hcju/proofwikips huggingface dataset:
Parity of Integer equals Parity of Positive Power
Let $p \in \Z$ be an integer. Let $n \in \Z_{>0}$ be a strictly positive integer. Then $p$ is even  $p^n$ is even. That is, the parity of an integer equals the parity of all its (strictly) positive powers.
Two divides Power Plus One iff Odd
Let $q, n \in \Z_{>0}$. Then: :$2 \divides \paren {q^n + 1}$  $q$ is odd. In the above, $\divides$ denotes divisibility.
Primes of form Power plus One
Let $q, n \in \Z_{>0}$ such that $q > 1$. Then $q^n + 1$ is prime only if: :$(1): \quad q$ is even and :$(2): \quad n$ is of the form $2^m$ for some positive integer $m$.
Fermat Number whose Index is Sum of Integers
Let $F_n = 2^{\left({2^n}\right)} + 1$ be the $n$th Fermat number. Let $k \in \Z_{>0}$. Then: :$F_{n + k} - 1 = \left({F_n - 1}\right)^{2^k}$
Goldbach's Theorem
Let $F_m$ and $F_n$ be Fermat numbers such that $m \ne n$. Then $F_m$ and $F_n$ are coprime.
Number of Boolean Interpretations for Finite Set of Variables
Let $\mathcal P_0$ be the vocabulary of language of propositional logic. Let $S \subseteq \mathcal P_0$ be a finite set of $n$ letters from $\mathcal P_0$. Then there are $2^n$ different partial boolean interpretations for $S$.
Rational Numbers under Addition form Monoid
The set of rational numbers under addition $\struct {\Q, +}$ forms a monoid.
Real Numbers form Field
The set of real numbers $\R$ forms a field under addition and multiplication: $\struct {\R, +, \times}$.
Rational Numbers form Field
Consider the algebraic structure $\struct {\Q, +, \times}$, where: :$\Q$ is the set of all rational numbers :$+$ is the operation of rational addition :$\times$ is the operation of rational multiplication. Then $\struct {\Q, +, \times}$ forms a field.
Integers Modulo m under Addition form Abelian Group
Let $\Z_m$ is the set of integers modulo $m$ Let $+_m$ be the operation of addition modulo $m$. Then the structure $\struct {\Z_m, +_m}$ is an abelian group.
Rational Numbers under Multiplication form Monoid
The set of rational numbers under multiplication $\struct {\Q, \times}$ forms a monoid.
Real Numbers under Multiplication form Monoid
The set of real numbers under multiplication $\struct {\R, \times}$ forms a monoid.
Complex Numbers under Multiplication form Monoid
The set of complex numbers under multiplication $\struct {\C, \times}$ forms a monoid.
Non-Zero Natural Numbers under Addition do not form Monoid
Let $\N_{>0}$ be the set of natural numbers without zero, i.e. $\N_{>0} = \N \setminus \left\{{0}\right\}$. The structure $\left({\N_{>0}, +}\right)$ does ''not'' form a monoid.
Natural Numbers Bounded Below under Addition form Commutative Semigroup
Let $m \in \N$ where $\N$ is the set of natural numbers. Let $M \subseteq \N$ be defined as: :$M := \set {x \in \N: x \ge m}$ That is, $M$ is the set of all natural numbers greater than or equal to $m$. Then the algebraic structure $\struct {M, +}$ is a commutative semigroup.
Positive Real Numbers under Max Operation form Monoid
Let $\R_{\ge 0}$ be the set of positive (that is, non-negative) real numbers. Let $\max: \R_{\ge 0}^2 \to \R_{\ge 0}$ be the max operation on $\R_{\ge 0}$. Then $\left({\R_{\ge 0}, \max}\right)$ is a monoid whose identity is $0$.
Free Commutative Monoid on One Element is Isomorphic to Positive Integers under Addition
Let $X = \left\{{x}\right\}$ be a singleton. Let $M$ be the free commutative monoid on $X$. Then $M$ is isomorphic to the additive monoid of natural numbers.
Finite Monoid with Right Cancellable Operation is Group
Let $\struct {S, \circ}$ be a finite monoid. Let $\circ$ be a right cancellable operation. Then $\struct {S, \circ}$ is a group.
