hcju/proofwikips huggingface dataset:
Cyclicity Condition for Units of Ring of Integers Modulo m
Let $n \in \Z_{\ge 0}$ be a positive integer. Let $\struct {\Z / n \Z, +, \times}$ be the ring of integers modulo $n$. Let $U = \struct {\paren {\Z / n \Z}^\times, \times}$ denote the group of units of $\struct {\Z / n \Z, +, \times}$. Then $U$ is cyclic  either: :$n = p^\alpha$ or: :$n = 2 p^\alpha$ where $p \ge 3$ is prime and $\alpha \ge 0$.
Chinese Remainder Theorem/Corollary
Let $n_1, n_2, \ldots, n_r$ be positive integers such that $n_i \perp n_j$ for all $i \ne j$ (that is, $\gcd \left\{{n_i, n_j}\right\} = 1$). Let $N = n_1 \cdots n_r$. For an integer $k$, let $\Z / k \Z$ denote the ring of integers modulo $k$. Then we have a ring isomorphism: :$\Z / N \Z \simeq \Z / n_1 \Z \times \cdots \times \Z / n_r \Z$
Minkowski's Theorem
Let $L$ be a lattice in $\R^n$. Let $d$ be the covolume of $L$. Let $\mu$ be a translation invariant measure on $\R^n$ Let $S$ be a convex subset of $\R^n$ that is symmetric about the origin, i.e. such that: :$\forall p \in S : -p \in S$  Let the volume of $S$ be greater than $2^n d$. Then $S$ contains a non-zero point of $L$.
Component of Finite Union in Ultrafilter
Let $S$ be a non-empty set. Let $\mathcal U$ be a ultrafilter on $S$. Let $\left\{{Y_1, \dots, Y_n}\right\}$ be a pairwise disjoint set of subsets of $S$ such that $S = Y_1 \cup \cdots \cup Y_n$. Then there is a unique $k \in \left\{{1, \dots, n}\right\}$ such that $Y_k \in \mathcal U$.
Restricted Tukey-Teichmüller Theorem/Weak Form
Let $X$ be a set. Let $\mathcal A$ be a non-empty set of subsets of $X$. Let $'$ be a unary operation on $X$. Let $\mathcal A$ have finite character. For all $A \in \mathcal A$ and all $x \in X$, let either: :$A \cup \set x \in \mathcal A$ or: :$A \cup \set {x'} \in \mathcal A$ Then there exists a $B \in \mathcal A$ such that for all $x \in X$, either $x \in B$ or $x' \in B$.
Restricted Tukey-Teichmüller Theorem/Strong Form
Let $X$ be a set. Let $\mathcal A$ be a non-empty set of subsets of $X$. Let $'$ be a unary operation on $X$. Let $\mathcal A$ have finite character. For all $A \in \mathcal A$ and all $x \in X$, let either: : $A \cup \left\{ {x}\right\} \in \mathcal A$ or: :$A \cup \left\{ {x'}\right\} \in \mathcal A$ Then for each $A \in \mathcal A$ there exists a $C \in \mathcal A$ such that: :$A \subseteq C$ and: :for all $x \in X$, either $x \in C$ or $x' \in C$.
Boolean Prime Ideal Theorem/Extension Lemma
Let $\struct {B, \vee, \wedge, \neg, \bot, \top}$ be a Boolean algebra. Let $J \subseteq B$ have the finite join property. Let $z \in B$. Then either $J \vee z$ or $J \vee \neg z$ also has the finite join property.
