, d} \iff a = c \land b = d$


Identity is Unique

Let $\left({S, \circ}\right)$ be an algebraic structure that has an identity element $e \in S$. Then $e$ is unique.


Cancellation Laws

Let $G$ be a group. Let $a, b, c \in G$. Then the following hold: ;Right cancellation law :$b a = c a \implies b = c$ ;Left cancellation law :$a b = a c \implies b = c$


Center of Group is Normal Subgroup

The center $\map Z G$ of any group $G$ is a normal subgroup of $G$ which is abelian.


Composition of Relations is Associative

The composition of relations is an associative binary operation: :$\paren {\RR_3 \circ \RR_2} \circ \RR_1 = \RR_3 \circ \paren {\RR_2 \circ \RR_1}$


Intersection of Subsemigroups

Let $\left({S, \circ}\right)$ be a semigroup. Let $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$ be subsemigroups of $\left({S, \circ}\right)$. Then the intersection of $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$ is itself a subsemigroup of that $\left({S, \circ}\right)$. If $\left({T, \circ}\right)$ is that intersection of $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$, it follows that $\left({T, \circ}\right)$ is also a subsemigroup of both $\left({T_1, \circ}\right)$ and $\left({T_2, \circ}\right)$.


Cancellable Elements of Semigroup form Subsemigroup

Let $\struct {S, \circ}$ be a semigroup. Let $C$ be the set of cancellable elements of $\struct {S, \circ}$. Then $\struct {C, \circ}$ is a subsemigroup of $\struct {S, \circ}$.


Subgroup of Cyclic Group is Cyclic

Let $G$ be a cyclic group. Let $H$ be a subgroup of $G$. Then $H$ is cyclic.


Empty Set is Subset of All Sets

The empty set $\O$ is a subset of every set (including itself). That is: :$\forall S: \O \subseteq S$


Relation Reflexivity

Every relation has exactly one of these properties: it is either: :reflexive, :antireflexive or :non-reflexive.


Union is Idempotent

Set union is idempotent: :$S \cup S = S$


Intersection is Idempotent

Set intersection is idempotent: :$S \cap S = S$


Set is Subset of Union

The union of two sets is a superset of each: :$S \subseteq S \cup T$ :$T \subseteq S \cup T$


Set is Subset of Itself

Every set is a subset of itself: :$\forall S: S \subseteq S$ Thus, by definition, the relation '''is a subset of''' is reflexive.


Subset of Set with Propositional Function

Let $S$ be a set. Let $P: S \to \set {\text{true}, \text{false} }$ be a propositional function on $S$. Then: :$\set {x \in S: \map P x} \subseteq S$


Universal Generalisation

Let $\mathbf a$ be any arbitrarily selected object in the universe of discourse. Then: :$\map P {\mathbf a} \vdash \forall x: \map P x$ In natural language: :''Suppose $P$ is true of any arbitrarily selected $\mathbf a$'' in the universe of discourse.'' :''Then $P$ is true of everything in the universe of discourse.''


Existential Instantiation

:$\exists x: P \left({x}\right), P \left({\mathbf a}\right) \implies y \vdash y$ Suppose we have the following: : From our universe of discourse, ''any'' arbitrarily selected object $\mathbf a$ which has the property $P$ implies a conclusion $y$ : $\mathbf a$ is not free in $y$ : It is known that there ''does'' actually exists an object that has $P$. Then we may infer $y$. This is called the '''Rule of Existential Instantiation''' and often appears in a proof with its abbreviation '''EI'''. When using this rule of existential instantiation: :$\exists x: P \left({x}\right), P \left({\mathbf a}\right) \implies y \vdash y$ the instance of $P \left({\mathbf a}\right)$ is referred to as the '''typical disjunct'''.


Singleton of Element is Subset

Let $S$ be a set. Let $\set x$ be the singleton of $x$. Then: :$x \in S \iff \set x \subseteq S$


Subset Relation is Transitive

The relation "is a subset of" is transitive: :$\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$


Set Inequality

:$S \ne T \iff \left({S \nsubseteq T}\right) \lor \left({T \nsubseteq S}\right)$


Set Equals Itself

All sets are equal to themselves: :$\forall S: S = S$


Union is Smallest Superset

Let $S_1$ and $S_2$ be sets. Then $S_1 \cup S_2$ is the smallest set containing both $S_1$ and $S_2$. That is: :$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$


Union with Empty Set

The union of any set with the empty set is the set itself: :$S \cup \O = S$


Intersection is Subset

The intersection of two sets is a subset of each: :$S \cap T \subseteq S$ :$S \cap T \subseteq T$


Intersection with Empty Set

The intersection of any set with the empty set is itself the empty set: :$S \cap \O = \O$


Intersection is Largest Subset

Let $T_1$ and $T_2$ be sets. Then $T_1 \cap T_2$ is the largest set contained in both $T_1$ and $T_2$. That is: :$S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$


Intersection is Subset of Union

The intersection of two sets is a subset of their union: :$S \cap T \subseteq S \cup T$


Absorption Laws (Set Theory)/Union with Intersection

:$S \cup \paren {S \cap T} = S$


Absorption Laws (Set Theory)/Intersection with Union

:$S \cap \paren {S \cup T} = S$


Union Distributes over Intersection

Set union is distributive over set intersection: :$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$


Intersection Distributes over Union

Set intersection is distributive over set union: :$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$


Set Difference is Subset

:$S \setminus T \subseteq S$


Set Difference with Empty Set is Self

The set difference between a set and the empty set is the set itself: :$S \setminus \O = S$


Set Difference with Self is Empty Set

The set difference of a set with itself is the empty set: :$S \setminus S = \O$


Set Difference Equals First Set iff Empty Intersection

:$S \setminus T = S \iff S \cap T = \O$


Equal Set Differences iff Equal Intersections

:$R \setminus S = R \setminus T \iff R \cap S = R \cap T$


Euler's Number is Irrational

Euler's number $e$ is irrational.


Results Concerning Set Difference with Union

Let: * $S \setminus T$ denote set difference * $S \cup T$ denote set union * $S \cap T$ denote set intersection.


Set Difference with Intersection

Let $S$ and $T$ be sets.


Results Concerning Set Difference with Intersection

Let: :$S \setminus T$ denote set difference :$S \cap T$ denote set intersection.


Pascal's Rule

For positive integers $n, k$ with $1 \le k \le n$: :$\dbinom n {k - 1} + \dbinom n k = \dbinom {n + 1} k$ This is also valid for the real number definition: :$\forall r \in \R, k \in \Z: \dbinom r {k - 1} + \dbinom r k = \dbinom {r + 1} k$


Set Difference Union Intersection

:$S = \paren {S \setminus T} \cup \paren {S \cap T}$


Absorption Laws (Logic)

For any two propositions $p$ and $q$, we have:


Set Difference with Set Difference

:$S \setminus \paren {S \setminus T} = S \cap T = T \setminus \paren {T \setminus S}$


Dirichlet's Test for Uniform Convergence

Let $D$ be a set. Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space. Let $a_i, b_i$ be mappings from $D \to M$. Let the following conditions be satisfied: :$(1): \quad$ The sequence of partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty \map {a_n} x$ be bounded on $D$ :$(2): \quad \sequence {\map {b_n} x}$ be monotonic for each $x \in D$ :$(3): \quad \map {b_n} x \to 0$ converge uniformly on $D$. Then: :$\displaystyle \sum_{n \mathop = 1}^\infty \map {a_n} x \, \map {b_n} x$ converges uniformly on $D$.


