hcju/proofwikips huggingface dataset:
GO-Space Embeds as Closed Subspace of Linearly Ordered Space
Let $(X, \preceq_X, \tau_X)$ be a generalized ordered space. Then there is a linearly ordered space $(Y, \preceq_Y, \tau_Y)$ and a mapping $\phi: X \to Y$ such that $\phi$ is a topological embedding and an order embedding, and $\phi(X)$ is closed in $Y$.
Union of Total Ordering with Lower Sets is Total Ordering
Let $\left({Y, \preceq}\right)$ be a totally ordered set. Let $X$ be the disjoint union of $Y$ with the set of lower sets of $Y$. Define a relation $\preceq'$ on $X$ extending $\preceq$ by letting: :$y_1 \preceq' y_2 \iff y_1 \preceq y_2$ :$y \preceq' L \iff y \in L$ :$L_1 \preceq' L_2 \iff L_1 \subseteq L_2$ :$L \preceq' y \iff y \in Y \setminus L$ Then $\preceq'$ is a total ordering.
Lower Closure is Lower Set
Let $\left({S, \preceq}\right)$ be an ordered set. Let $T$ be a subset of $S$. Let $L$ be the lower closure of $T$. Then $L$ is a lower set.
Ordered Set is Lower Set in Itself
Let $(S, \preceq)$ be an ordered set. Then $S$ is a lower set in $S$.
Ordered Set is Convex in Itself
Let $\struct {S, \preceq}$ be an ordered set. Then $S$ is a convex set in $S$.
Equivalence of Definitions of Generalized Ordered Space/Definition 3 implies Definition 1
Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space by Definition 3: {{:Definition:Generalized Ordered Space/Definition 3}} Then $\left({S, \preceq, \tau}\right)$ is a generalized ordered space by Definition 1: {{:Definition:Generalized Ordered Space/Definition 1}}
Strict Upper Closure is Upper Set
Let $(S, \preceq)$ be an ordered set. Let $p \in S$. Then $p^\succ$, the strict upper closure of $p$, is an upper set.
Strict Lower Closure is Lower Set
Let $\struct {S, \preceq}$ be an ordered set. Let $p \in S$. Then $p^\prec$, the strict lower closure of $p$, is a lower set.
Topology Discrete iff All Singletons Open
Let $(X, \tau)$ be a topological space. Then $\tau$ is the discrete topology on $X$ iff: : For all $x \in X$: $\{ x \} \in \tau$ That is, iff every singleton of $X$ is $\tau$-open.
Characteristic Function of Universe
Let $S$ be a set. Let $\chi_S: S \to \left\{ {0, 1}\right\}$ be its characteristic function (in itself). Then: :$\chi_S = f_1$ where $f_1: S \to \left\{ {0, 1}\right\}$ is the constant mapping with value $1$.
Faltings' Theorem
Let $C$ be a curve over $\Q$ of genus $g > 1$. Then $C$ has only finitely many rational points.
Supremum of Lower Closure of Set
Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$. Let $L = T^\preceq$ be the lower closure of $T$ in $S$. Let $s \in S$ Then $s$ is the supremum of $T$  it is the supremum of $L$.
Upper Closure is Smallest Containing Upper Set
Let $\struct {S, \preceq}$ be an ordered set. Let $T \subseteq S$. Let $U = T^\succeq$ be the upper closure of $T$. Then $U$ is the smallest upper set containing $T$ as a subset.
Upper Closure is Closure Operator
Let $\left({S, \preceq}\right)$ be an ordered set. Let $T^\succeq$ be the upper closure of $T$ for each $T \subseteq S$. Then $\cdot^\succeq$ is a closure operator.
Equivalence of Definitions of Upper Set
Let $\left({S, \preceq}\right)$ be an ordered set. Let $U \subseteq S$. {{TFAE|def = Upper Set}}
Topological Closure is Closure Operator
The topological closure operator is a closure operator.
Reflexive Closure is Closure Operator
Let $S$ be a set. Let $R$ be the set of all endorelations on $S$. Then the reflexive closure operator on $R$ is a closure operator.