Multiplicative Group of Field is Abelian Group
Let $\struct {F, +, \times}$ be a field. Let $F^* := F \setminus \set 0$ be the set $F$ less its zero. The algebraic structure $\struct {F^*, \times}$ is an abelian group.
One and Minus One form Subgroup of Multiplicative Group of Rational Numbers
Let $\left({\Q_{\ne 0}, \times}\right)$ be the multiplicative group of rational numbers. Let $S \subseteq \Q$ where $S = \left\{{1, -1}\right\}$. Then $\left({S, \times}\right)$ is a subgroup of $\left({\Q_{\ne 0}, \times}\right)$.
Symmetric Group is not Abelian
Let $S_n$ be the symmetric group of order $n$ where $n \ge 3$. Then $S_n$ is not abelian.
Möbius Function is Multiplicative/Corollary
Let $\gcd \set {m, n} > 1$. Then: :$\map \mu {m n} = 0$
Identity of Affine Group of One Dimension
The $1$-dimensional affine group on $\R$ $\operatorname{Af}_1 \left({\R}\right)$ has $f_{1, 0}$ as an identity element.
Inverse in Affine Group of One Dimension
Let $\map {\operatorname {Af}_1} \R$ denote the $1$-dimensional affine group on $\R$. Let $f_{a b} \in \map {\operatorname {Af}_1} \R$. Let $c = \dfrac 1 a$ and $d = \dfrac {-b} a$. Then $f_{c d} \in \map {\operatorname {Af}_1} \R$ is the inverse of $f_{a b}$.
Commutation with Group Elements implies Commuation with Product with Inverse
Let $G$ be a group. Let $a, b, c \in G$ such that $a$ commutes with both $b$ and $c$. Then $a$ commutes with $b c^{-1}$.
Centralizer of Subset is Intersection of Centralizers of Elements
Let $\struct {G, \circ}$ be a group. Let $S \subseteq G$. Let $\map {C_G} S$ be the centralizer of $S$ in $G$. Then: :$\ds \map {C_G} S = \bigcap_{x \mathop \in S} \map {C_G} x$ where $\map {C_G} z$ is the centralizer of $x$ in $G$.
Group Action of Symmetric Group
Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$. Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$. The mapping $*: S_n \times \N_n \to \N_n$ defined as: :$\forall \pi \in S_n, \forall n \in \N_n: \pi * n = \map \pi n$ is a group action.
Trivial Group Action is Group Action
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $S$ be a set. Let $*: G \times S \to S$ be the trivial group action: :$\forall \tuple {g, s} \in G \times S: g * s = s$ Then $*$ is indeed a group action.
Right Regular Representation by Inverse is Group Action
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $*: G \times G \to G$ be the operation: :$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$ where $\rho_g$ is the right regular representation of $G$ with respect to $g$. Then $*$ is a group action.
Left Regular Representation is Group Action
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $*: G \times G \to G$ be the operation: :$\forall g, h \in G: g * h = \map {\lambda_g} h$ where $\lambda_g$ is the left regular representation of $G$ with respect to $g$. Then $*$ is a group action.
Right Regular Representation by Inverse is Transitive Group Action
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $*: G \times G \to G$ be the group action: :$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$ where $\rho_g$ is the right regular representation of $G$ with respect to $g$. Then $*$ is a transitive group action.
Conjugacy Action is not Transitive
Let $\struct {G, \circ}$ be a non-trivial group whose identity is $e$. Let $*: G \times G \to G$ be the conjugacy group action: : $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$ Then $*$ is ''not'' a transitive group action.
Group Action on Subgroup by Right Regular Representation
Let $G$ be a group. Let $H$ be a subgroup of $G$. Let $*: H \times G \to G$ be the operation defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$ where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$. Then $*$ is a group action.
Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset
Let $G$ be a group. Let $H$ be a proper subgroup of $G$. Let $*: H \times G \to G$ be the group action defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$ where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$. Let $x \in G$. Then the orbit of $x$ under $*$ is given by: :$\forall x \in G: \Orb x = H x$ where $H x$ is the right coset of $H$ by $x$.
Left Coset by Identity
Then: : $e H = H$ where $e H$ is the left coset of $H$ by $e$.