Relative Complement of Empty Set

The relative complement of the empty set is the set itself: :$\relcomp S \O = S$


Relative Complement with Self is Empty Set

The relative complement of a set in itself is the empty set: :$\relcomp S S = \O$


Relative Complement of Relative Complement

:$\relcomp S {\relcomp S T} = T$


Intersection with Relative Complement is Empty

The intersection of a set and its relative complement is the empty set: :$T \cap \relcomp S T = \O$


Union with Relative Complement

The union of a set $T$ and its relative complement in $S$ is the set $S$: :$\relcomp S T \cup T = S$


Set Difference as Intersection with Relative Complement

Let $A, B \subseteq S$. Then the set difference between $A$ and $B$ can be expressed as the intersection with the relative complement with respect to $S$: :$A \setminus B = A \cap \relcomp S B$


Equivalence of Axiom Schemata for Groups

In the definition of a group, the axioms for the existence of an identity element and for closure under taking inverses can be replaced by the following two axioms: : Given a group $G$, there exists at least one element $e \in G$ such that $e$ is a '''left identity'''; : For any element $g$ in a group $G$, there exists at least one '''left inverse''' of $g$. Alternatively, we can also replace the aforementioned axioms with the following two: : Given a group $G$, there exists at least one element $e \in G$ such that $e$ is a '''right identity'''; : For any element $g$ in a group $G$, there exists at least one '''right inverse''' of $g$. Thus we can formulate the group axioms as either of the following:


Group has Latin Square Property

Let $\struct {G, \circ}$ be a group. Then $G$ satisfies the Latin square property. That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$. Similarly, there exists a unique $h \in G$ such that $h \circ a = b$.


Symmetric Difference is Commutative

Symmetric difference is commutative: :$S * T = T * S$


Symmetric Difference of Equal Sets

The symmetric difference of two equal sets is the empty set: :$S = T \iff S * T = \O$


Union is Empty iff Sets are Empty

:$S \cup T = \O \iff S = \O \land T = \O$


Symmetric Difference with Empty Set

:$S * \O = S$ where $*$ denotes the symmetric difference.


Intersection Distributes over Symmetric Difference

Intersection is distributive over symmetric difference: :$\paren {R * S} \cap T = \paren {R \cap T} * \paren {S \cap T}$ :$T \cap \paren {R * S} = \paren {T \cap R} * \paren {T \cap S}$


Symmetric Difference of Unions

Let $R$, $S$ and $T$ be sets. Then: :$\paren {R \cup T} * \paren {S \cup T} = \paren {R * S} \setminus T$ where: :$*$ denotes the symmetric difference :$\setminus$ denotes set difference :$\cup$ denotes set union


Intersection with Universe

The intersection of a set with the universe is the set itself: :$\mathbb U \cap S = S$


Union with Universe

The union of a set with the universe is the universe: :$\mathbb U \cup S = \mathbb U$


Complement of Empty Set is Universe

The complement of the empty set is the universe: :$\map \complement \O = \mathbb U$


Complement of Universe is Empty Set

The complement of the universe is the empty set: :$\map \complement {\mathbb U} = \O$


Complement of Complement

The complement of the complement of a set is the set itself: :$\map \complement {\map \complement S} = S$


Intersection with Complement

The intersection of a set and its complement is the empty set: :$S \cap \map \complement S = \O$


Union with Complement

The union of a set and its complement is the universe: :$S \cup \map \complement S = \mathbb U$


Set Difference as Intersection with Complement

Set difference can be expressed as the intersection with the set complement: :$A \setminus B = A \cap \map \complement B$


Set Difference of Complements

:$\complement \left({S}\right) \setminus \complement \left({T}\right) = T \setminus S$


Symmetric Difference of Complements

The symmetric difference of two sets equals the symmetric difference of their complements: :$\map \complement S * \map \complement T = S * T$


Symmetric Difference with Self is Empty Set

The symmetric difference of a set with itself is the empty set: :$S * S = \O$


Symmetric Difference with Complement

The symmetric difference of a set with its complement is the universe: :$S * \relcomp {} S = \mathbb U$


Symmetric Difference is Associative

Symmetric difference is associative: :$R * \paren {S * T} = \paren {R * S} * T$


Cartesian Product is Empty iff Factor is Empty

:$S \times T = \O \iff S = \O \lor T = \O$ Thus: :$S \times \O = \O = \O \times T$


Cartesian Product of Subsets

Let $A, B, S, T$ be sets such that $A \subseteq B$ and $S \subseteq T$. Then: :$A \times S \subseteq B \times T$ In addition, if $A, S \ne \O$, then: :$A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$


Cartesian Product of Intersections

:$\paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} = \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2}$ where $S_1, S_2, T_1, T_2$ are sets.


Cartesian Product Distributes over Union

Cartesian product is distributive over union: :$A \times \paren {B \cup C} = \paren {A \times B} \cup \paren {A \times C}$ :$\paren {B \cup C} \times A = \paren {B \times A} \cup \paren {C \times A}$


Cartesian Product of Unions

:$\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2} = \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \cup \paren {S_1 \times T_2} \cup \paren {S_2 \times T_1}$


Cartesian Product Distributes over Set Difference

Cartesian product is distributive over set difference: :$(1): \quad S \times \paren {T_1 \setminus T_2} = \paren {S \times T_1} \setminus \paren {S \times T_2}$ :$(2): \quad \paren {T_1 \setminus T_2} \times S = \paren {T_1 \times S} \setminus \paren {T_2 \times S}$


Set Difference of Cartesian Products

:$\paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2} = \paren {S_1 \times \paren {S_2 \setminus T_2} } \cup \paren {\paren {S_1 \setminus T_1} \times S_2}$


Inverse of Inverse Relation

The inverse of an inverse relation is the relation itself: :$\paren {\RR^{-1} }^{-1} = \RR$


Diagonal Relation is Equivalence

The diagonal relation $\Delta_S$ on a set $S$ is always an equivalence in $S$.


Trivial Relation is Equivalence

The trivial relation on $S$: :$\RR = S \times S$ is always an equivalence in $S$.


Equality of Relations

Let $\RR_1$ and $\RR_2$ be relations on $S_1 \times T_1$ and $S_2 \times T_2$ respectively. Then $\RR_1$ and $\RR_2$ are '''equal''' {{iff}}: :$S_1 = S_2$ :$T_1 = T_2$ :$\tuple {s, t} \in \RR_1 \iff \tuple {s, t} \in \RR_2$ It is worth labouring the point that for two relations to be '''equal''', not only must their domains be equal, but so must their codomains.


Image of Singleton under Relation

Let $\RR \subseteq S \times T$ be a relation. Then the image of an element of $S$ is equal to the image of a singleton containing that element, the singleton being a subset of $S$: :$\forall s \in S: \map \RR s = \RR \sqbrk {\set s}$


Image of Subset under Relation is Subset of Image

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation from $S$ to $T$. Let $A, B \subseteq S$ such that $A \subseteq B$. Then the image of $A$ is a subset of the image of $B$: :$A \subseteq B \implies \RR \sqbrk A \subseteq \RR \sqbrk B$ In the notation of direct image mappings, this can be written: :$A \subseteq B \implies \map {\RR^\to} A \subseteq \map {\RR^\to} B$


Image of Element is Subset

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $A \subseteq S$. Then: :$s \in A \implies \map \RR s \subseteq \RR \sqbrk A$


Image is Subset of Codomain

Let $\RR = S \times T$ be a relation. For all subsets $A$ of the domain of $\RR$, the image of $A$ is a subset of the codomain of $\RR$: :$\forall A \subseteq \Dom \RR: \RR \sqbrk A \subseteq T$ In the notation of direct image mappings, this can be written as: :$\forall A \in \powerset S: \map {\RR^\to} A \in \powerset T$


Image of Empty Set is Empty Set

Let $\RR \subseteq S \times T$ be a relation. The image of the empty set is the empty set: :$\RR \sqbrk \O = \O$


Domain of Composite Relation

Let $\RR_2 \circ \RR_1$ be a composite relation. Then the domain of $\RR_2 \circ \RR_1$ is the domain of $\RR_1$: :$\Dom {\RR_2 \circ \RR_1} = \Dom {\RR_1}$


Codomain of Composite Relation

Let $\RR_2 \circ \RR_1$ be a composite relation. Then the codomain of $\RR_2 \circ \RR_1$ is the codomain of $\RR_2$: :$\Cdm {\RR_2 \circ \RR_1} = \Cdm {\RR_2}$


Preimage of Mapping equals Domain

The preimage of a mapping is the same set as its domain: :$\Preimg f = \Dom f$


Preimage of Relation is Subset of Domain

Let $\mathcal R \subseteq S \times T$ be a relation. Then the preimage of $\mathcal R$ is a subset of its domain: :$\Preimg {\mathcal R} \subseteq S$


Preimage of Image under Left-Total Relation is Superset

Let $\RR \subseteq S \times T$ be a left-total relation. Then: :$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$ where: :$\RR \sqbrk A$ denotes the image of $A$ under $\RR$ :$\RR^{-1} \sqbrk A$ denotes the preimage of $A$ under $\RR$ :$\RR^{-1} \circ \RR$ denotes composition of $\RR^{-1}$ and $\RR$. This can be expressed in the language and notation of direct image mappings and inverse image mappings as: :$\forall A \in \powerset S: A \subseteq \map {\paren {\RR^\gets \circ \RR^\to} } A$


Inverse of Composite Relation

Let $\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$ be the composite of the two relations $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$. Then: :$\paren {\RR_2 \circ \RR_1}^{-1} = \RR_1^{-1} \circ \RR_2^{-1}$


Image of Union under Relation

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $S_1$ and $S_2$ be subsets of $S$. Then: :$\RR \sqbrk {S_1 \cup S_2} = \RR \sqbrk {S_1} \cup \RR \sqbrk {S_2}$ That is, the image of the union of subsets of $S$ is equal to the union of their images.