Set Closure is Smallest Closed Set/Closure Operator
Let $S$ be a set. Let $\cl: \powerset S \to \powerset S$ be a closure operator. Let $T \subseteq S$. Then $\map \cl T$ is the smallest closed set (with respect to $\cl$) containing $T$ as a subset.
Equivalence of Definitions of Lower Set
{{TFAE|def = Lower Set}} Let $\left({S, \preceq}\right)$ be an ordered set. Let $U \subseteq S$. Then the following are equivalent: {{begin-axiom}} {{axiom | n = 1         | m = \forall u \in U: \forall s \in S: s \preceq u \implies s \in U }} {{axiom | n = 2         | m = U^\preceq \subseteq U }} {{axiom | n = 3         | m = U^\preceq = U }} {{end-axiom}} where $U^\preceq$ is the lower closure of $U$.
Zero and One are the only Consecutive Perfect Squares/Proof 1
{{:Zero and One are the only Consecutive Perfect Squares}} <section title="Proof" name="proof"> Let $x$ and $h$ be integers such that $x^2 + 1 = \paren {x - h}^2$ {{begin-eqn}} {{eqn | l = x^2 + 1       | r = \left({x - h}\right)^2 }} {{eqn | l = 1       | r = -2 x h + h^2 }} {{eqn | l = 2 x h       | r = h^2 - 1 }} {{eqn | l = 2 x h       | r = \paren {h - 1} \paren {h + 1} }} {{end-eqn}} We have that Consecutive Integers are Coprime. However, both sides must have the same unique prime factorization by the Fundamental Theorem of Arithmetic Therefore $h$ cannot have any prime factors since they cannot be shared by $\paren {h - 1} \paren {h + 1}$. This leaves $h = -1$, $h = 0$, or $h = 1$ as the only possibilities since they are the only integers with no prime factors. If $h = -1$ then $h + 1 = 0$, so $2 x h = 0$. It follows that $x = 0$. If $h = 1$ then $h - 1 = 0$, so $2 x h = 0$. It follows that $x = 0$. If $h = 0$, then $2 x \cdot 0 = \paren {-1} \paren 1$, which is a contradiction. Therefore the only pairs of consecutive perfect squares are: :$0^2 = 0$ and $\paren {0 + \paren {-1} }^2 = \paren {-1}^2 = 1$ and: :$0^2 = 0$ and $\paren {0 + 1}^2 = 1^2 = 1$  </section>
Zero and One are the only Consecutive Perfect Squares/Proof 2
{{:Zero and One are the only Consecutive Perfect Squares}} <section title="Proof" name="proof"> Suppose that $k, l \in \Z$ are such that their squares are consecutive, i.e.: :$l^2 - k^2 = 1$ Then we can factor the left-hand side as: :$l^2 - k^2 = \left({l + k}\right) \left({l - k}\right)$ By Invertible Integers under Multiplication, it follows that: :$l + k = \pm 1 = l - k$ Therefore, it must be that: :$\left({l + k}\right) - \left({l - k}\right) = 0$ That is, $2 k = 0$, from which we conclude $k = 0$. So if $n$ and $n + 1$ are squares, then necessarily $n = 0$. The result follows.  </section>
Convergent Series of Natural Numbers
Let $\left({a_n}\right)_{n \in \N}$ be a sequence of natural numbers. Then the following are equivalent: $(1): \quad \displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges $(2): \quad \exists N \in \N: \forall n \ge N: a_n = 0$ That is, $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges  only finitely many of the $a_n$ are non-zero.
Closure Operator from Closed Sets
Let $S$ be a set. Let $\CC$ be a set of subsets of $S$. Let $\CC$ be closed under arbitrary intersections: :$\forall \KK \in \powerset \CC: \bigcap \KK \in \CC$ where $\bigcap \O$ is taken to be $S$. Define $\cl: \powerset S \to \CC$ by letting: :$\map \cl T = \bigcap \set {C \in \CC: T \subseteq C}$ Then $\cl$ is a closure operator whose closed sets are the elements of $\CC$.
