hcju/proofwikips huggingface dataset:
De Morgan's Laws (Set Theory)/Set Complement/Complement of Union
:$\overline {T_1 \cup T_2} = \overline T_1 \cap \overline T_2$
De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection/Proof
Let $\mathbb T = \set {T_i: i \mathop \in I}$, where each $T_i$ is a set and $I$ is some finite indexing set. Then: :$\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection
:$\displaystyle S \setminus \bigcap \mathbb T = \bigcup_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$ where: :$\displaystyle \bigcap \mathbb T := \set {x: \forall T' \in \mathbb T: x \in T'}$ that is, the intersection of $\mathbb T$
De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union
:$\displaystyle S \setminus \bigcup \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$ where: :$\displaystyle \bigcup \mathbb T := \set {x: \exists T' \in \mathbb T: x \in T'}$ that is, the union of $\mathbb T$.
De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection
:$S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$
De Morgan's Laws (Set Theory)/Set Difference/Difference with Union
:$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$
Isomorphism between Ring of Integers Modulo 2 and Parity Ring
The ring of integers modulo $2$ and the parity ring are isomorphic.
Parity Addition is Associative
Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring. The operation $+$ is associative: :$\forall a, b, c \in R: \paren {a + b} + c = a + \paren {b + c}$
Parity Multiplication is Commutative
Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring. The operation $\times$ is commutative: :$\forall a, b \in R: a \times b = b \times a$
Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition
Let $n \in \Z_{>0}$ be a (strictly) positive integer. Let $\struct {R_n, \times}$ be the complex $n$th roots of unity under complex multiplication. Let $\struct {\Z_n, +_n}$ be the integers modulo $n$ under modulo addition. Then $\struct {R_n, \times}$ and $\struct {\Z_n, +_n}$ are isomorphic algebraic structures.
Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition
Let $\struct {U_\C, \times}$ be the group of Gaussian integer units under complex multiplication. Let $\struct {\Z_n, +_4}$ be the integers modulo $4$ under modulo addition. Then $\struct {U_\C, \times}$ and $\struct {\Z_4, +_4}$ are isomorphic algebraic structures.
Gaussian Integer Units are 4th Roots of Unity
The units of the ring of Gaussian integers: :$\set {1, i, -1, -i}$ are the (complex) $4$th roots of $1$.
Isomorphism by Cayley Table
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures whose underlying sets are both finite. Then $\struct {S, \circ}$ and $\struct {T, *}$ are isomorphic : :a bijection $f: S \to T$ can be found such that: ::the Cayley table of $\struct {T, *}$ can be generated from the Cayley table of $\struct {S, \circ}$ by replacing each entry of $S$ with its image under $f$.
Bijection between Integers and Even Integers
Let $\Z$ be the set of integers. Let $2 \Z$ be the set of even integers. Then there exists a bijection $f: \Z \to 2 \Z$ between the two.
Set of Integers under Addition is Isomorphic to Set of Even Integers under Addition
Let $\struct {\Z, +}$ be the algebraic structure formed by the set of integers under the operation of addition. Let $\struct {2 \Z, +}$ be the algebraic structure formed by the set of even integers under the operation of addition. Then $\struct {\Z, +}$ and $\struct {2 \Z, +}$ are isomorphic.
Natural Numbers under Addition do not form Group
The algebraic structure $\struct {\N, +}$ consisting of the set of natural numbers $\N$ under addition $+$ is not a group.
Natural Numbers under Multiplication do not form Group
The algebraic structure $\struct {\N, \times}$ consisting of the set of natural numbers $\N$ under multiplication $\times$ is not a group.
Symmetry Group of Square is Group
The symmetry group of the square is a non-abelian group.
Dihedral Group is Group
Let $D_n$ be the dihedral group of order $2 n$. Then $D_n$ is indeed a group.
Symmetric Group is Subgroup of Monoid of Self-Maps
Let $S$ be a set. Let $S^S$ be the set of all mappings from $S$ to itself Let $\struct {\Gamma \paren S, \circ}$ denote the symmetric group on $S$. Let $\struct {S^S, \circ}$ be the monoid of self-maps under composition of mappings. Then $\struct {\Gamma \paren S, \circ}$ is a subgroup of $\struct {S^S, \circ}$.