Right Coset by Identity
Then: : $H = H e$ where $H e$ is the right coset of $H$ by $e$.
Inversion Mapping is Permutation
Let $\struct {G, \circ}$ be a group. Let $\iota: G \to G$ be the inversion mapping on $G$. Then $\iota$ is a permutation on $G$.
Inversion Mapping is Mapping
Let $\left({G, \circ}\right)$ be a group. Let $\iota: G \to G$ be the inversion mapping on $G$. Then $\iota$ is indeed a mapping.
Cartesian Product of Group Actions
Let $\struct {G, \circ}$ be a group. Let $S$ and $T$ be sets. Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be group actions. Then the operation $*: G \times \paren {S \times T} \to S \times T$ defined as: :$\forall \tuple {g, \tuple {s, t} } \in G \times \paren {S \times T}: g * \tuple {s, t} = \tuple {g *_S s, g *_T t}$ is a group action.
Index in Subgroup
Let $G$ be a group. Let $H, K$ be subgroups of finite index of $G$. Then: :$\index H {H \cap K} \le \index G K$ where $\index G K$ denotes the index of $K$ in $G$. Equality happens  $G = H K$.
Finite Cyclic Group has Euler Phi Generators
Let $C_n$ be a (finite) cyclic group of order $n$. Then $C_n$ has $\map \phi n$ generators, where $\map \phi n$ denotes the Euler $\phi$ function.
Extendability Theorem for Derivatives Continuous on Open Intervals
Let $f$ be a continuous real function defined on an interval $\closedint a b$ where $a < b$. Then $f$ is continuously differentiable on $\closedint a b$ : :$f$ is continuously differentiable on $\openint a b$ and: :$\displaystyle \lim_{x \mathop \to a^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to b^-} \map {f'} x$ exist.
Universal Affirmative and Universal Negative are Contrary iff First Predicate is not Vacuous
Consider the categorical statements: :$\mathbf A: \quad$ The universal affirmative: $\forall x: \map S x \implies \map P x$ :$\mathbf E: \quad$ The universal negative: $\forall x: \map S x \implies \neg \map P x$ Then: :$\mathbf A$ and $\mathbf E$ are contrary : :$\exists x: \map S x$ Using the symbology of predicate logic: :$\exists x: \map S x \iff \neg \paren {\paren {\forall x: \map S x \implies \map P x} \land \paren {\forall x: \map S x \implies \neg \map P x} }$
Particular Affirmative and Particular Negative are Subcontrary iff First Predicate is not Vacuous
Consider the categorical statements: :$\mathbf I: \quad$ The particular affirmative: $\exists x: \map S x \land \map P x$ :$\mathbf O: \quad$ The particular negative: $\exists x: \map S x \land \neg \map P x$ Then: :$\mathbf I$ and $\mathbf O$ are subcontrary : :$\exists x: \map S x$ Using the symbology of predicate logic: :$\exists x: \map S x \iff \paren {\paren {\exists x: \map S x \land \map P x} \lor \paren {\exists x: \map S x \land \neg \map P x} }$
Universal Affirmative and Particular Negative are Contradictory
Consider the categorical statements: :$\mathbf A: \quad$ The universal affirmative: $\forall x: \map S x \implies \map P x$ :$\mathbf O: \quad$ The particular negative: $\exists x: \map S x \land \neg \map P x$ Then $\mathbf A$ and $\mathbf O$ are contradictory. Using the symbology of predicate logic: :$\neg \paren {\paren {\forall x: \map S x \implies \map P x} \iff \paren {\exists x: \map S x \land \neg \map P x} }$
Particular Affirmative and Universal Negative are Contradictory
Consider the categorical statements: :$\mathbf I: \quad$ The particular affirmative: $\exists x: \map S x \land \map P x$ :$\mathbf E: \quad$ The universal negative: $\forall x: \map S x \implies \neg \map P x$ Then $\mathbf I$ and $\mathbf E$ are contradictory. Using the symbology of predicate logic: :$\neg \paren {\paren {\exists x: \map S x \land \map P x} \iff \paren {\forall x: \map S x \implies \neg \map P x} }$
Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous
Consider the categorical statements: :$\map {\mathbf A} {S, P}: \quad$ The universal affirmative: $\forall x: \map S x \implies \map P x$ :$\map {\mathbf I} {S, P}: \quad$ The particular affirmative: $\exists x: \map S x \land \map P x$ Then: :$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$ : :$\exists x: \map S x$ Using the symbology of predicate logic: :$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map S x \land \map P x} }$
Socrates is Mortal
:$(1): \quad$ ''All humans are mortal.'' :$(2): \quad$ ''{{AuthorRef|Socrates}} is human.'' :$(3): \quad$ ''Therefore {{AuthorRef|Socrates}} is mortal.''