Intersection is Decreasing
Let $U$ be a set. Let $\mathcal F$ and $\mathcal G$ be sets of subsets of $U$. Then $\mathcal F \subseteq \mathcal G \implies \bigcap \mathcal G \subseteq \bigcap \mathcal F$, where by convention $\bigcap \varnothing = U$. That is, $\bigcap$ is a decreasing mapping from $(\mathcal P(\mathcal P(U)), \subseteq)$ to $(\mathcal P(U), \subseteq)$, where $\mathcal P(U)$ is the power set of $U$.
Open Ray is Open in GO-Space
Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space. Let $p \in S$. Then: : $p^\prec$ and $p^\succ$ are $\tau$-open where: : $p^\prec$ is the strict lower closure of $p$ : $p^\succ$ is the strict upper closure of $p$.
Union is Increasing
Let $U$ be a set. Let $\mathcal F$ and $\mathcal G$ be sets of subsets of $U$. Then $\mathcal F \subseteq \mathcal G \implies \bigcup \mathcal F \subseteq \bigcup \mathcal G$. That is, $\bigcup$ is an increasing mapping from $(\mathcal P(\mathcal P(U)), \subseteq)$ to $(\mathcal P(U), \subseteq)$, where $\mathcal P(U)$ is the power set of $U$.
Intersection is Idempotent/Indexed Family
Let $\family {F_i}_{i \mathop \in I}$ be a non-empty indexed family of sets. Suppose that all the sets in the $\family {F_i}_{i \mathop \in I}$ are the same. That is, suppose that for some set $S$: :$\forall i \in I: F_i = S$ Then: :$\displaystyle \bigcap_{i \mathop \in I} F_i = S$ where $\displaystyle \bigcap_{i \mathop \in I} F_i$ is the intersection of $\family {F_i}_{i \mathop \in I}$.
Factor Principles/Conjunction on Left/Formulation 1/Proof 2
:$p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$
Factor Principles/Conjunction on Left/Formulation 1/Proof 1
: $p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$
Constructive Dilemma for Join Semilattices
Let $\struct {S, \vee, \preceq}$ be a join semilattice. Let $a, b, c, d \in S$. Let $a \preceq b$. Let $c \preceq d$. Then $\paren {a \vee c} \preceq \paren {b \vee d}$.
Praeclarum Theorema for Meet Semilattices
Let $(S, \wedge, \preceq)$ be a meet semilattice. Let $a, b, c, d \in S$. Let $a \preceq b$. Let $c \preceq d$. Then $(a \wedge c) \preceq (b \wedge d)$.
Supremum is Increasing relative to Product Ordering
Let $(S, \preceq)$ be an ordered set. Let $I$ be a set. Let $f, g: I \to S$. Let $f \left[{I}\right]$ denote the image of $I$ under $f$. Let: :$\forall i \in I: f \left({i}\right) \preceq g \left({i}\right)$ That is, let $f \preceq g$ in the product ordering. Let $f \left[{I}\right]$ and $g \left[{I}\right]$ admit suprema. Then: : $\sup f \left[{I}\right] \preceq \sup g \left[{I}\right]$
Reflexive Closure of Strict Ordering is Ordering
Let $S$ be a set. Let $\prec$ be a strict ordering on $S$. Let $\preceq$ be the reflexive closure of $\prec$. Then $\preceq$ is an ordering.
Reflexive Closure is Reflexive
Let $\mathcal R$ be a relation on a set $S$. Then $\mathcal R^=$, the reflexive closure of $\mathcal R$, is reflexive.
Reflexive Closure is Inflationary
Let $S$ be a set. Let $R$ denote the set of all endorelations on $S$. Then the reflexive closure operator is an inflationary mapping on $R$.
Reflexive Closure is Order Preserving
Let $S$ be a set. Let $R$ denote the set of all endorelations on $S$. Then the reflexive closure operator is an order preserving mapping on $R$. That is: :$\forall \RR, \SS \in R: \RR \subseteq \SS \implies \mathcal R^= \subseteq \SS^=$ where $\RR^=$ and $\SS^=$ denote the reflexive closure of $\RR$ and $\SS$ respectively.