Cancellability by Cayley Table
Let $\left({S, \circ}\right)$ be a finite algebraic structure. Let $\mathcal T$ be the Cayley table for $\left({S, \circ}\right)$. Let $a \in S$ be an element of $S$. Then $a$ is cancellable for $\circ$ : :$(1): \quad$ no element of $S$ is repeated in $\mathcal T$ in the row headed by $a$ and: :$(2): \quad$ no element of $S$ is repeated in $\mathcal T$ in the column headed by $a$.
Piecewise Continuous Function does not necessarily have Improper Integrals
Let $f$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$, $a < b$.  Let $f$ be a piecewise continuous function: {{:Definition:Piecewise Continuous Function}} Then it is not necessarily the case that $f$ is a piecewise continuous function with improper integrals: {{:Definition:Piecewise Continuous Function with Improper Integrals}}
Right Cancellable Element is Right 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 right cancellable in $S$. Then $x$ is also right cancellable in $T$.
Left Cancellable Element is Left Cancellable in Subset
Let $\struct {S, \circ}$ be an algebraic structure. Let $\struct {T, \circ} \subseteq \struct {S, \circ}$. Let $x \in T$ be left cancellable in $S$. Then $x$ is also left cancellable in $T$.
Left Cancellable Elements of Semigroup form Subsemigroup
Let $\struct {S, \circ}$ be a semigroup. Let $C_\lambda$ be the set of left cancellable elements of $\struct {S, \circ}$. Then $\struct {C_\lambda, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Right Cancellable Elements of Semigroup form Subsemigroup
Let $\struct {S, \circ}$ be a semigroup. Let $C_\rho$ be the set of right cancellable elements of $\struct {S, \circ}$. Then $\struct {C_\rho, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Surjection iff Right Cancellable/Necessary Condition
Let $f$ be a surjection. Then $f$ is right cancellable.
Surjection iff Right Cancellable/Sufficient Condition
Let $f$ be a mapping which is right cancellable. Then $f$ is a surjection.
Absolute Value Function on Integers induces Equivalence Relation
Let $\Z$ be the set of integers. Let $\mathcal R$ be the relation on $\Z$ defined as: :$\forall x, y \in \Z: \left({x, y}\right) \in \mathcal R \iff \left\vert{x}\right\vert = \left\vert{y}\right\vert$ where $\left\vert{x}\right\vert$ denotes the absolute value of $x$. Then $\mathcal R$ is an equivalence relation.
Absolute Value induces Equivalence not Compatible with Integer Addition
Let $\Z$ be the set of integers. Let $\RR$ be the relation on $\Z$ defined as: :$\forall x, y \in \Z: \tuple {x, y} \in \RR \iff \size x = \size y$ where $\size x$ denotes the absolute value of $x$. Then $\RR$ is not a congruence relation for integer addition.
Equivalence Relation is Congruence for Right Operation
Every equivalence is a congruence for the right operation.
Equivalence Relation is Congruence for Left Operation
Every equivalence is a congruence for the left operation.
Left Coset Space forms Partition
The left coset space of $H$ forms a partition of its group $G$, and hence: {{begin-eqn}} {{eqn | l = x \equiv^l y \pmod H       | o = \iff       | r = x H = y H }} {{eqn | l = \neg \paren {x \equiv^l y} \pmod H       | o = \iff       | r = x H \cap y H = \O }} {{end-eqn}}
Right Coset Space forms Partition
The right coset space of $H$ forms a partition of its group $G$: {{begin-eqn}} {{eqn | l = x \equiv^r y \pmod H       | o = \iff       | r = H x = H y }} {{eqn | l = \neg \paren {x \equiv^r y} \pmod H       | o = \iff       | r = H x \cap H y = \O }} {{end-eqn}}
Right Cosets are Equal iff Product with Inverse in Subgroup
Let $H x$ denote the right coset of $H$ by $x$. Then: :$H x = H y \iff x y^{-1} \in H$
Left Cosets are Equal iff Product with Inverse in Subgroup
Let $x H$ denote the left coset of $H$ by $x$. Then: :$x H = y H \iff x^{-1} y \in H$
Left Congruence Class Modulo Subgroup is Left Coset
Let $\mathcal R^l_H$ be the equivalence defined as left congruence modulo $H$. The equivalence class $\eqclass g {\mathcal R^l_H}$ of an element $g \in G$ is the left coset $g H$. This is known as the '''left congruence class of $g \bmod H$'''.