Right-Hand Differentiable Function is Right-Continuous
Let $f$ be a real function defined on an interval $I$. Let $a$ be a point in $I$ where $f$ is right-hand differentiable. Then $f$ is right-continuous at $a$.
Left-Hand Differentiable Function is Left-Continuous
Let $f$ be a real function defined on an interval $I$. Let $a$ be a point in $I$ where $f$ is left-hand differentiable. Then $f$ is left-continuous at $a$.
Left-Hand and Right-Hand Differentiable Function is Continuous
Let $f$ be a real function defined on an interval $I$. Let $a$ be a point in $I$ where $f$ is left- and right-hand differentiable. Then $f$ is continuous at $a$.
Universal Affirmative and Negative are both False iff Particular Affirmative and Negative are both True
Consider the categorical statements: {{begin-axiom}} {{axiom | q = \mathbf A \left({S, P}\right)         | lc= The universal affirmative:         | ml= \forall x: S \left({x}\right)         | mo= \implies         | mr= P \left({x}\right) }} {{axiom | q = \mathbf E \left({S, P}\right)         | lc= The universal negative:         | ml= \forall x: S \left({x}\right)         | mo= \implies         | mr= \neg P \left({x}\right) }} {{axiom | q = \mathbf I \left({S, P}\right)         | lc= The particular affirmative:         | ml= \exists x: S \left({x}\right)         | mo= \land          | mr= P \left({x}\right) }} {{axiom | q = \mathbf O \left({S, P}\right)         | lc= The particular negative:         | ml= \exists x: S \left({x}\right)         | mo= \land          | mr= \neg P \left({x}\right) }} {{end-axiom}} Then: :$\mathbf A \left({S, P}\right)$ and $\mathbf E \left({S, P}\right)$ are both false : :$\mathbf I \left({S, P}\right)$ and $\mathbf O \left({S, P}\right)$ are both true.
Law of Simple Conversion of I
Consider the particular affirmative categorical statement ''Some $S$ is $P$'': :$\map {\mathbf I} {S, P}: \exists x: \map S x \land  \map P x$ Then ''Some $P$ is $S$'': :$\map {\mathbf I} {P, S}$
Law of Simple Conversion of E
Consider the universal negative categorical statement ''No $S$ is $P$'': :$\mathbf E \left({S, P}\right): \forall x: S \left({x}\right) \implies \neg P \left({x}\right)$ Then ''No $P$ is $S$'': :$\mathbf E \left({P, S}\right)$
Conversion per Accidens
Consider the categorical statements: :$\map {\mathbf A} {S, P}: \quad$ The universal affirmative: $\forall x: \map S x \implies \map P x$ :$\map {\mathbf I} {P, S}: \quad$ The particular affirmative: $\exists x: \map P x \land \map S x$ Then: :$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {P, S}$ : :$\exists x: \map S x$ Using the symbology of predicate logic: :$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map P x \land \map S x} }$ This law has the traditional name '''conversion per accidens of $\mathbf A$'''. Thus the $\mathbf A$ form '''converts per accidens''' to the $\mathbf I$ form.
Number of Standard Instances of Categorical Syllogism
There are $256$ distinct standard instances of the categorical syllogism.
No Valid Categorical Syllogism contains two Negative Premises
No categorical syllogism of which both premises are negative categorical statements is valid.