Finite Character for Sets of Mappings
Let $S$ and $T$ be sets. Let $\mathcal F$ be a set of mappings from subsets of $S$ to $T$. That is, let $\mathcal F$ be a set of partial mappings from $S$ to $T$. Then the following are equivalent: {{begin-axiom}} {{axiom | n = 1         | t = $\mathcal F$ has finite character in the sense of Definition:Finite Character/Mappings. }} {{axiom | n = 2         | t = $\mathcal F$ has finite character as a set of subsets of $S \times T$ in the sense of Definition:Finite Character. }} {{end-axiom}}
Cowen-Engeler Lemma
Let $X$ and $Y$ be sets. Let $M$ be a set of mappings from subsets of $X$ to $Y$. Thus each element of $M$ is a partial mapping from $X$ to $Y$. Define a mapping $\Phi: X \to \powerset Y$ thus: :$\map \Phi x = \set {\map f x: f \in M \text{ and } x \in \Dom f}$ That is: :$\map \Phi x = \set {y \in Y: \exists f \in M: \tuple {x, y} \in f}$ Suppose that the following hold: {{begin-axiom}} {{axiom | n = 1         | t = $\map \Phi x$ is finite for each $x \in X$. }} {{axiom | n = 2         | t = For each finite subset $F$ of $X$, there exists an element $f \in M$ such that the domain of $f$ is $F$. }} {{axiom | n = 3         | t = $M$ has finite character. }} {{axiom | t = That is, a mapping $f$ from some subset of $X$ to $Y$ is an element of $M$  its restriction to each finite subset of $\Dom f$ is in $M$. }} {{end-axiom}} Then there exists an element of $M$ with domain $X$.
Ultrafilter Lemma/Corollary
Let $S$ be a non-empty set. Let $\mathcal A$ be a set of subsets of $S$. Suppose that $\mathcal A$ has the finite intersection property. Then there is an ultrafilter $\mathcal U$ on $S$ such that $\mathcal A \subseteq \mathcal U$.
Order-Extension Principle/Strict
Let $S$ be a set. Let $\prec$ be a strict ordering on $S$. Then there exists a strict total ordering $<$ on $S$ such that: :$\forall a, b \in S: a \prec b \implies a < b$
Order-Extension Principle/Strict/Finite Set
Let $T$ be a finite set. Let $\prec$ be a strict ordering on $T$. Then there exists a strict total ordering $<$ on $T$ such that: :$\forall a, b \in T: \left({a \prec b \implies a < b}\right)$
Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma
Let $T$ be a finitely satisfiable $\mathcal L$-theory. Let $\phi$ be an $\mathcal L$-sentence. Then either $T \cup \left\{ {\phi}\right\}$ or $T \cup \left\{ {\neg \phi}\right\}$ is finitely satisfiable.
Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Proof 2
Let $T$ be a finitely satisfiable $\mathcal L$-theory. There is a finitely satisfiable $\mathcal L$-theory $T'$ which contains $T$ as a subset such that for all $\mathcal L$-sentences $\phi$, either $\phi \in T'$ or $\neg\phi \in T'$. {{explain|Does this actually mean the same as the statement of the theorem in Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension? If so, replace it. If not, then it is a different proof altogether. If the two statements are equivalent, then this needs to be demonstrated.}}
Reflexive Closure of Strict Total Ordering is Total Ordering
Let $S$ be a set. Let $\prec$ be a strict total ordering of $S$. Let $\preceq$ be the reflexive closure of $\prec$. Then $\preceq$ is an total ordering of $S$.
Node of Rooted Tree is on Branch
Let $T$ be a rooted tree with root node $r_T$. Let $t$ be a node of $T$. Then there exists a branch $\Gamma$ of $T$ such that $t \in \Gamma$.
Leaf of Rooted Tree is on One Branch
Let $T$ be a rooted tree with root node $r_T$. Let $t$ be a leaf node of $T$. Then there exists a unique branch $\Gamma$ of $T$ such that $t \in \Gamma$.
Node of Rooted Tree with Multiple Children is on Multiple Branches
Let $T$ be a rooted tree with root node $r_T$. Let $t$ be a node of $T$. Let $t$ have more than one child nodes. Then $t$ is on more than one branch of $T$.
Branch of Finite Tree is Finite
Let $T$ be a finite rooted tree with root node $r_T$. Let $\Gamma$ be a branch of $T$. Then $\Gamma$ is a finite branch.
Same Dimensional Vector Spaces are Isomorphic
Let $K$ be a division ring. Let $V$, $W$ be finite dimensional $K$-vector spaces. Suppose that $\dim_K V = \dim_K W$. Then: :$V \cong W$ That is, $V$ and $W$ are isomorphic.