Reflexive Closure is Idempotent
Let $S$ be a set. Let $R$ denote the set of all endorelations on $S$. Then the reflexive closure operator is an idempotent mapping on $R$. That is: :$\forall \RR \in R: \RR^= = \paren {\RR^=}^=$ where $\RR^=$ denotes the reflexive closure of $\RR$.
Transitive Closure Always Exists (Relation Theory)/Proof 1
Let $\mathcal R$ be a relation on a set $S$. Then the transitive closure $\mathcal R^+$ of $\mathcal R$ always exists.
Transitive Closure Always Exists (Relation Theory)/Proof 2
Let $\mathcal R$ be a relation on a set $S$. Then the transitive closure $\mathcal R^+$ of $\mathcal R$ always exists.
Equivalence of Definitions of Transitive Closure (Relation Theory)/Union of Compositions is Smallest
Let $\RR$ be a relation on a set $S$. Let: :$\RR^n := \begin{cases} \RR & : n = 0 \\ \RR^{n - 1} \circ \RR & : n > 0 \end{cases}$ where $\circ$ denotes composition of relations. {{explain|Really? I would have thought $\RR^1 {{=}} \RR$, not $\RR^0 {{=}} \RR$. If anything, the diagonal relation $\Delta_S$ should be $\RR^0$.}} Finally, let: :$\displaystyle \RR^+ = \bigcup_{i \mathop \in \N} \RR^i$ Then $\RR^+$ is the smallest transitive relation on $S$ that contains $\RR$.
Equivalence of Definitions of Transitive Closure (Relation Theory)/Finite Chain Equivalent to Union of Compositions
The finite chain and union of compositions definitions of '''transitive closure''' are equivalent.
Chasles' Relation
Let $\mathcal E$ be an affine space. Let $p, q, r \in \mathcal E$ be points. Then: :$\vec{p q} = \vec{p r} + \vec{r q}$
Affine Coordinates are Well-Defined
Let $\mathcal E$ be an affine space with difference space $V$ over a field $k$. Let $\mathcal R = \left({p_0, e_1, \ldots, e_n}\right)$ be an affine frame in $\mathcal E$. Define a mapping $\Theta_{\mathcal R} : k^n \to \mathcal E$ by: :$\displaystyle \Theta_\mathcal R \left({\lambda_1, \ldots, \lambda_n}\right) = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$ Then $\Theta_\mathcal R$ is a bijection.
Transitive Chaining
Let $\mathcal R$ be a transitive relation on a set $S$. Let $n \in \N$ be a natural number. Let $n \ge 2$. Let $\langle x_k \rangle_{k \in \left\{ {1, 2, \dots, n}\right\} }$ be a sequence of $n$ terms. For each $k \in \left\{ {1, 2, \dots, n-1}\right\}$, let $x_k \mathrel {\mathcal R} x_{k+1}$. That is, let $x_1 \mathrel {\mathcal R} x_2$, $x_2 \mathrel {\mathcal R} x_3, \dotsc, x_n-1 \mathrel {\mathcal R} x_n$. Then $x_1 \mathrel {\mathcal R} x_n$
Szpilrajn Extension Theorem
Let $\struct {S, \prec}$ be a strictly ordered set. {{Disambiguate|Definition:Strictly Ordered Set}} Then there is a strict total ordering on $S$ of which $\prec$ is a subset.
Strict Ordering can be Expanded to Compare Additional Pair
Let $\left({S, \prec}\right)$ be an ordered set. Let $a$ and $b$ be distinct, $\prec$-incomparable elements of $S$. That is, let: :$a \not \prec b$ and $b \not \prec a$. Let ${\prec'} = {\prec} \cup \left\{ {\left({a, b}\right)} \right\}$. Define a relation $\prec'^+$ by letting $p \prec'^+ q$ : : $p \prec q$ or: : $p \preceq a$ and $b \preceq q$ where $\preceq$ is the reflexive closure of $\prec$. Then: : $\prec'^+$ is a strict ordering : $\prec^+$ is the transitive closure of $\prec'$.