Valid Patterns of Categorical Syllogism
The following categorical syllogisms are valid: :$\begin{array}{rl} \text{I} & AAA \\ \text{I} & AII \\ \text{I} & EAE \\ \text{I} & EIO \\ * \text{I} & AAI \\ * \text{I} & EAO \\ \end{array} \qquad \begin{array}{rl} \text{II} & EAE \\ \text{II} & AEE \\ \text{II} & AOO \\ \text{II} & EIO \\ * \text{II} & EAO \\ * \text{II} & AEO \\ \end{array} \qquad \begin{array}{rl} \dagger \text{III} & AAI \\ \text{III} & AII \\ \text{III} & IAI \\ \dagger \text{III} & EAO \\ \text{III} & EIO \\ \text{III} & OAO \\ \end{array} \qquad \begin{array}{rl} \S \text{IV} & AAI \\ \text{IV} & AEE \\ \dagger \text{IV} & EAO \\ \text{IV} & EIO \\ \text{IV} & IAI \\ * \text{IV} & AEO \\ \end{array}$ In the above: :$\text{I}, \text{II}, \text{III}, \text{IV}$ denote the four figures of the categorical syllogisms :$A, E, I, O$ denote the universal affirmative, universal negative, particular affirmative and particular negative respectively: see Shorthand for Categorical Syllogism :Syllogisms marked $*$ require the assumption that $\exists x: \map S x$, that is, that there exists an object fulfilling the secondary predicate :Syllogisms marked $\dagger$ require the assumption that $\exists x: \map M x$, that is, that there exists an object fulfilling the middle predicate :Syllogisms marked $\S$ require the assumption that $\exists x: \map P x$, that is, that there exists an object fulfilling the primary predicate
Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise
Let $T$ be a term of the conclusion $C$ of a valid categorical syllogism $Q$. Let $T$ be distributed in $C$. Then $T$ is also distributed in whichever premise of $Q$ in which it appears.
Middle Term of Valid Categorical Syllogism is Distributed at least Once
Let $M$ be the middle term of a valid categorical syllogism $Q$. Then $M$ is distributed in at least one of the premises of $Q$ in which it appears.
Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative
The conclusion of a valid categorical syllogism is negative  one of the premises is also negative.
No Valid Categorical Syllogism contains two Particular Premises
Let $Q$ be a valid categorical syllogism. Then at least one of the premises of $Q$ is universal.
No Valid Categorical Syllogism with Particular Premise has Universal Conclusion
Let $Q$ be a valid categorical syllogism. Let one of the premises of $Q$ be particular. Then the conclusion of $Q$ is also particular.
Elimination of all but 48 Categorical Syllogisms as Invalid
Of the $256$ different types of categorical syllogism, all but $48$ can immediately be identified as invalid by consideration of the Rules of Quantity and the Rules of Quality.
Conjunction implies Disjunction
:$\vdash \paren {p \land q} \implies \paren {p \lor q}$
Angle Bisector Vector
Let $\mathbf u$ and $\mathbf v$ be vectors of non-zero length. Let $\left\Vert{\mathbf u}\right\Vert$ and $\left\Vert{\mathbf v}\right\Vert$ be their respective lengths. Then $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ is the angle bisector of $\mathbf u$ and $\mathbf v$.
Injection from Finite Set to Itself is Surjection/Corollary
Let $S$ be a finite set. Let $f: S \to S$ be an injection. Then $f$ is a permutation.
Surjection from Finite Set to Itself is Permutation
Let $S$ be a finite set. Let $f: S \to S$ be an surjection. Then $f$ is a permutation.
Order of Power of Group Element
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $g \in G$ be an element of $G$ such that: :$\order g = n$ where $\order g$ denotes the order of $g$. Then: :$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n} }$ where $\gcd \set {m, n}$ denotes the greatest common divisor of $m$ and $n$.
Existence of Group of Finite Order
Let $n \in \Z_{>0}$. Then there exists at least one group whose order is $n$.
Vitali Theorem
There exists a set of real numbers which is not Lebesgue measurable.
Finite Number of Groups of Given Finite Order
Let $n \in \Z_{>0}$ be a (strictly) positive integer. Then there exists a finite number of types of group of order $n$.
Nu of Prime Number is 1
Let $p$ be a prime number. Then: :$\map \nu p = 1$ where $\nu$ denotes the $\nu$ function: the number of types of group of a given order.