Five Lemma
Let $A$ be a commutative ring with unity. Let: ::$\begin{xy}\xymatrix@L+2mu@+1em{  M_1 \ar[r]^*{\alpha_1}      \ar[d]^*{\phi_1} &  M_2 \ar[r]^*{\alpha_2}      \ar[d]^*{\phi_2} &  M_3 \ar[r]^*{\alpha_3}      \ar[d]^*{\phi_3} &  M_4 \ar[r]^*{\alpha_4}      \ar[d]^*{\phi_4} &  M_5 \ar[d]^*{\phi_5} \\  N_1 \ar[r]_*{\beta_1} &  N_2 \ar[r]_*{\beta_2} &  N_3 \ar[r]_*{\beta_3} &  N_4 \ar[r]_*{\beta_4} &  N_5 }\end{xy}$ be a commutative diagram of $A$-modules. Suppose that the rows are exact. Then: :If $\phi_2$ and $\phi_4$ are surjective and $\phi_5$ is injective then $\phi_3$ is surjective. :If $\phi_2$ and $\phi_4$ are injective and $\phi_1$ is surjective then $\phi_3$ is injective.
Snake Lemma
Let $A$ be a commutative ring with unity. Let: ::$\begin{xy}\xymatrix@L+2mu@+1em{ &  M_1 \ar[r]_*{\alpha_1}      \ar[d]^*{\phi_1} &  M_2 \ar[r]_*{\alpha_2}      \ar[d]^*{\phi_2} &  M_3 \ar[d]^*{\phi_3}      \ar[r] &   0 \\  0 \ar[r] &  N_1 \ar[r]_*{\beta_1} &  N_2 \ar[r]_*{\beta_2} &  N_3 & }\end{xy}$ be a commutative diagram of $A$-modules. Suppose that the rows are exact. Then we have a commutative diagram: ::$\begin{xy}\xymatrix@L+2mu@+1em{ &  \ker \phi_1 \ar[r]_*{\tilde\alpha_1}      \ar[d]^*{\iota_1} &  \ker \phi_2 \ar[r]_*{\tilde\alpha_2}      \ar[d]^*{\iota_2} &  \ker \phi_3      \ar[d]^*{\iota_3} & \\ &  M_1 \ar[r]_*{\alpha_1}      \ar[d]^*{\phi_1} &  M_2 \ar[r]_*{\alpha_2}      \ar[d]^*{\phi_2} &  M_3 \ar[d]^*{\phi_3}      \ar[r] &   0 \\  0 \ar[r] &  N_1 \ar[r]_*{\beta_1}      \ar[d]^*{\pi_1} &  N_2 \ar[r]_*{\beta_2}      \ar[d]^*{\pi_2} &  N_3 \ar[d]^*{\pi_3} \\ &  \operatorname{coker} \phi_1 \ar[r]_*{\bar\beta_1} &  \operatorname{coker} \phi_2 \ar[r]_*{\bar\beta_2} &  \operatorname{coker} \phi_3 & }\end{xy}$ where: *For $i=1,2,3$, $\iota_i$ is the inclusion mapping *For $i=1,2,3$, $\pi_i$ is the canonical epimorphism *For $i = 1,2$, $\tilde\alpha_i = \left({ \alpha_i }\right)_{| \ker \phi_i}$ *For $i = 1,2$, $\bar\beta_i$ is defined by: ::$\forall\; n_i + \operatorname{im}\phi_i \in \operatorname{coker}\phi_i : \bar\beta_i\left({ n_i + \operatorname{im}\phi_i }\right) = \beta_i\left({ n_i }\right) + \operatorname{im} \phi_{i+1}$ Moreover there exists a morphism $\delta : \ker\phi_3 \to \operatorname{coker}\phi_1$ such that we have an exact sequence: ::$\begin{xy}\xymatrix@L+2mu@+1em{  \ker \phi_1 \ar[r]_*{\tilde\alpha_1} &  \ker \phi_2 \ar[r]_*{\tilde\alpha_2} &  \ker\phi_3 \ar[r]_*{\delta} &   \operatorname{coker}\phi_1 \ar[r]_*{\bar\beta_1} &  \operatorname{coker}\phi_2 \ar[r]_*{\bar\beta_2} &  \operatorname{coker}\phi_3 }\end{xy}$
Hall's Marriage Theorem/Finite Set
Let $\SS = \sequence {S_k}_{k \in I}$ be a finite indexed family of finite sets. For each $F \subseteq I$, let $\displaystyle Y_F = \bigcup_{k \mathop \in F} S_k$. Let $Y = Y_I$. Then the following are equivalent: :$(1): \quad \SS$ satisfies the '''marriage condition''': for each subset $F$ of $I$, $\card F \le \card {Y_F}$. :$(2): \quad$ There exists an injection $f: I \to Y$ such that $\forall k \in I: \map f k \in S_k$.