Left Congruence Modulo Subgroup is Equivalence Relation
Let $x \equiv^l y \pmod H$ denote the relation that $x$ is left congruent modulo $H$ to $y$. Then the relation $\equiv^l$ is an equivalence relation.
Right Congruence Modulo Subgroup is Equivalence Relation
Let $x \equiv^r y \pmod H$ denote the relation that $x$ is right congruent modulo $H$ to $y$ Then the relation $\equiv^r$ is an equivalence relation.
Group Epimorphism is Isomorphism iff Kernel is Trivial
Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be groups. Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a group epimorphism. Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively. Let $K = \map \ker \phi$ be the kernel of $\phi$. Then: :the epimorphism $\phi$ is an isomorphism  :$K = \set {e_G}$
Quotient Group of Quotient Group is Isomorphic to Quotient Group by Preimage under Quotient Mapping
Let $G$ be a group. Let $H \lhd G$ where $\lhd$ denotes that $H$ is a normal subgroup of $G$. Let $K \lhd G/H$ and $L = q_H^{-1} \left[{K}\right]$, where: :$q_H: G \to G/H$ is the quotient epimorphism from $G$ to the quotient group $G/H$ :$q_H^{-1} \left[{K}\right]$ is the preimage of $K$ under $q_H$. Then there exists a group isomorphism $\phi: \left({G / H}\right) / K \to G / L$ defined as: ::$\phi \circ q_K \circ q_H = q_L$
Real Numbers are not Well-Ordered under Conventional Ordering
Let $\left({\R, \leqslant}\right)$ be the ordered structure consisting of the real numbers under the conventional ordering. Then $\left({\R, \leqslant}\right)$ is not a well-ordered set.
Supremum is Unique
Let $\struct {S, \preceq}$ be an ordered set. Let $T$ be a non-empty subset of $S$. Then $T$ has at most one supremum in $S$.
Infimum is Unique
Let $\struct {S, \preceq}$ be an ordered set. Let $T$ be a non-empty subset of $S$. Then $T$ has at most one infimum in $S$.
Infimum of Subgroups in Lattice
Then: :$\inf \set {H, K} = H \cap K$
Supremum of Subgroups in Lattice
Let either $H$ or $K$ be normal in $G$. Then: :$\sup \left\{{H, K}\right\} = H \circ K$ where $H \circ K$ denotes subset product.
Lexicographic Order on Totally Ordered Sets is Total Ordering
Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be ordered sets. Let $\preccurlyeq$ be the lexicographic order on $S_1 \times S_2$''': :$\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2} \iff \paren {x_1 \prec_1 y_1} \lor \paren {x_1 = y_1 \land x_2 \preceq_2 y_2}$ Then: :$\preccurlyeq$ is a total ordering on $S_1 \times S_2$  :both $\preceq_1$ and $\preceq_2$ are total orderings.
Lexicographic Order is Ordering
Let $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ be ordered sets. Let $\preccurlyeq$ be the lexicographic order on $S_1 \times S_2$''': :$\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right) \iff \left({x_1 \prec_1 y_1}\right) \lor \left({x_1 = y_1 \land x_2 \preceq_2 y_2}\right)$ Then $\preccurlyeq$ is an ordering on $S_1 \times S_2$.
Powers of Commuting Elements of Monoid Commute
:$\forall m, n \in \N: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
Power of Product of Commuting Elements in Monoid equals Product of Powers
:$\forall n \in \N: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
Index Laws/Product of Indices/Monoid
Let $\struct {S, \circ}$ be a monoid whose identity element is $e$. For $a \in S$, let $\circ^n a = a^n$ be the $n$th power of $a$. Then: :$\forall m, n \in \N: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$
Cardinality of Proper Subset of Finite Set
Let $A$ and $B$ be finite sets such that $A \subsetneqq B$. Let $\card B = n$, where $\card {\, \cdot \,}$ denotes cardinality. Then $\card A < n$.