Characterization of Interior of Triangle
Let $\triangle$ be a triangle embedded in $\R^2$. Denote the vertices of $\triangle$ as $A_1, A_2, A_3$. For $i \in \set {1, 2, 3}$, put $j = i \bmod 3 + 1$, $k = \paren {i + 1} \bmod 3 + 1$, and: :$U_i = \set {A_i + s t \paren {A_j - A_i} + \paren {1 - s} t \paren {A_k - A_i} : s \in \openint 0 1, t \in \R_{>0} }$ Then: :$\displaystyle \Int \triangle = \bigcap_{i \mathop = 1}^3 U_i$ where $\Int \triangle$ denotes the interior of the boundary of $\triangle$.
Equivalence of Definitions of Reflexive Transitive Closure
Let $\mathcal R$ be a relation on a set $S$. {{TFAE|def = Reflexive Transitive Closure}}
Intersection of Relation with Inverse is Symmetric Relation
Let $\mathcal R$ be a relation on a set $S$. Then $\mathcal R \cap \mathcal R^{-1}$, the intersection of $\mathcal R$ with its inverse, is symmetric.
Intersection of Closed Sets is Closed/Closure Operator
Let $S$ be a set. Let $f: \powerset S \to \powerset S$ be a closure operator on $S$. Let $\CC$ be the set of all subsets of $S$ that are closed with respect to $f$. Let $\AA \subseteq \CC$. Then $\bigcap \AA \in \CC$.
Closure is Closed/Power Set
Let $S$ be a set. Let $\powerset S$ be the power set of $S$. Let $\cl: \powerset S \to \powerset S$ be a closure operator. Let $T \subseteq S$. Then $\map \cl T$ is a closed set with respect to $\cl$.
Relation Intersection Inverse is Greatest Symmetric Subset of Relation
Let $\RR$ be a relation on a set $S$. Let $\powerset \RR$ be the power set of $\RR$. By definition, $\powerset \RR$ is the set of all relations on $S$ that are subsets of $\RR$. Then the greatest element of $\powerset \RR$ that is symmetric is: :$\RR \cap \RR^{-1}$
Composition of Compatible Closure Operators
Let $S$ be a set. Let $f, g: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right)$ be closure operators on $S$. Let $\mathcal C_f$ and $\mathcal C_g$ be the sets of closed sets of $S$ with respect to $f$ and $g$ respectively. For each subset $T$ of $S$, let the following hold: : $(1): \quad$ If $T$ is closed with respect to $g$, then $f \left({T}\right)$ is closed with respect to $g$. :: That is, if $T \in \mathcal C_g$ then $f \left({T}\right) \in \mathcal C_g$. : $(2): \quad$ If $T$ is closed with respect to $f$, then $g \left({T}\right)$ is closed with respect to $f$. :: That is, if $T \in \mathcal C_f$ then $g \left({T}\right) \in \mathcal C_f$. Let $\mathcal C_h = \mathcal C_f \cap \mathcal C_g$. Then: : $\mathcal C_h$ induces a closure operator $h$ on $S$ : $f \circ g = g \circ f = h$, where $\circ$ represents composition of mappings.
Closure Operator from Closed Elements
Let $\struct {S, \preceq}$ be an ordered set. Let $C \subseteq S$. Suppose that $C$ is a subset of $S$ with the property that every element of $S$ has a smallest successor in $C$. Let $\cl: S \to S$ be defined as follows: For $x \in S$: :$\map \cl x = \map \min {C \cap x^\succeq}$ where $x^\succeq$ is the upper closure of $x$. That is, let $\map \cl x$ be the smallest successor of $x$ in $C$. Then: :$\cl$ is a closure operator on $S$ :The closed elements of $\cl$ are precisely the elements of $C$.
Product of Affine Spaces is Affine Space
Let $\mathcal E, \mathcal F$ be affine spaces. Let $\mathcal G = \mathcal E \times \mathcal F$ be the product of $\mathcal E$ and $\mathcal F$. Then $\mathcal G$ is an affine space.
Intersection of Complete Meet Subsemilattices invokes Closure Operator
Let $\struct {S, \preccurlyeq}$ be an ordered set. Let $f_i$ be a closure operator on $S$ for each $i \in I$. Let $C_i = \map {f_i} S$ be the set of closed elements with respect to $f_i$ for each $i \in I$. Suppose that for each $i \in I$, $C_i$ is a '''complete meet subsemilattice''' of $S$ in the following sense: :For each $D \subseteq C_i$, $D$ has an infimum in $S$ such that $\inf D \in C_i$. Then $C = \displaystyle \bigcap_{i \mathop \in I} C_i$ induces a closure operator on $S$.