Image of Subset under Relation equals Union of Images of Elements
Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation on $S \times T$. Let $X \subseteq S$ be a subset of $S$. Then: :$\displaystyle \RR \sqbrk X = \bigcup_{x \mathop \in X} \map \RR x$ where: :$\RR \sqbrk X$ is the image of the subset $X$ under $\RR$ :$\map \RR x$ is the image of the element $x$ under $\RR$.
Preimage of Subset under Relation equals Union of Preimages of Elements
Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation on $S \times T$. Let $\RR^{-1} \subseteq T \times S$ be the inverse relation to $\RR$ Let $Y \subseteq T$ be a subset of $T$. Then: :$\RR^{-1} \sqbrk Y = \displaystyle \bigcup_{y \mathop \in Y} \map {\RR^{-1} } y$ where: :$\RR^{-1} \sqbrk Y$ is the preimage of the subset $Y$ under $\RR$ :$\RR^{-1} \sqbrk Y$ is the preimage of the element $y$ under $\RR$.
Preimage of Subset under Mapping equals Union of Preimages of Elements
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping from $S$ to $T$. Let $f^{-1} \subseteq T \times S$ be the inverse of $f$, defined as: :$f^{-1} = \set {\tuple {t, s}: \map f s = t}$ Let $Y \subseteq T$ be a subset of $T$. Then: :$\displaystyle f^{-1} \sqbrk Y = \bigcup_{y \mathop \in Y} \map {f^{-1} } y$ where: :$f^{-1} \sqbrk Y$ is the preimage of the subset $Y$ under $f$ :$\map {f^{-1} } y$ is the preimage of the element $y$ under $f$.
Image of Domain of Mapping is Image Set
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. The image of $S$ is the image set of $f$: :$f \sqbrk S = \Img f$
Image of Singleton under Mapping
Let $f: S \to T$ be a mapping. 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 $S$: :$\forall s \in S: \set {\map f s} = f \sqbrk {\set s}$
Image of Subset under Mapping equals Union of Images of Elements
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping from $S$ to $T$. Let $X \subseteq S$ be a subset of $S$. Then: :$\displaystyle f \sqbrk X = \bigcup_{x \mathop \in X} \map f x$ where: :$f \sqbrk X$ is the image of the subset $X$ under $f$ :$\map f x$ is the image of the element $x$ under $f$.
Mapping Images are Disjoint only if Domains are Disjoint
Let $S$ and $T$ be sets. Let: :$f \sqbrk S \cap f \sqbrk T = \O$ where $f \sqbrk S$ denotes the image set of $S$. Then: :$S \cap T = \O$
Image of Relation is Domain of Inverse Relation
Let $\RR \subseteq S \times T$ be a relation. Let $\RR^{-1} \subseteq T \times S$ be the inverse of $\RR$. Then: :$\Img \RR = \Dom {\RR^{-1} }$ That is, the image of a relation is the domain of its inverse.
Domain of Relation is Image of Inverse Relation
Let $\RR \subseteq S \times T$ be a relation. Let $\RR^{-1} \subseteq T \times S$ be the inverse of $\RR$. Then: :$\Dom \RR = \Img {\RR^{-1} }$ That is, the domain of a relation is the image of its inverse.
Image of Set Difference under Injection
Let $f: S \to T$ be a mapping. Let $S_1$ and $S_2$ be subsets of $S$. Let $S_1 \setminus S_2$ denote the set difference between $S_1$ and $S_2$. Then: :$\forall S_1, S_2 \subseteq S: f \left[{S_1}\right] \setminus f \left[{S_2}\right] = f \left[{S_1 \setminus S_2}\right]$  $f$ is an injection.
De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Union
:$\displaystyle \map \complement {\bigcup_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \map \complement {S_i}$
De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection
:$\displaystyle \map \complement {\bigcap_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \map \complement {S_i}$
Closure of Intersection and Symmetric Difference imply Closure of Set Difference
Let $\mathcal R$ be a system of sets such that for all $A, B \in \mathcal R$: : $(1): \quad A \cap B \in \mathcal R$ : $(2): \quad A * B \in \mathcal R$ where $\cap$ denotes set intersection and $*$ denotes set symmetric difference. Then: :$\forall A, B \in \mathcal R: A \setminus B \in \mathcal R$ where $\setminus$ denotes set difference.