First-Countable Space is Hausdorff iff All Convergent Sequences have Unique Limit
Let $T = \struct {S, \tau}$ be a first-countable topological space. Then $T$ is Hausdorff  all convergent sequences on $T$ have a unique limit point.
Topologically Distinguishable Points are Distinct
Let $T = \left({X, \tau}\right)$ be a topological space. Let $x, y \in X$ be topologically distinguishable. Then the singleton sets $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are disjoint and so: : $x \ne y$
Restriction of Strict Total Ordering is Strict Total Ordering
Let $\left({S, \prec}\right)$ be a strict total ordering. Let $T \subseteq S$. Let $\prec \restriction_T$ be the restriction of $\prec$ to $T$. Then $\prec \restriction_T$ is a strict total ordering of $T$.
Principle of Recursive Definition
Let $\N$ be the natural numbers. Let $T$ be a set. Let $a \in T$. Let $g: T \to T$ be a mapping. Then there exists exactly one mapping $f: \N \to T$ such that: :$\forall x \in \N: \map f x = \begin{cases} a & : x = 0 \\ \map g {\map f n} & : x = n + 1 \end{cases}$
Principle of Recursive Definition/General Result
Let $p \in \N$. Let $p^\ge$ be the upper closure of $p$ in $\N$: :$p^\ge := \set {x \in \N: x \ge p} = \set {p, p + 1, p + 2, \ldots}$ Then there exists exactly one mapping $f: p^\ge \to T$ such that: :$\forall x \in p^\ge: \map f x = \begin{cases} a & : x = p \\ \map g {\map f n} & : x = n + 1 \end{cases}$
Real Zero is Zero Element
:$\forall x \in \R: 0 \times x = 0$
Negative of Real Zero equals Zero
Let $0$ denote the identity for addition in the real numbers $\R$. Then: :$-0 = 0$
Negative of Negative Real Number
:$\forall x \in \R: -\paren {-x} = x$
Multiplication by Negative Real Number
:$\forall x, y \in \R: x \times \paren {-y} = -\paren {x \times y} = \paren {-x} \times y$
Negative of Sum of Real Numbers
:$\forall x, y \in \R: -\paren {x + y} = -x - y$
Real Multiplication Identity is One/Corollary
:$\forall x \in \R_{\ne 0}: x \times y = x \implies y = 1$
Real Number Divided by Itself
:$\forall x \in \R_{\ne 0}: \dfrac x x = 1$
Product of Reciprocals of Real Numbers
:$\forall x, y \in \R_{\ne 0}: \dfrac 1 x \times \dfrac 1 y = \dfrac 1 {x \times y}$
Product of Quotients of Real Numbers
:$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y \times \dfrac w z = \dfrac {x \times w} {y \times z}$
Sum of Quotients of Real Numbers
:$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y + \dfrac w z = \dfrac {\paren {x \times z} + \paren {y \times w} } {y \times z}$
Reciprocal of Quotient of Real Numbers
:$\forall x, y \in \R_{\ne 0}: \dfrac 1 {x / y} = \dfrac y x$
Quotient of Quotients of Real Numbers
:$\forall x \in \R, y, w, z \in \R_{\ne 0}: \dfrac {x / y} {w / z} = \dfrac {x \times z} {y \times w}$
Product of Real Number with Quotient
:$\forall a, x \in \R, y \in \R_{\ne 0}: \dfrac {a \times x} y = a \times \dfrac x y$
Neighborhood in Topological Subspace
Let $\struct {X, \tau}$ be a topological space. Let $S \subseteq X$ be a subset of $X$. Let $\tau_S$ denote the subspace topology on $S$. Let $x \in S$ be an arbitrary point of $S$. Let $E \subseteq S$. Then: :$E$ is a neighborhood of $x$ in $\struct {S, \tau_S}$ : :$\exists D \subseteq X$ such that: ::$D$ is a neighborhood of $x$ in $X$ ::$E = D \cap S$.