Image of Intersection under Relation

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $S_1$ and $S_2$ be subsets of $S$. Then: :$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ That is, the image of the intersection of subsets of $S$ is a subset of the intersection of their images.


Image of Set Difference under Relation

Let $\RR \subseteq S \times T$ be a relation. Let $A$ and $B$ be subsets of $S$. Then: :$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$ where: :$\setminus$ denotes set difference :$\RR \sqbrk A$ denotes image of $A$ under $\RR$.


Preimage of Union under Relation

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $T_1$ and $T_2$ be subsets of $T$. Then: :$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$


Preimage of Intersection under Relation

Let $S$ and $T$ be sets. Let $\mathcal R \subseteq S \times T$ be a relation. Let $C$ and $D$ be subsets of $T$. Then: :$\mathcal R^{-1} \left[{C \cap D}\right] \subseteq \mathcal R^{-1} \left[{C}\right] \cap \mathcal R^{-1} \left[{D}\right]$


Restriction is Subset of Relation

Let $\mathcal R \subseteq S \times T$ be a relation. Let $X \subseteq S$. Then the restriction of $\mathcal R$ to $X$ is a subset of $\mathcal R$.


Null Relation is Antireflexive, Symmetric and Transitive

Let $S$ be a set which is non-empty. Let $\RR \subseteq S \times S$ be the null relation. Then $\RR$ is antireflexive, symmetric and transitive. If $S = \O$ then Relation on Empty Set is Equivalence applies.


Symmetry of Relations is Symmetric

Let $\mathcal R$ be a relation on $S$ which is symmetric. Then: :$\left({x, y}\right) \in \mathcal R \iff \left({y, x}\right) \in \mathcal R$.


Relation is Symmetric and Antisymmetric iff Coreflexive

Let $S$ be a set. Let $\RR \subseteq S \times S$ be a relation in $S$. Then: :$\RR$ is both symmetric and antisymmetric {{iff}}: :$\RR$ is coreflexive.


Relation both Symmetric and Asymmetric is Null

Let $\RR$ be a relation in $S$ which is both symmetric and asymmetric. Then: :$\RR = \O$


Equivalent Characterizations of Abelian Group

Let $G$ be a group. {{TFAE}} {{begin-eqn}} {{eqn | n = 1       | o =       | c = $G$ is abelian }} {{eqn | n = 2       | o =       | c = $\forall a, b \in G: \paren {a b}^{-1} = a^{-1} b^{-1}$ }} {{eqn | n = 3       | o =       | c = Cross cancellation property: $\forall a, b, c \in G: a b = c a \implies b = c$ }} {{eqn | n = 4       | o =       | c = Middle cancellation property: $\forall a, b, c, d, x \in G: a x b = c x d \implies a b = c d$  }} {{end-eqn}}


Asymmetric Relation is Antisymmetric

Let $\RR$ be an asymmetric relation. Then $\RR$ is also antisymmetric.


Asymmetric Relation is Antireflexive

Let $S$ be a set. Let $\RR \subseteq S \times S$ be a relation on $S$. Let $\RR$ be asymmetric. Then $\RR$ is also antireflexive.


Antireflexive and Transitive Relation is Asymmetric

Let $\mathcal R \subseteq S \times S$ be a relation which is not null. Let $\mathcal R$ be antireflexive and transitive. Then $\mathcal R$ is also asymmetric.


Antitransitive Relation is Antireflexive

Let $S$ be a set. Let $\mathcal R \subseteq S \times S$ be a relation on $S$. Let $\mathcal R$ be antitransitive. Then $\mathcal R$ is also antireflexive.


Symmetric Transitive and Serial Relation is Reflexive

Let $\RR$ be a relation which is: :symmetric :transitive :serial. Then $\RR$ is reflexive. Thus such a relation is an equivalence.


Inverse Relation Properties

Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ has any of the properties: * Reflexive * Antireflexive * Non-reflexive * Symmetric * Asymmetric * Antisymmetric * Non-symmetric * Transitive * Antitransitive * Non-transitive ... then its inverse $\mathcal R^{-1}$ has the same properties.


Properties of Restriction of Relation

Let $S$ be a set. Let $\mathcal R \subseteq S \times S$ be a relation on $S$. Let $T \subseteq S$ be a subset of $S$. Let $\mathcal R \restriction_T \ \subseteq T \times T$ be the restriction of $\mathcal R$ to $T$. If $\mathcal R$ on $S$ has any of the properties: * Reflexive * Antireflexive * Symmetric * Antisymmetric * Asymmetric * Transitive * Antitransitive * Connected ... then $\mathcal R \restriction_T$ on $T$ has the same properties.


Relation is Antireflexive iff Disjoint from Diagonal Relation

Let $\mathcal R \subseteq S \times S$ be a relation on a set $S$. Then: : $\mathcal R$ is antireflexive iff : $\Delta_S \cap \mathcal R = \varnothing$ where $\Delta_S$ is the diagonal relation.


Inverse Relation Equal iff Subset

If a relation $\RR$ is a subset or superset of its inverse, then it equals its inverse. That is, the following are equivalent: {{begin-axiom}} {{axiom | n = 1         | m = \RR \subseteq \RR^{-1} }} {{axiom | n = 2         | m = \RR^{-1} \subseteq \RR }} {{axiom | n = 3         | m = \RR = \RR^{-1} }} {{end-axiom}}


Relation is Symmetric iff Inverse is Symmetric

A relation $\mathcal R$ is symmetric {{iff}} its inverse $\mathcal R^{-1}$ is also symmetric.


Inverse of Many-to-One Relation is One-to-Many

The inverse of a many-to-one relation is a one-to-many relation, and vice versa.


Many-to-One Relation Composite with Inverse is Transitive

Let $\mathcal R \subseteq S \times T$ be a relation which is many-to-one. Then the composites (both ways) of $\mathcal R$ and its inverse $\mathcal R^{-1}$, that is, both $\mathcal R^{-1} \circ \mathcal R$ and $\mathcal R \circ \mathcal R^{-1}$, are transitive.


One-to-Many Relation Composite with Inverse is Coreflexive

Let $\mathcal R \subseteq S \times S$ be a relation which is one-to-many. Then the composite of $\mathcal R$ with its inverse is a coreflexive relation: :$\mathcal R^{-1} \circ \mathcal R \subseteq \Delta_X$ {{refactor|This is a separate result.}} That is, by Relation is Symmetric and Antisymmetric iff Coreflexive, $\mathcal R^{-1} \circ \mathcal R$ is both symmetric and antisymmetric.


Element in its own Equivalence Class

Let $\mathcal R$ be an equivalence relation on a set $S$. Then every element of $S$ is in its own $\mathcal R$-class: :$\forall x \in S: x \in \eqclass x {\mathcal R}$


Equivalence Class of Element is Subset

Let $\mathcal R$ be an equivalence relation on a set $S$. The $\mathcal R$-class of every element of $S$ is a subset of the set the element is in: :$\forall x \in S: \eqclass x {\mathcal R} \subseteq S$


Equivalence Class is not Empty

Let $\mathcal R$ be an equivalence relation on a set $S$. Then no $\mathcal R$-class is empty.


Handshake Lemma

Let $G$ be a $\tuple {p, q}$-undirected graph, which may be a multigraph or a loop-graph, or both. Let $V = \set {v_1, v_2, \ldots, v_p}$ be the vertex set of $G$. Then: :$\displaystyle \sum_{i \mathop = 1}^p \map {\deg_G} {v_i} = 2 q$ where $\map {\deg_G} {v_i}$ is the degree of vertex $v_i$. That is, the sum of all the degrees of all the vertices of an graph is equal to twice its size. This result is known as the '''Handshake Lemma''' or '''Handshaking Lemma'''.


