hcju/proofwikips huggingface dataset:
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 $c$ : :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 $\struct {G, \circ}$ : : $(1): \quad H \ne \O$, 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 $\struct {G, \circ}$ : :$(1): \quad H \ne \O$, 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 $\struct {G, \circ}$  $\struct {H, \circ}$ 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 $9$  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 $9$  : :$a_0 + a_1 + \ldots + a_n$ 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 $G$ : :$\forall x \in G: x H x^{-1} \subseteq H$.
Equality of Ordered Pairs
Two ordered pairs are equal  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''' : :$S_1 = S_2$ :$T_1 = T_2$ :$\tuple {s, t} \in \RR_1 \iff \tuple {s, t} \in \RR_2$ 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: :$\forall A \in \powerset S: \map {\RR^\to} A \in \powerset T$