Negative of Quotient of Real Numbers
:$\forall x \in \R, y \in \R_{\ne 0}: \dfrac {-x} y = -\dfrac x y = \dfrac x {-y}$
Product of Strictly Positive Real Numbers is Strictly Positive
:$x, y \in \R_{>0} \implies x \times y \in \R_{>0}$
Real Number is Greater than Zero iff its Negative is Less than Zero
:$\forall x \in \R: x > 0 \iff \paren {-x} < 0$
Order of Real Numbers is Dual of Order of their Negatives
:$\forall x, y \in \R: x > y \iff \paren {-x}< \paren {-y}$
Square of Non-Zero Real Number is Strictly Positive
:$\forall x \in \R: x \ne 0 \implies x^2 > 0$
Real Zero is Less than Real One
The real number $0$ is less than the real number $1$: :$0 < 1$
Product of Real Numbers is Positive iff Numbers have Same Sign
The product of two real numbers is greater than $0$  either both are greater than $0$ or both are less than $0$. :$\forall x, y \in \R: x \times y > 0 \iff \paren {x, y \in \R_{>0} } \lor \paren {x, y \in \R_{<0} }$
Reciprocal of Strictly Positive Real Number is Strictly Positive
:$\forall x \in \R: x > 0 \implies \dfrac 1 x > 0$
Order of Strictly Positive Real Numbers is Dual of Order of their Reciprocals
:$\forall x, y \in \R: x > y > 0 \implies \dfrac 1 x < \dfrac 1 y$
Mean of Unequal Real Numbers is Between them
:$\forall x, y \in \R: x < y \implies x < \dfrac {x + y} 2 < y$
Intersection of Inductive Set as Subset of Real Numbers is Inductive Set
Let $\AA$ be a set of inductive sets defined as subsets of real numbers. Then their intersection is an inductive set.
Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element
Let $n \in \N_{>0}$ be a non-zero natural number. Let $\N^*_n$ denote the Initial segment $\set {1, 2, \ldots, n}$ of the non-zero natural numbers. Then every non-empty subset of $\N^*_n$ has a greatest element.
Not every Non-Empty Subset of Natural Numbers has Greatest Element
Let $S \subseteq \N_{>0}$. Then, despite Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element, it is not necessarily the case that $S$ has a greatest element.
Existence of Integral on Union of Adjacent Intervals
Let $f$ be a real function defined on a closed interval $\closedint a b$ where $a < b$. Let $c$ be a point in $\openint a b$. Then: :$f$ is Darboux integrable on $\closedint a c$ and $\closedint c b$ : :$f$ is Darboux integrable on $\closedint a b$.
Cartesian Product of Bijections is Bijection
Let $S_1 \times S_2$ be the Cartesian product of two sets $S_1$ and $S_2$. Let $T_1 \times T_2$ be the Cartesian product of two sets $T_1$ and $T_2$. Let $f_1: S_1 \to T_1$ and $f_2: S_2 \to T_2$ be bijections. Let $f_1 \times f_2: S_1 \times S_2 \to T_1 \times T_2$ be the Cartesian product of $f_1$ and $f_2$ defined as: :$\forall \left({s_1, s_2}\right) \in S_1 \times S_2: f_1 \times f_2 \left({s_1, s_2}\right) := \left({f_1 \left({s_1}\right), f_2 \left({s_2}\right)}\right)$ Then $f_1 \times f_2$ is a bijection.