Equivalence Class holds Equivalent Elements

Let $\RR$ be an equivalence relation on a set $S$. Then: :$\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y \RR$


Equivalence Classes are Disjoint

Let $\RR$ be an equivalence relation on a set $S$. Then all $\RR$-classes are pairwise disjoint: :$\tuple {x, y} \notin \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$


Fundamental Theorem on Equivalence Relations

Let $\RR \subseteq S \times S$ be an equivalence on a set $S$. Then the quotient $S / \RR$ of $S$ by $\RR$ forms a partition of $S$.


Equivalence Class is Unique

Let $\mathcal R$ be an equivalence relation on $S$. For each $x \in S$, the one and only one $\mathcal R$-class to which $x$ belongs is $\eqclass x {\mathcal R}$.


Equivalence Class Equivalent Statements

Let $\mathcal R$ be an equivalence on $S$. {{refactor|Use TFAE template}} Then $\forall x, y \in S$, the following statements are all equivalent: :$(1): \quad x$ and $y$ are in the same $\mathcal R$-class :$(2): \quad \eqclass x {\mathcal R} = \eqclass y {\mathcal R}$ :$(3): \quad x \mathrel {\mathcal R} y$ :$(4): \quad x \in \eqclass y {\mathcal R}$ :$(5): \quad y \in \eqclass x {\mathcal R}$ :$(6): \quad \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O$


Relation Induced by Partition is Equivalence

Let $\mathbb S$ be a partition of a set $S$. Let $\RR$ be the relation induced by $\mathbb S$. Then: :$(1): \quad \RR$ is unique :$(2): \quad \RR$ is an equivalence relation on $S$. Hence $\mathbb S$ is the quotient set of $S$ by $\RR$, that is: :$\mathbb S = S / \RR$


Relation Partitions Set iff Equivalence

Let $\RR$ be a relation on a set $S$. Then $S$ can be partitioned into subsets by $\RR$ {{iff}} $\RR$ is an equivalence relation on $S$. The partition of $S$ defined by $\RR$ is the quotient set $S / \RR$.


Intersection of Equivalences

The intersection of two equivalence relations is itself an equivalence relation.


Equivalence of Definitions of Equivalence Relation

Let $\RR$ be a relation on a set $S$. {{TFAE|def = Equivalence Relation}}


Equivalence iff Diagonal and Inverse Composite

Let $\RR$ be a relation on $S$. Then $\RR$ is an equivalence relation on $S$ {{iff}}: :$\Delta_S \subseteq \RR$ and: :$\RR = \RR \circ \RR^{-1}$


Equality of Elements in Range of Mapping

Let $f: S \to T$ be a mapping. Then: :$\exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f \iff \map f {x_1} = \map f {x_2}$


Inverse of Mapping is One-to-Many Relation

Let $f$ be a mapping. Then its inverse $f^{-1}$ is a one-to-many relation.


Equality of Mappings

Two mappings $f_1: S_1 \to T_1, f_2: S_2 \to T_2$ are equal {{iff}}: :$(1): \quad S_1 = S_2$ :$(2): \quad T_1 = T_2$ :$(3): \quad \forall x \in S_1: \map {f_1} x = \map {f_2} x$


Mapping is Constant iff Image is Singleton

A mapping is a constant mapping {{iff}} its image is a singleton.


Diagonal Relation is Right Identity

Let $\RR \subseteq S \times T$ be a relation on $S \times T$. Then: :$\RR \circ \Delta_S = \RR$ where $\Delta_S$ is the diagonal relation on $S$, and $\circ$ signifies composition of relations.


Identity Mapping is Left Identity

Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Then: :$I_T \circ f = f$ where $I_T$ is the identity mapping on $T$, and $\circ$ signifies composition of mappings.


Image of Element under Inverse Mapping

Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping such that its inverse $f^{-1}: T \to S$ is also a mapping. Then: :$\forall x \in S, y \in T: \map f x = y \iff \map {f^{-1} } y = x$


Injection iff Left Cancellable

A mapping $f$ is an injection {{iff}} $f$ is left cancellable.


Identity Mapping is Injection

On any set $S$, the identity mapping $I_S: S \to S$ is an injection.


Composite of Injections is Injection

A composite of injections is an injection. That is: :If $f$ and $g$ are injections, then so is $f \circ g$.


Injection if Composite is Injection

Let $f$ and $g$ be mappings such that their composite $g \circ f$ is an injection. Then $f$ is an injection.


Injection iff Left Inverse

A mapping $f: S \to T, S \ne \O$ is an injection {{iff}}: :$\exists g: T \to S: g \circ f = I_S$ where $g$ is a mapping. That is, {{iff}} $f$ has a left inverse.


Inclusion Mapping is Injection

Let $S, T$ be sets such that $S$ is a subset of $T$. Then the inclusion mapping $i_S: S \to T$ defined as: :$\forall x \in S: \map {i_S} x = x$ is an injection. For this reason the inclusion mapping can be known as the '''canonical injection of $S$ to $T$'''.


Surjection iff Right Cancellable

Let $f$ be a mapping. Then $f$ is a surjection {{iff}} $f$ is right cancellable.


Identity Mapping is Surjection

On any set $S$, the identity mapping $I_S: S \to S$ is a surjection.


Restriction of Mapping to Image is Surjection

Let $f: S \to T$ be a mapping. Let $g: S \to \Img f$ be the restriction of $f$ to $S \times \Img f$. Then $g$ is a surjective restriction of $f$.


Composite of Surjections is Surjection

A composite of surjections is a surjection. That is: :If $g$ and $f$ are surjections, then so is $g \circ f$.


Surjection if Composite is Surjection

Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings such that $g \circ f$ is a surjection. Then $g$ is a surjection.


Surjection iff Right Inverse

A mapping $f: S \to T, S \ne \O$ is a surjection {{iff}}: :$\exists g: T \to S: f \circ g = I_T$ where: :$g$ is a mapping :$I_T$ is the identity mapping on $T$. That is, {{iff}} $f$ has a right inverse.


Identity Mapping is Bijection

The identity mapping $I_S: S \to S$ on the set $S$ is a bijection.


Identity Mapping is Permutation

The identity mapping $I_S: S \to S$ on the set $S$ is a permutation.


Bijection iff Left and Right Inverse

Let $f: S \to T$ be a mapping. $f$ is a bijection {{iff}}: :$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$ :$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$ where both $g_1$ and $g_2$ are mappings. It also follows that it is necessarily the case that $g_1 = g_2$ for such to be possible.


Inverse of Bijection is Bijection

Let $f: S \to T$ be a bijection in the sense that: :$(1): \quad f$ is an injection :$(2): \quad f$ is a surjection. Then the inverse $f^{-1}$ of $f$ is itself a bijection by the same definition.


Inverse Element of Bijection

Let $S$ and $T$ be sets. Let $f: S \to T$ be a bijection. Then: :$\map {f^{-1} } y = x \iff \map f x = y$ where $f^{-1}$ is the inverse mapping of $f$.


Composite of Bijection with Inverse is Identity Mapping

Let $f: S \to T$ be a bijection. Then: : $f^{-1} \circ f = I_S$ : $f \circ f^{-1} = I_T$ where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.


Inverse of Inverse of Bijection

Let $f: S \to T$ be a bijection. Then: :$\paren {f^{-1} }^{-1} = f$ where $f^{-1}$ is the inverse of $f$.


Composite of Bijections is Bijection

Let $f$ and $g$ be mappings such that $\Dom f = \Cdm g$. Then: :If $f$ and $g$ are both bijections, then so is $f \circ g$ where $f \circ g$ is the composite mapping of $f$ with $g$.


Inverse of Composite Bijection

Let $f$ and $g$ be bijections such that $\Dom g = \Cdm f$. Then: :$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$ and $f^{-1} \circ g^{-1}$ is itself a bijection.


Composite of Permutations is Permutation

Let $f, g$ are permutations of a set $S$. Then their composite $g \circ f$ is also a permutation of $S$.


Inverse of Permutation is Permutation

If $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.