Connecting Homomorphism is Functorial
Let $A$ be a commutative ring with unity. Let: ::$\begin{xy}\xymatrix{ &&& M_1 \ar@{->}[rr]      \ar@{->}[dl]      \ar@{->}[dd]|!{[d];[d]}\hole && M_2 %      \ar@{->}[rr]      \ar@{->}[dl]      \ar@{->}[dd]|!{[d];[d]}\hole && M_3 \ar@{->}[dl]       \ar@{->}[dd]|!{[d];[d]}\hole       \ar@{->}[rr] && 0 \\ && M_1' \ar@{->}[rr]    \ar@{->}[dd] && M_2' \ar@{->}[rr]      \ar@{->}[dd] && M_3' \ar@{->}[dd]        \ar@{->}[rr] && 0 \\ & 0 \ar@{->}[rr]|!{[r];[r]}\hole && N_1 \ar@{->}[rr]|!{[r];[r]}\hole      \ar@{->}[dl]_{F’} && N_2 \ar@{->}[rr]|!{[r];[r]}\hole      \ar@{->}[dl] && N_3 \ar@{->}[dl]_{F’’} \\ % 0 \ar@{->}[rr] && N_1' \ar@{->}^(.65){e’}[rr] && N_2' \ar@{->}[rr] && N_3' }\end{xy}$ be a commutative diagram of $A$-modules. Suppose that the rows are exact.
Law of Cosines/Proof 3
=== Lemma: Right Triangle === {{:Law of Cosines/Right Triangle}} === Acute Triangle === {{:Law of Cosines/Proof 3/Acute Triangle}} === Obtuse Triangle === {{:Law of Cosines/Proof 3/Obtuse Triangle}}
Hall's Marriage Theorem/General Set
Let $\SS = \sequence {S_k}_{k \in I}$ be an indexed family of finite sets. For each $F \subseteq I$, let $\displaystyle Y_F = \bigcup_{k \mathop \in F} S_k$. Let $Y = Y_I$. Then the following are equivalent: :$(1): \quad \SS$ satisfies the '''marriage condition''': for each finite subset $F$ of $I$, $\card F \le \card {Y_F}$. :$(2): \quad$ There exists an injection $f: I \to Y$ such that $\forall k \in I: \map f k \in S_k$.
Branch of Finite Propositional Tableau is Finite
Let $T$ be a finite propositional tableau. Let $\Gamma$ be a branch of $T$. Then $\Gamma$ is a finite branch.
Tableau Confutation is Finished
Let $T$ be a tableau confutation. Then $T$ is a finished tableau.
Expression of Vector as Linear Combination from Basis is Unique/General Result
Let $V$ be a vector space over a division ring $R$. Let $B$ be a basis for $V$. Let $x \in V$. Then there is a unique finite subset $C$ of $R \times B$ such that: :$\displaystyle x = \sum_{\left({r, v}\right) \mathop \in C} r \cdot v$ :$\forall \left({r, v}\right) \in C: r \ne 0_R$
Axiom:Axiom of Choice for Finite Sets/Proof from Ordering Principle
Suppose that the Ordering Principle holds. Let $\SS$ be a non-empty set of finite non-empty sets. Then there exists a choice function for $\SS$.
Axiom:Axiom of Choice for Finite Sets/Proof from Hall's Marriage Theorem
Suppose that Hall's Marriage Theorem holds for all sets of finite sets. Let $\SS$ be a non-empty set of finite, non-empty sets. Then there exists a choice function for $\SS$.