First Order ODE/y' - f (y) phi' (x) over f' (y) = phi (x) phi' (x) over f' (y)
Let $\map f y$ and $\map \phi x$ be known real functions of $y$ and $x$ respectively. The general solution of: :$(1): \quad \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x = \dfrac {\map \phi x \, \map {\phi'} x} {\map {f'} y}$ is: :$\map f y = C e^{\map \phi x} - \map \phi x - 1$
Cardinality of Set of All Mappings/Finite Sets
Let $S$ and $T$ be finite sets. The cardinality of the set of all mappings from $S$ to $T$ (that is, the total number of mappings from $S$ to $T$) is: :$\card {T^S} = \card T^{\card S}$
Cardinality of Set of All Mappings/Infinite Sets
Let $S$ and $T$ be sets such that either $S$ or $T$ is infinite. The cardinality of the set of all mappings from $S$ to $T$ (that is, the total number of mappings from $S$ to $T$) is: :$\card {T^S} = \card T^{\card S}$
Cardinality of Set Union/General Case
Let $S_1, S_2, \ldots$ be sets. Then: {{begin-eqn}}      {{eqn | l = \card {\bigcup_{i \mathop = 1}^n S_i}       | r = \sum_{i \mathop = 1}^n \card {S_i}       | c = }} {{eqn | o = -       | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \card {S_i \cap S_j}       | c = }} {{eqn | o = +       | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \card {S_i \cap S_j \cap S_k}       | c = }} {{eqn | o = \cdots       | c = }} {{eqn | o = +       | r = \paren {-1}^{n - 1} \card {\bigcap_{i \mathop = 1}^n S_i}       | c = }} {{end-eqn}}
Structure of Inverse Completion of 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: :$T = S \circ' C^{-1}$ where: :$C^{-1}$ is the inverse of $C$ in $T$ :$S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$.
Inverse in Monoid is Unique
Let $\struct {S, \circ}$ be a monoid. Then an element $x \in S$ can have at most one inverse for $\circ$.
Inverse of Inverse/Monoid
Let $\struct {S, \circ}$ be a monoid. Let $x \in S$ be invertible, and let its inverse be $x^{-1}$. Then $x^{-1}$ is also invertible, and: :$\paren {x^{-1} }^{-1} = x$
Inverse of Product/Monoid
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e$. Let $a, b \in S$ be invertible for $\circ$, with inverses $a^{-1}, b^{-1}$. Then $a \circ b$ is invertible for $\circ$, and: : $\left({a \circ b}\right)^{-1} = b^{-1} \circ a^{-1}$
Right Identity Element is Idempotent
Let $\struct {S, \circ}$ be an algebraic structure. Let $e_R \in S$ be a right identity with respect to $\circ$. Then $e_R$ is idempotent under $\circ$.
More than one Right Identity then no Left Identity
Let $\left({S, \circ}\right)$ be an algebraic structure. If $\left({S, \circ}\right)$ has more than one right identity, then it has no left identity.
Inverse Completion is Commutative Monoid
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 a commutative monoid.
Inverse Completion 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 $T = S \circ' C^{-1}$, and is a commutative semigroup.
Complement of G-Delta Set is F-Sigma Set
Let $T = \struct {S, \tau}$ be a topological space. Let $X$ be an $G_\delta$ set of $T$. Then its complement $S \setminus X$ is an $F_\sigma$ set of $T$.
Bounded Piecewise Continuous Function has Improper Integrals
Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$.  Let $f$ be piecewise continuous and bounded on $\closedint a b$. Then $f$ is a piecewise continuous function with improper integrals.
F-Sigma Set is not necessarily Closed Set
Let $T = \struct {S, \tau}$ be a topological space. Let $X$ be an $F_\sigma$ set of $T$. Then it is not necessarily the case that $X$ is a closed set of $T$.
Not every Closed Set is G-Delta Set
Let $T = \struct {S, \tau}$ be a topological space. Let $V$ be a closed set of $T$. Then it is not necessarily the case that $V$ is a $G_\delta$ set of $T$.
Open Set of Uncountable Finite Complement Topology is not F-Sigma
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$. Let $U \in \tau$ be an open set of $T$. Then $U$ is not an $F_\sigma$ set.
Closed Set of Uncountable Finite Complement Topology is not G-Delta
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$. Let $V \in \tau$ be a closed set of $T$. Then $V$ is not a $G_\delta$ set.
G-Delta Set is not necessarily Open Set
Let $T = \struct {S, \tau}$ be a topological space. Let $X$ be a $G_\delta$ set of $T$. Then it is not necessarily the case that $X$ is a open set of $T$.
Limit Point of Sequence may only be Adherent Point of Range
Let $T = \struct {S, \tau}$ be a topological space. Let $A \subseteq S$. Let $\sequence {x_n}$ be a sequence in $A$. Let $\alpha$ be a limit point of $\sequence {x_n}$. Then $\alpha$ may be only an adherent point of $A$ and not a limit point of $A$.