Left Inverse Mapping is Surjection

Let $f: S \to T$ be an injection. Let $g: T \to S$ be a left inverse of $f$. Then $g$ is a surjection.


Right Inverse Mapping is Injection

Let $f: S \to T$ be a mapping. Let $g: T \to S$ be a right inverse of $f$. Then $g$ is an injection.


Set Equivalence is Equivalence Relation

Set equivalence is an equivalence relation.


Set Equivalence Less One Element

Let $S$ and $T$ be sets which are equivalent: :$S \sim T$ Let $a \in S$ and $b \in T$. Then: : $S \setminus \left\{{a}\right\} \sim T \setminus \left\{{b}\right\}$ where $\setminus$ denotes set difference.


Image of Union under Mapping

Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $A$ and $B$ be subsets of $S$. Then: :$f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$ This can be expressed in the language and notation of direct image mappings as: :$\forall A, B \in \powerset S: \map {f^\to} {A \cup B} = \map {f^\to} A \cup \map {f^\to} B$


Image of Intersection under Mapping

Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $S_1$ and $S_2$ be subsets of $S$. Then: :$f \sqbrk {S_1 \cap S_2} \subseteq f \sqbrk {S_1} \cap f \sqbrk {S_2}$ This can be expressed in the language and notation of direct image mappings as: :$\forall S_1, S_2 \in \powerset S: \map {f^\to} {S_1 \cap S_2} \subseteq \map {f^\to} {S_1} \cap \map {f^\to} {S_2}$


Image of Intersection under Injection

Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Then: :$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$ {{iff}} $f$ is an injection.


Image of Intersection under One-to-Many Relation

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Then: :$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$  {{iff}} $\RR$ is one-to-many.


One-to-Many Image of Set Difference

Let $\mathcal R \subseteq S \times T$ be a relation. Let $A$ and $B$ be subsets of $S$. Then: :$(1): \quad \mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right] = \mathcal R \left[{A \setminus B}\right]$ {{iff}} $\mathcal R$ is one-to-many.


Preimage of Union under Mapping

Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $T_1$ and $T_2$ be subsets of $T$. Then: :$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$ This can be expressed in the language and notation of inverse image mappings as: :$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cup T_2} = \map {f^\gets} {T_1} \cup \map {f^\gets} {T_2}$


Preimage of Intersection under Mapping

Let $f: S \to T$ be a mapping. Let $T_1$ and $T_2$ be subsets of $T$. Then: :$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$ This can be expressed in the language and notation of inverse image mappings as: :$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cap T_2} = \map {f^\gets} {T_1} \cap \map {f^\gets} {T_2}$


Preimage of Set Difference under Mapping

Let $f: S \to T$ be a mapping. Let $T_1$ and $T_2$ be subsets of $T$. Then: :$f^{-1} \sqbrk {T_1 \setminus T_2} = f^{-1} \sqbrk {T_1} \setminus f^{-1} \sqbrk {T_2}$ where: :$\setminus$ denotes set difference :$f^{-1} \sqbrk {T_1}$ denotes preimage.


Projection is Surjection

Let $S$ and $T$ be non-empty sets. Let $S \times T$ be the Cartesian product of $S$ and $T$. Let $\pr_1: S \times T \to T$ and $\pr_2: S \times T \to T$ be the first projection and second projection respectively on $S \times T$. Then $\pr_1$ and $\pr_2$ are both surjections.


Sum Rule for Derivatives

Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$. Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable. Let $\map f x = \map j x + \map k x$. Then $f$ is differentiable at $\xi$ and: :$\map {f'} \xi = \map {j'} \xi + \map {k'} \xi$ It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then: :$\forall x \in I: \map {f'} x = \map {j'} x + \map {k'} x$


Quotient Mapping is Surjection

Let $\RR$ be an equivalence relation on $S$. Then the quotient mapping $q_\RR: S \to S / \RR$ is a surjection.


Trivial Quotient is a Bijection

Let $\Delta_S$ be the diagonal relation on a set $S$. Let $q_{\Delta_S}: S \to S / \Delta_S$ be the trivial quotient of $S$. Then $q_{\Delta_S}: S \to S / \Delta_S$ is a bijection.


Renaming Mapping is Well-Defined

Let $f: S \to T$ be a mapping. Let $r: S / \RR_f \to \Img f$ be the renaming mapping,  defined as: :$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$ where: :$\RR_f$ is the equivalence induced by the mapping $f$ :$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$ :$\eqclass x {\RR_f}$ is the equivalence class of $x$ under $\RR_f$. The renaming mapping is always well-defined.


Powers of Group Elements

Let $\left({G, \circ}\right)$ be a group whose identity is $e$. Let $a \in G$. Then the following results hold:


Renaming Mapping is Bijection

Let $f: S \to T$ be a mapping. Let $r: S / \RR_f \to \Img f$ be the renaming mapping, defined as: :$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$ where: :$\RR_f$ is the equivalence induced by the mapping $f$ :$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$ :$\eqclass x {\RR_f}$ is the equivalence class of $x$ under $\RR_f$. The renaming mapping is a bijection.


Composite of Mapping with Inverse

Let $f: S \to T$ be a mapping. Then: :$\forall x \in S: \map {f^{-1} \circ f} x = \eqclass x {\RR_f}$ where: :$\RR_f$ is the equivalence induced by $f$ :$\eqclass x {\RR_f}$ is the $\RR_f$-equivalence class of $x$.


Factoring Mapping into Surjection and Inclusion

Every mapping $f:S \to T$ can be uniquely '''factored into''' a surjection $g$ followed by the inclusion mapping $i_T$. That is, $f = i_T \circ g$ where: :$g: S \to \Img f: \map g x = \map f x$ :$i_T: \Img f \to T: \map {i_T} x = x$ This can be illustrated using a commutative diagram as follows: ::$\begin{xy}\xymatrix@L+2mu@+1em {  S \ar@{-->}[r]^*{g}      \ar[rd]_*{f = i_T \circ g} &  \Img f \ar@{-->}[d]^*{i_T} \\ &  T }\end{xy}$


Quotient Theorem for Surjections

Let $f: S \to T$ be a surjection. Then there is one and only one bijection $r: S / \RR_f \to T$ such that: :$r \circ q_{\RR_f} = f$ where: :$\RR_f$ is the equivalence induced by $f$ :$r: S / \RR_f \to T$ is the renaming mapping :$q_{\RR_f}: S \to S / \RR_f$ is the quotient mapping induced by $\RR_f$. This can be illustrated using a commutative diagram as follows: ::$\begin {xy} \xymatrix@L + 2mu@ + 1em {  S \ar@{-->}[rr]^*{f = r \circ q_{\RR_f} }    \ar[dd]_*{q_{\RR_f} } && T \\ \\ S / \RR_f \ar[urur]_*{r}  } \end {xy}$


Quotient Theorem for Sets

A mapping $f: S \to T$ can be uniquely '''factored into''' a surjection, followed by a bijection, followed by an injection. Thus: :$f = i \circ r \circ q_{\RR_f}$ where: {{begin-eqn}} {{eqn | ll= q_{\RR_f}:       | lo= S \to S / \RR_f:       | l = \map {q_{\RR_f} } s       | r = \eqclass s {\RR_f}       | c = Quotient Mapping }} {{eqn | ll= r:       | lo= S / \RR_f \to \Img f:       | l = \map r {\eqclass s {\RR_f} }       | r = \map f s       | c = Renaming Mapping }} {{eqn | ll= i:       | lo= \Img f \to T:       | l = \map i t       | r = t       | c = Inclusion Mapping }} {{end-eqn}} where: :$\RR_f$ is the equivalence induced by $f$ :$S / \RR_f$ is the quotient set of $S$ induced by $\RR_f$ This can be illustrated using a commutative diagram as follows: ::$\begin {xy} \xymatrix@L + 2mu@ + 1em {  S \ar@{-->}[rrr]^*{f = i_T \circ r \circ q_{\RR_f} }    \ar[d]_*{q_{\RR_f} } & & & T \\ S / \RR_f \ar[rrr]_*{r} & & & \Img f \ar[u]_*{i_T} } \end {xy}$


Empty Set is Element of Power Set

The empty set is an element of all power sets: :$\forall S: \O \in \powerset S$


Direct Image Mapping of Relation is Mapping

Let $S$ and $T$ be sets. Let $\mathcal R \subseteq S \times T$ be a relation on $S \times T$. Let $\mathcal R^\to: \powerset S \to \powerset T$ be the direct image mapping of $\mathcal R$: :$\forall X \in \powerset S: \map {\mathcal R^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \mathcal R}$ Then $\mathcal R^\to$ is indeed a mapping.