Dependent Choice for Finite Sets
Let $\RR$ be a binary relation on a non-empty set $S$. For each $a \in S$, let $C_a = \set {b \in S: a \mathrel \RR b }$ Suppose that: :For all $a \in S$, $C_a$ is a non-empty finite set. Let $s \in S$. Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that: :$x_0 = s$ :$\forall n \in \N: x_n \mathrel \RR x_{n+1}$
Dependent Choice (Fixed First Element)
Let $\mathcal R$ be a binary relation on a non-empty set $S$. Suppose that: :$\forall a \in S: \exists b \in S: a \mathrel{\mathcal R} b$ that is, that $\mathcal R$ is a left-total relation (specifically a ''serial relation''). Let $s \in S$. Then there exists a sequence $\left\langle{x_n}\right\rangle_{n \in \N}$ in $S$ such that: :$x_0 = s$ :$\forall n \in \N: x_n \mathrel{\mathcal R} x_{n+1}$
Locally Finite Connected Graph is Countable
Let $G = \struct {V, E}$ be a graph which is connected and locally finite. Then $G$ has countably many vertices and countably many edges.
Law of Cosines/Right Triangle
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that: : $a$ is opposite $A$ : $b$ is opposite $B$ : $c$ is opposite $C$. Let $\triangle ABC$ be a right triangle such that $\angle A$ is right. Then: :$c^2 = a^2 + b^2 - 2 a b \cos C$
Minimal Infinite Successor Set is Minimal
The minimal infinite successor set $\omega$ is a subset of every infinite successor set.
Law of Cosines/Proof 3/Acute Triangle
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that: : $a$ is opposite $A$ : $b$ is opposite $B$ : $c$ is opposite $C$. Let $\triangle ABC$ be an acute triangle. Then: :$c^2 = a^2 + b^2 - 2a b \cos C$
Law of Cosines/Proof 3/Obtuse Triangle
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that: : $a$ is opposite $A$ : $b$ is opposite $B$ : $c$ is opposite $C$. Let $\triangle ABC$ be an obtuse triangle such that $A$ is obtuse Then: :$c^2 = a^2 + b^2 - 2a b \cos C$
König's Lemma/Countable
Let $G = \struct {V, E}$ be a graph with countably infinitely many vertices which is connected and is locally finite. Then every vertex of $G$ lies on a path of infinite length.
Schröder Rule/Proof 2
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 Logical Consistence
Let $\mathbf H$ be a countable set (either finite or infinite) of WFFs of propositional logic. The following statements are logically equivalent: :$(1): \quad$ $\mathbf H$ has a model. :$(2): \quad$ $\mathbf H$ is consistent for the proof system of propositional tableaus. :$(3): \quad$ $\mathbf H$ has no tableau confutation.
Paths from Vertex of Minimal Length form Tree
Let $G = \left({V, E}\right)$ be a simple graph. Let $r \in V$ be a vertex in $G$. Let $P$ be the set of finite open paths beginning at $r$ that are of minimal length to reach their endpoints. {{refactor|Extract the below sentence and write it as a definition of a "minimal length path".}} That is, an open path $p$ beginning at $r$ and ending at some vertex $s$ is in $P$  there is no path $q$ beginning at $r$ and ending at $s$ such that the length of $q$ is less than the length of $p$. Let $p, q \in P$. Let $\left\{{p, q}\right\} \in E'$  either: :$q$ is formed by extending $p$ with one edge and one vertex of $G$ or: :$p$ is formed by extending $q$ with one edge and one vertex of $G$. Then $T = \left({P, E'}\right)$ is a tree.
Finite Sequences in Set Form Acyclic Graph
Let $S$ be a set. Let $V$ be the set of finite sequences in $S$. Let $E$ be the set of unordered pairs $\{p, q\}$ of elements of $V$ such that either: : $q$ is formed by extending $p$ by one element or : $p$ is formed by extending $q$ by one element. That is: : $| \operatorname{Dom}(p) * \operatorname{Dom}(q) | = 1$, where $*$ is symmetric difference and : $p \restriction D = q \restriction D$, where $D = \operatorname{Dom}(p) \cap \operatorname{Dom}(q)$ Then $T = (V, E)$ is an acyclic graph.
Connected Vertices are Connected by Path
Let $G = \struct {V, E}$ be a simple graph. Let $x, y \in V$. Let there exist a walk $w: \N_n \to V$ from $x$ to $y$. Then there exists a subsequence $z_n$ of $w$ which is a path from $x$ to $y$.