Reflexive Reduction of Ordering is Strict Ordering
Let $\mathcal R$ be an ordering on a set $S$. Let $\mathcal R^\ne$ be the reflexive reduction of $\mathcal R$. Then $\mathcal R^\ne$ is a strict ordering on $S$.
Vectorialization of Affine Space is Vector Space
Let $\mathcal E$ be an affine space over a field $k$ with difference space $E$. Let $\mathcal R = \tuple {p_0, e_1, \ldots, e_n}$ be an affine frame in $\mathcal E$. Let $\struct {\mathcal E, +, \cdot}$ be the vectorialization of $\mathcal E$. Then $\struct {\mathcal E, +, \cdot}$ is a vector space.
Reflexive Reduction is Antireflexive
Let $\mathcal R$ be a relation on a set $S$. Let $\mathcal R^\ne$, be the reflexive reduction of $\mathcal R$. Then $\mathcal R^\ne$ is antireflexive.
Subband iff Idempotent under Induced Operation
Let $\left({S, \circ}\right)$ be a band. Let $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ be the algebraic structure consisting of the power set of $S$ and the operation induced on $\mathcal P \left({S}\right)$ by $\circ$. Let $X \in \mathcal P \left({S}\right)$. Then $X$ is idempotent  $\left({X, \circ}\right)$ is a subband of $\left({S, \circ}\right)$.
Restriction of Idempotent Operation is Idempotent
Let $\left({S, \circ}\right)$ be an algebraic structure. Let $T \subseteq S$. Let the operation $\circ$ be idempotent. Then $\circ$ is also idempotent upon restriction  to $\left({T, \circ \restriction_T}\right)$.
Composition of Commuting Idempotent Mappings is Idempotent
Let $S$ be a set. Let $f, g: S \to S$ be idempotent mappings from $S$ to $S$. Let: :$f \circ g = g \circ f$ where $\circ$ denotes composition. Then $f \circ g$ is idempotent.
Compositions of Closure Operators are both Closure Operators iff Operators Commute
Let $\left({S, \preceq}\right)$ be an ordered set. Let $f$ and $g$ be closure operators on $S$. Then the following are equivalent: :$(1): \quad f \circ g$ and $g \circ f$ are both closure operators. :$(2): \quad f$ and $g$ commute (that is, $f \circ g = g \circ f$). :$(3): \quad \operatorname{img}\left({f \circ g}\right) = \operatorname{img}\left({g \circ f}\right)$ where $\operatorname{img}$ represents the image of a mapping.
Composition of Increasing Mappings is Increasing
Let $\struct {S, \preceq_S}$, $\struct {T, \preceq_T}$ and $\struct {U, \preceq_U}$ be ordered sets. Let $g: S \to T$ and $f: T \to U$ be increasing mappings. Then their composition $f \circ g: S \to U$ is also increasing.
Composition of Inflationary Mappings is Inflationary
Let $\left({S, \preceq}\right)$ be an ordered set. Let $f, g: S \to S$ be inflationary mappings. Then $f \circ g$, the composition of $f$ and $g$, is also inflationary.
Fixed Point of Idempotent Mapping
Let $S$ be a set. Let $f: S \to S$ be an idempotent mapping. Let $f \left[{S}\right]$ be the image of $S$ under $f$. Let $x \in S$. Then $x$ is a fixed point of $f$  $x \in f \left[{S}\right]$.
Symmetric Closure of Ordering may not be Transitive
Let $\left({S, \preceq}\right)$ be an ordered set. Let $\preceq^\leftrightarrow$ be the symmetric closure of $\preceq$. Then it is not necessarily the case that $\preceq^\leftrightarrow$ is transitive.
Composition of Idempotent Mappings
Let $S$ be a set. Let $f, g: S \to S$ be idempotent mappings. Suppose that $f \circ g$ and $g \circ f$ have the same images. That is, suppose that $f \sqbrk {g \sqbrk S} = g \sqbrk {f \sqbrk S}$. Then $f \circ g$ and $g \circ f$ are idempotent.