Inverse Image Mapping of Relation is Mapping

Let $S$ and $T$ be sets. Let $\mathcal R \subseteq S \times T$ be a relation on $S \times T$. Let $\mathcal R^\gets$ be the inverse image mapping of $\mathcal R$: :$\mathcal R^\gets: \powerset T \to \powerset S: \map {\mathcal R^\gets} Y = \mathcal R^{-1} \sqbrk Y$ Then $\mathcal R^\gets$ is indeed a mapping.


Direct Image Mapping of Injection is Injection

Let $f: S \to T$ be an injection. Then the direct image mapping of $f$: :$f^\to: \powerset S \to \powerset T$ is an injection.


Direct Image Mapping of Surjection is Surjection

Let $f: S \to T$ be a surjection. Then the direct image mapping of $f$: :$f^\to: \powerset S \to \powerset T$ is a surjection.


Cantor's Theorem

There is no surjection from a set $S$ to its power set for any set $S$. That is, $S$ is strictly smaller than its power set.


Surjection Induced by Powerset is Induced by Surjection

Let $\mathcal R \subseteq S \times T$ be a relation. Let $\mathcal R^\to: \powerset S \to \powerset T$ be the direct image mapping $\mathcal R$. Let $\mathcal R^\to$ be a surjection. Let $X = \Preimg {\mathcal R}$, that is, the preimage of $\mathcal R$. Then $\mathcal R {\restriction_X} \subseteq X \times T$, that is, the restriction of $\mathcal R$ to $X$, is a surjection.


Mapping is Injection if its Direct Image Mapping is Injection

Let $f: S \to T$ be a mapping. Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$. Let $f^\to$ be an injection. Then $f: S \to T$ is also an injection.


Direct Image Mapping is Bijection iff Mapping is Bijection

Let $\mathcal R \subseteq S \times T$ be a relation. Let $\mathcal R^\to: \powerset S \to \powerset T$ be the direct image mapping of $\mathcal R$. Then $\mathcal R \subseteq S \times T$ is a bijection {{iff}} $\mathcal R^\to: \powerset S \to \powerset T$ is a bijection.


No Bijection from Set to its Power Set

Let $S$ be a set. Let $\powerset S$ denote the power set of $S$. There is no bijection $f: S \to \powerset S$.


Union of Power Sets

The union of the power sets of two sets $S$ and $T$ is a subset of the power set of their union: :$\powerset S \cup \powerset T \subseteq \powerset {S \cup T}$


Intersection of Power Sets

The intersection of the power sets of two sets $S$ and $T$ is equal to the power set of their intersection: :$\powerset S \cap \powerset T = \powerset {S \cap T}$


Cantor-Bernstein-Schröder Theorem

If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent: :$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$


Trivial Ordering is Universally Compatible

Let $S$ be a set. Let $\RR$ be the trivial ordering on $S$. Then $\RR$ is universally compatible.


Identity Mapping is Order Isomorphism

Let $\left({S, \preceq}\right)$ be an ordered set. The identity mapping $I_S$ is an order isomorphism from $\left({S, \preceq}\right)$ to itself.


Inverse of Order Isomorphism is Order Isomorphism

Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets. Let $\phi$ be a bijection from $\left({S, \preceq_1}\right)$ to $\left({T, \preceq_2}\right)$. Then: : $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is an order isomorphism {{iff}}: : $\phi^{-1}: \left({T, \preceq_2}\right) \to \left({S, \preceq_1}\right)$ is also an order isomorphism.


Composite of Order Isomorphisms is Order Isomorphism

Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ and $\left({S_3, \preceq_3}\right)$ be ordered sets. Let: : $\phi: \left({S_1, \preceq_1}\right) \to \left({S_2, \preceq_2}\right)$ and: : $\psi: \left({S_2, \preceq_2}\right) \to \left({S_3, \preceq_3}\right)$ be order isomorphisms. Then $\psi \circ \phi: \left({S_1, \preceq_1}\right) \to \left({S_3, \preceq_3}\right)$ is also an order isomorphism.


Order Isomorphism is Equivalence Relation

Order isomorphism between ordered sets is an equivalence relation. So any given family of ordered sets can be partitioned into disjoint classes of isomorphic sets.


Dual Ordering is Ordering

Let $\left({S, \preceq}\right)$ be an ordered set. Let $\succeq$ denote the dual ordering of $\preceq$. Then $\succeq$ is an ordering on $S$.


Power Set is Complete Lattice

Let $S$ be a set. Let $\left({\mathcal P \left({S}\right), \subseteq}\right)$ be the relational structure defined on $\mathcal P \left({S}\right)$ by the relation $\subseteq$. Then $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is a complete lattice.


Totally Ordered Set is Lattice

Every totally ordered set is a lattice.


Trichotomy Law (Ordering)

Let $\struct {S, \preceq}$ be an ordered set. Then $\preceq$ is a total ordering {{iff}}: :$\forall a, b \in S: \paren {a \prec b} \lor \paren {a = b} \lor \paren {a \succ b}$ That is, every element either strictly precedes, is the same as, or strictly succeeds, every other element. In other words, {{iff}} $\prec$ is a trichotomy.


Complement of Reflexive Relation

Let $\RR \subseteq S \times S$ be a relation. Then $\RR$ is reflexive {{iff}} its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is antireflexive. Likewise, $\RR$ is antireflexive {{iff}} its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is reflexive.


Strictly Increasing Mapping is Increasing

A mapping that is strictly increasing is an increasing mapping.


Strictly Decreasing Mapping is Decreasing

A mapping that is strictly decreasing is a decreasing mapping.


Strictly Monotone Mapping with Totally Ordered Domain is Injective

Let $\struct {S, \preceq_1}$ be a totally ordered set. Let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be a strictly monotone mapping. Then $\phi$ is injective.


Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing

Let $\struct {S, \preceq_1}$ be a totally ordered set. Let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: S \to T$ be a mapping. Then $\phi$ is an order embedding {{iff}} $\phi$ is strictly increasing. That is: :$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$ {{iff}}: :$\forall x, y \in S: x \prec_1 y \implies \map \phi x \prec_2 \map \phi y$


Finite Totally Ordered Set is Well-Ordered

Every finite totally ordered set is well-ordered.


Subset of Well-Ordered Set is Well-Ordered

Every non-empty subset of a well-ordered set is itself well-ordered.


Element Commutes with Product of Commuting Elements

Let $(S, \circ)$ be a semigroup. Let $x, y, z \in S$. If $x$ commutes with both $y$ and $z$, then $x$ commutes with $y \circ z$.


Associative Idempotent Anticommutative

Let $\circ$ be a binary operation on a set $S$. Let $\circ$ be associative. Then $\circ$ is anticommutative {{iff}}: : $(1): \quad \circ$ is idempotent and: : $(2): \quad \forall a, b \in S: a \circ b \circ a = a$.


Associative and Anticommutative

Let $\circ$ be a binary operation on a set $S$. Let $\circ$ be both associative and anticommutative. Then: :$\forall x, y, z \in S: x \circ y \circ z = x \circ z$


Constant Operation is Commutative

Let $S$ be a set. Let $x \sqbrk c y = c$ be a constant operation on $S$. Then $\sqbrk c$ is a commutative operation: :$\forall x, y \in S: x \sqbrk c y = y \sqbrk c x$


Constant Operation is Associative

Let $S$ be a set. Let $x \left[{c}\right] y = c$ be a constant operation on $S$. Then $\left[{c}\right]$ is an associative operation: :$\forall x, y, z \in S: \left({x \left[{c}\right] y}\right) \left[{c}\right] z = x \left[{c}\right] \left({y \left[{c}\right] z}\right)$


Left Operation is Idempotent

The left operation is idempotent: :$\forall x: x \leftarrow x = x$


Left Operation is Associative

The left operation is associative: :$\forall x, y, z: \paren {x \leftarrow y} \leftarrow z = x \leftarrow \paren {y \leftarrow z}$


Right Operation is Associative

The right operation is associative: :$\forall x, y, z: \paren {x \rightarrow y} \rightarrow z = x \rightarrow \paren {y \rightarrow z}$


Max and Min are Commutative

The Max and Min operations are commutative: : $\max \left({x, y}\right) = \max \left({y, x}\right)$ : $\min \left({x, y}\right) = \min \left({y, x}\right)$


Max and Min are Associative

The Max and Min operations are associative: : $\max \left({\max \left({x, y}\right), z}\right) = \max \left({x, \max \left({y, z}\right)}\right)$ : $\min \left({\min \left({x, y}\right), z}\right) = \min \left({x, \min \left({y, z}\right)}\right)$ Thus we are justified in writing $\max \left({x, y, z}\right)$ and $\min \left({x, y, z}\right)$.


Max and Min are Idempotent

The Max and Min operations are idempotent: : $\max \left({x, x}\right) = x$ : $\min \left({x, x}\right) = x$


Max and Min Operations are Distributive over Each Other

The Max and Min operations are distributive over each other: :$\max \set {x, \min \set {y, z} } = \min \set {\max \set {x, y}, \max \set {x, z} }$ :$\max \set {\min \set {x, y}, z} = \min \set {\max \set {x, z}, \max \set {y, z} }$ :$\min \set {x, \max \set {y, z} } = \max \set {\min \set {x, y}, \min \set {x, z} }$ :$\min \set {\max \set {x, y}, z} = \max \set {\min \set {x, z}, \min \set {y, z} }$


Equality of Algebraic Structures

Two algebraic structures $\struct {S, \circ}$ and $\struct {T, *}$ are equal {{iff}}: :$S = T$ :$\forall a, b \in S: a \circ b = a * b$


Restriction of Associative Operation is Associative

Let $\struct {S, \circ}$ be an semigroup. Let $T \subseteq S$. Let $T$ be closed under $\circ$. Then $\struct {T, \circ {\restriction_T} }$ is also a semigroup, where $\circ {\restriction_T}$ is the restriction of $\circ$ to $T$.


Restriction of Commutative Operation is Commutative

Let $\struct {S, \circ}$ be an algebraic structure. Let $T \subseteq S$. Let the operation $\circ$ be commutative on $\struct {S, \circ}$. Then the restriction $\circ {\restriction_T}$ of $\circ$ to $T$ is also commutative.


Idempotent Magma Element forms Singleton Submagma

Let $\struct {S, \circ}$ be a magma. Let $x \in S$ be an idempotent element of $\struct {S, \circ}$. Then  $\struct {\set x, \circ}$ is a submagma of $\struct {S, \circ}$.


Magma Subset Product with Self

Let $\left({S, \circ}\right)$ be a magma. Let $T \subseteq S$. Then $\left({T, \circ}\right)$ is a magma {{iff}} $T \circ T \subseteq T$, where $T \circ T$ is the subset product of $T$ with itself.


Subset Product within Semigroup is Associative

Let $\struct {S, \circ}$ be a semigroup. Then the operation $\circ_\PP$ induced on the power set of $S$ is also associative.


Subset Product within Commutative Structure is Commutative

Let $\struct {S, \circ}$ be a magma. If $\circ$ is commutative, then the operation $\circ_\mathcal P$ induced on the power set of $S$ is also commutative.


Subset of Subset Product

Let $\struct {S, \circ}$ be a magma. Let $\powerset S$ be the power set of $S$. Let $X, Y, Z \in \powerset S$. Then: :$X \subseteq Y \implies \paren {X \circ Z} \subseteq \paren {Y \circ Z}$ :$X \subseteq Y \implies \paren {Z \circ X} \subseteq \paren {Z \circ Y}$ where $X \circ Z$ etc. denotes subset product.


Cancellable Element is Cancellable in Subset

Let $\left ({S, \circ}\right)$ be an algebraic structure. Let $\left ({T, \circ}\right) \subseteq \left ({S, \circ}\right)$. Let $x \in T$ be cancellable in $S$. Then $x$ is also cancellable in $T$.


Subsemigroup Closure Test

To show that an algebraic structure $\left({T, \circ}\right)$ is a subsemigroup of a semigroup $\struct {S, \circ}$, we need to show only that: : $(1): \quad T \subseteq S$ : $(2): \quad \circ$ is a closed operation in $T$.


Cancellable iff Regular Representations Injective

Let $\struct {S, \circ}$ be an algebraic structure. Then $a \in S$ is cancellable {{iff}}: ::the left regular representation $\map {\lambda_a} x$ is injective and ::the right regular representation $\map {\rho_a} x$ is injective.


More than one Left Identity then no Right Identity

Let $\struct {S, \circ}$ be an algebraic structure. If $\struct {S, \circ}$ has more than one left identity, then it has no right identity.


Element under Left Operation is Right Identity

Let $\struct {S, \leftarrow}$ be an algebraic structure in which the operation $\leftarrow$ is the left operation. Then no matter what $S$ is, $\struct {S, \leftarrow}$ is a semigroup all of whose elements are right identities. Thus it can be seen that any right identity in a semigroup is not necessarily unique.


Element under Right Operation is Left Identity

Let $\struct {S, \rightarrow}$ be an algebraic structure in which the operation $\rightarrow$ is the right operation. Then $\struct {S, \rightarrow}$ is a semigroup all of whose elements are left identities. Thus it can be seen that any left identity in a semigroup is not necessarily unique.


Left and Right Identity are the Same

Let $\struct {S, \circ}$ be an algebraic structure. Let $e_L \in S$ be a left identity, and $e_R \in S$ be a right identity. Then: :$e_L = e_R$ that is, both the left identity and right identity are the same, and are therefore an identity $e$. Furthermore, $e$ is the only left identity and right identity for $\circ$.


Identity Property in Semigroup

Let $\struct {S, \circ}$ be a semigroup. Let $s \in S$ be such that: :$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$ Then $\struct {S, \circ}$ has an identity.


Identity of Monoid is Cancellable

The identity of a monoid is cancellable.


Identity is only Idempotent Cancellable Element

Let $e_S$ is the identity of an algebraic structure $\left({S, \circ}\right)$. Then $e_S$ is the only cancellable element of $\left({S, \circ}\right)$ that is idempotent.


Set of all Self-Maps is Monoid

Let $S$ be a set. Let $S^S$ be the set of all mappings from $S$ to itself. Let the operation $\circ$ represent composition of mappings. Then the algebraic structure $\struct {S^S, \circ}$ is a monoid whose identity element is the identity mapping on $S$.


Test for Submonoid

To show that $\left({T, \circ}\right)$ is a submonoid of a monoid $\left({S, \circ}\right)$, we need to show that: :$(1): \quad T \subseteq S$ :$(2): \quad \left({T, \circ}\right)$ is a magma (i.e. that it is closed) :$(3): \quad \left({T, \circ}\right)$ has an identity.


Identity of Cancellable Monoid is Identity of Submonoid

Let $\struct {S, \circ}$ be a monoid, all of whose elements are cancellable. Let $\struct {T, \circ}$ be a submonoid of $\struct {S, \circ}$. Then the identity of $\struct {T, \circ}$ is the same element as the identity of $\struct {S, \circ}$.


Cancellable Elements of Monoid form Submonoid

The cancellable elements of a monoid $\struct {S, \circ}$ form a submonoid of $\struct {S, \circ}$.


Product of Semigroup Element with Left Inverse is Idempotent

Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$. Let $x \in S$ such that $\exists x_L: x_L \circ x = e_L$, that is $x$ has a left inverse with respect to the left identity. Then: :$\paren {x \circ x_L} \circ \paren {x \circ x_L} = x \circ x_L$ That is, $x \circ x_L$ is idempotent.


Product of Semigroup Element with Right Inverse is Idempotent

Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$. Let $x \in S$ such that $\exists x_R: x \circ x_R = e_R$, i.e. $x$ has a right inverse with respect to the right identity. Then: :$\paren {x_R \circ x} \circ \paren {x_R \circ x} = x_R \circ x$ That is, $x_R \circ x$ is idempotent.


Left Inverse for All is Right Inverse

Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$ such that: :$\forall x \in S: \exists x_L: x_L \circ x = e_L$ That is, every element of $S$ has a left inverse with respect to the left identity. Then $x \circ x_L = e_L$, that is, $x_L$ is also a right inverse with respect to the left identity.