Composition of Inflationary and Idempotent Mappings
Let $\struct {S, \preceq}$ be an ordered set. Let $f$ and $g$ be inflationary and idempotent mappings on $S$. Then the following are equivalent: :$(1): \quad f \circ g$ and $g \circ f$ are both idempotent :$(2): \quad f$ and $g$ commute (that is, $f \circ g = g \circ f$) :$(3): \quad \Img {f \circ g} = \Img {g \circ f}$ where: :$\circ$ represents composition :$\Img f$ represents the image of a mapping $f$.
Closure is Smallest Closed Successor
Let $\struct {S, \preceq}$ be an ordered set. Let $f: S \to S$ be a closure operator on $S$. Let $x \in S$. Then $\map f x$ is the smallest closed element that succeeds $x$.
Closed Elements Uniquely Determine Closure Operator
Let $\left({S, \preceq}\right)$ be an ordered set. Let $f, g: S \to S$ be closure operators on $S$. Suppose that $f$ and $g$ have the same closed elements. Then $f = g$.
Square of Number Always Exists
Let $x$ be a number. Then its square $x^2$ is guaranteed to exist.
Schröder Rule
Let $A$, $B$ and $C$ be relations on a set $S$. Then the following are equivalent statements: :$(1): \quad A \circ B \subseteq C$ :$(2): \quad A^{-1} \circ \overline C \subseteq \overline B$ :$(3): \quad \overline C \circ B^{-1} \subseteq \overline A$ where: : $\circ$ denotes relation composition : $A^{-1}$ denotes the inverse of $A$ : $\overline A$ denotes the complement of $A$.
Equivalence of Definitions of Dual Relation
{{TFAE|def = Dual Relation}} Let $\RR \subseteq S \times T$ be a relation.
Trivial Gradation is Gradation
Let $\left({R, +, \circ}\right)$ be a ring. Let $\left({M, e,\cdot}\right)$ be a monoid. Let :$\displaystyle R = \bigoplus_{m \mathop \in M} R_m$ be the trivial $M$-gradation on $R$. This is a gradation on $R$.
Fixed Point of Mappings is Fixed Point of Composition
Let $S$ be a set. Let $f, g: S \to S$ be mappings. Let $x \in S$ be a fixed point of both $f$ and $g$. Then $x$ is also a fixed point of $f \circ g$, the composition of $f$ and $g$.
Fixed Point of Mappings is Fixed Point of Composition/General Result
Let $S$ be a set. Let $n \in \N$ be a strictly positive integer. Let $\N_n$ be the initial segment of $n$ in $\N$. That is, let $\N_n = \left\{{0, 1, \dots, n-1}\right\}$. For each $i \in \N_n$, let $f_i: S \to S$ be a mapping. Let $x \in S$ be a fixed point of $f_i$ for each $i \in \N_n$. Let $g = f_0 \circ f_1 \circ \dots \circ f_{n-1}$ be the composition of all the $f_i$s. Then $x$ is a fixed point of $g$.
Polynomial has Integer Coefficients iff Content is Integer
$f$ has integer coefficients  $\cont f$ is an integer.
Content of Scalar Multiple
:$\cont {q f} = q \cont f$
Fixed Point of Composition of Inflationary Mappings
Let $\left({S, \preceq}\right)$ be a ordered set. Let $f, g: S \to S$ be inflationary mappings. Let $x \in S$. Then: : $x$ is a fixed point of $f \circ g$ : : $x$ is a fixed point of both $f$ and $g$.
Group of Units is Group
Let $\struct {R, +, \circ}$ be a ring with unity. Then the set of units of $\struct {R, +, \circ}$ forms a group under $\circ$. Hence the justification for referring to the group of units of $\struct {R, +, \circ}$.
Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication
Let $\struct {D, +, \circ}$ be a principal ideal domain. Let $p$ be an irreducible element of $D$. Let $\ideal p$ be the principal ideal of $D$ generated by $p$. Then $\ideal p$ is a maximal ideal of $D$.