Left Inverse and Right Inverse is Inverse

Let $\struct {S, \circ}$ be a monoid with identity element $e_S$. Let $x \in S$ such that $x$ has both a left inverse and a right inverse. That is: :$\exists x_L \in S: x_L \circ x = e_S$ :$\exists x_R \in S: x \circ x_R = e_S$ Then $x_L = x_R$, that is, $x$ has an inverse. Furthermore, that element is the ''only'' inverse (both right and left) for $x$


Left and Right Inverses of Product

Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$. Let $x, y \in S$. Let: :$(1): \quad x \circ y$ have a left inverse for $\circ$ :$(2): \quad y \circ x$ have a right inverse for $\circ$. Then both $x$ and $y$ are invertible for $\circ$.


Equivalence of Definitions of Self-Inverse

Let $\left({S, \circ}\right)$ be a monoid whose identity is $e_S$. Let $x \in S$. {{TFAE|def = Self-Inverse Element|context = Abstract Algebra}}


Inverse of Identity Element is Itself

Let $\struct {S, \circ}$ be an algebraic structure with an identity element $e$. Let the inverse of $e$ be $e^{-1}$. Then: : $e^{-1} = e$ That is, $e$ is self-inverse.


Invertible Element of Associative Structure is Cancellable

Let $\left({S, \circ}\right)$ be an algebraic structure where $\circ$ is associative. Let $\left({S, \circ}\right)$ have an identity element $e_S$. An element of $\left({S, \circ}\right)$ which is invertible is also cancellable.


Regular Representation of Invertible Element is Permutation

Let $\struct {S, \circ}$ be a monoid. Let $a \in S$ be invertible. Then the left regular representation $\lambda_a$ and the right regular representation $\rho_a$ are permutations of $S$.


Group is Inverse Semigroup with Identity

A group is an inverse semigroup with an identity.


Invertible Elements of Monoid form Subgroup of Cancellable Elements

Let $\struct {S, \circ}$ be an monoid whose identity is $e_S$. Let $C$ be the set of all cancellable elements of $S$. Let $T$ be the set of all invertible elements of $S$. Then $\struct {T, \circ}$ is a subgroup of $\struct {C, \circ}$.


Structure Induced by Associative Operation is Associative

Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set. Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$. Let $\circ$ be associative. Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also associative.


Structure Induced by Commutative Operation is Commutative

Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set. Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$. Let $\circ$ be a commutative operation. Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also commutative.


Induced Structure Identity

Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set. Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$. Let $e$ be an identity for $\circ$. Then the constant mapping $f_e: S \to T$ defined as: :$\forall x \in S: \map {f_e} x = e$ is the identity for the pointwise operation $\oplus$ induced on $T^S$ by $\circ$.


Structure Induced by Abelian Group Operation is Abelian Group

Let $\struct {G, \circ}$ be an abelian group whose identity is $e$. Let $S$ be a set. Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$. Then $\struct {G^S, \oplus}$ is an abelian group.


Power Set with Union is Commutative Monoid

Let $S$ be a set and let $\powerset S$ be its power set. Then $\struct {\powerset S, \cup}$ is a commutative monoid whose identity is $\O$. The only invertible element of this structure is $\O$. Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cup}$ cannot be a group.


Power Set with Intersection is Commutative Monoid

Let $S$ be a set and let $\powerset S$ be its power set. Then $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$. The only invertible element of this structure is $S$. Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cap}$ cannot be a group.


Diagonal Relation is Universally Compatible

The diagonal relation $\Delta_S$ on a set $S$ is universally compatible with every operation on $S$.


Trivial Relation is Universally Congruent

The trivial relation $\mathcal R = S \times S$ on a set $S$ is universally congruent with every closed operation on $S$.


Quotient Structure is Well-Defined

Let $\struct {S, \circ}$ be an algebraic structure. Let $\RR$ be a congruence relation on $\struct {S, \circ}$. Let $S / \RR$ be the quotient set of $S$ by $\RR$. Let $\circ_\RR$ be the operation induced on $S / \RR$ by $\circ$. Then $\circ_\RR$ is a well-defined operation in the quotient structure $\struct {S / \RR, \circ_\RR}$.


External Direct Product Closure

Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures. Let $\struct {S \times T, \circ}$ be the external direct product of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed. Then $\struct {S \times T, \circ}$ is also closed.


External Direct Product Associativity

Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. Let $\circ_1$ and $\circ_2$ be associative. Then $\circ$ is also associative.


External Direct Product of Semigroups

The external direct product of two semigroups is itself a semigroup.


External Direct Product Commutativity

Let $\left({S \times T, \circ}\right)$ be the external direct product of the two algebraic structures $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$. Let $\circ_1$ and $\circ_2$ be commutative operations. Then $\circ$ is also a commutative operation.


External Direct Product Identity

Let $\struct {S \times T, \circ}$ be the external direct product of the two monoids $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. Let: : $e_S$ be the identity for $\struct {S, \circ_1}$ and: : $e_T$ be the identity for $\struct {T, \circ_2}$. Then $\tuple {e_S, e_T}$ is the identity for $\struct {S \times T, \circ}$.


External Direct Product Inverses

Let $\struct {S \times T, \circ}$ be the external direct product of the two monoids $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. Let: :$s^{-1}$ be an inverse of $s \in \struct {S, \circ_1}$ and: :$t^{-1}$ be an inverse of $t \in \struct {T, \circ_2}$. Then $\tuple {s^{-1}, t^{-1} }$ is an inverse of $\tuple {s, t} \in \struct {S \times T, \circ}$.


Morphism Property Preserves Closure

Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ be a mapping from one algebraic structure $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ to another $\left({T, *_1, *_2, \ldots, *_n}\right)$. Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$. Then the following properties hold: : If $S' \subseteq S$ is closed under $\circ_k$, then $\phi \left[{S'}\right]$ is closed under $*_k$ : If $T' \subseteq T$ is closed under $*_k$, then $\phi^{-1} \left[{T'}\right]$ is closed under $\circ_k$ where $\phi \left[{S'}\right]$ denotes the image of $S'$.


Morphism Property Preserves Cancellability

Let: : $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ be a mapping from one algebraic structure: : $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ to another: : $\left({T, *_1, *_2, \ldots, *_n}\right)$ Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$. Then if an element $a \in S$ is either left cancellable or right cancellable under $\circ_k$, then $\phi \left({a}\right)$ is correspondingly left cancellable or right cancellable under $*_k$. Thus, the morphism property is seen to preserve cancellability.


Quotient Mapping on Structure is Canonical Epimorphism

Let $\mathcal R$ be a congruence relation on an algebraic structure $\struct {S, \circ}$. Then the quotient mapping from $\struct {S, \circ}$ to the quotient structure $\struct {S / \mathcal R, \circ_\mathcal R}$ is an epimorphism: :$q_\mathcal R: \struct {S, \circ} \to \struct {S / \mathcal R, \circ_\mathcal R}: \forall x, y \in S: \map {q_\mathcal R} {x \circ y} = \map {q_\mathcal R} x \circ_\mathcal R \map {q_\mathcal R} y$ This is sometimes called the '''canonical epimorphism''' from $\struct {S, \circ}$ to $\struct {S / \mathcal R, \circ_\mathcal R}$.


Restriction of Homomorphism to Image is Epimorphism

Let $S$ and $T$ be algebraic structures. Let $\phi: S \to T$ be a homomorphism. Then a surjective restriction of $\phi$ can be produced by limiting the codomain of $\phi$ to its image $\operatorname{Im} \left({\phi}\right)$.


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 {{iff}} $x \circ y$ 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 {{iff}} $x \circ y \circ x^{-1} = y$.


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 {{iff}} $\phi$ 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 {{iff}} 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 {{iff}} $\struct {H, *}$ 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 $G$ {{iff}} $H$ and $K$ are permutable. That is: $H \circ K$ is a subgroup of $G$ {{iff}}: :$H \circ K = K \circ H$ 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 {{iff}} $z$ 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 {{iff}} $\struct {R^*, \circ}$ is a semigroup.


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 $\struct {R, +, \circ}$ {{iff}} these all hold: :$(1): \quad S \ne \O$ :$(2): \quad \forall x, y \in S: x + \paren {-y} \in S$ :$(3): \quad \forall x, y \in S: x \circ y \in S$