Negation as Implication of Bottom
$p \implies \bot \dashv\vdash \neg p$
Clavius's Law/Formulation 1/Proof 2
From Peirce's Law: :$\left({p \implies q}\right) \implies p \vdash p$ follows Clavius's Law: :$\neg p \implies p \vdash p$
Clavius's Law implies Law of Excluded Middle
From Clavius's Law: :$\neg p \implies p \vdash p$ follows the Law of Excluded Middle: :$\vdash p \lor \neg p$
Parity of Integer equals Parity of its Square/Even
Let $p \in \Z$ be an integer. Let $p$ be even. Then $p^2$ is also even.
Parity of Integer equals Parity of its Square/Odd
Let $p \in \Z$ be an integer. Let $p$ be odd. Then $p^2$ is also odd.
Negation of Excluded Middle is False/Form 1
$\neg (p \lor \neg p) \vdash \bot$
Negation of Excluded Middle is False/Form 2
$\vdash \neg \neg (p \lor \neg p)$
Double Negation Elimination implies Law of Excluded Middle/Proof 2
Let the Law of Double Negation Elimination be supposed to hold: :$\neg \neg p \vdash p$ Then the Law of Excluded Middle likewise holds: :$\vdash p \lor \neg p$
Injection iff Left Cancellable/Necessary Condition
Let $f: Y \to Z$ be an injection. Then $f$ is left cancellable.
Injection iff Left Cancellable/Sufficient Condition
Let $f: Y \to Z$ be a mapping which is left cancellable. Then $f$ is an injection.
Peirce's Law implies Law of Excluded Middle
From Peirce's Law: :$\left({p \implies q}\right) \implies p \vdash p$ follows the Law of Excluded Middle: :$\vdash p \lor \neg p$
Pseudocomplemented Lattice is Bounded
Let $(L, \wedge, \vee, \preceq)$ be a pseudocomplemented lattice. Then $(L, \wedge, \vee, \preceq)$ is a bounded lattice.
Topology forms Complete Lattice
Let $\struct {X, \tau}$ be a topological space. Then $\struct {\tau, \subseteq}$ is a complete lattice.
Law of Excluded Middle implies Peirce's Law
From the Law of Excluded Middle follows Peirce's Law: :$\left({p \lor \neg p}\right) \vdash \left({\left({p \implies q}\right) \implies p}\right) \implies p$
Inverse of Composite Bijection/Proof 2
Let $f$ and $g$ be bijections. Then: :$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$ and $f^{-1} \circ g^{-1}$ is itself a bijection.
Sierpiński's Theorem
Let $\left({S, \tau}\right)$ be a compact connected Hausdorff space. Let $\left\{{F_n: n \in \N}\right\}$ be a pairwise disjoint closed cover of $S$. {{explain|In context it's obvious, but worth mentioning that $F_n$ is a finite cover as well?}} Then $F_n = S$ for some $n \in \N$.
Sierpiński's Theorem/Lemma 1
Let $\struct {S, \tau}$ be a compact connected Hausdorff space. Let $A$ be a closed, non-empty proper subset of $S$. Let $C$ be a component of $A$. Then: :$C \cap \partial A \ne \O$ where $\partial A$ denotes the boundary of $A$.
Quasicomponent of Compact Hausdorff Space is Connected
Let $\left({X, \tau}\right)$ be a compact Hausdorff space. Let $C$ be a quasicomponent of $\left({X, \tau}\right)$. Then $C$ is connected.
Quasicomponents and Components are Equal in Compact Hausdorff Space
Let $T = \left({S, \tau}\right)$ be a compact Hausdorff space. Then for each $A \subseteq S$: $A$ is a component of $S$  $A$ is a quasicomponent of $S$.
Derived Set in T1 Space is Closed
Let $\struct {X, \tau}$ be a $T_1$ space. Let $S \subseteq X$. Let $S'$ be the derived set of $S$. Then $S'$ is closed.
Multiplicative Inverse in Monoid of Integers Modulo m
Let $\struct {\Z_m, \times_m}$ be the multiplicative monoid of integers modulo $m$. Then: :$\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ : :$k \perp m$
Element Commutes with Square in Group
Let $\left({G, \circ}\right)$ be a group. Let $x \in G$. Then $x$ commutes with $x \circ x$.
Element Commutes with Square in Semigroup
Let $\left({S, \circ}\right)$ be a semigroup. Let $x \in S$. Then $x$ commutes with $x \circ x$.
Subsets of Disjoint Sets are Disjoint
Let $S$ and $T$ be disjoint sets. Let $S' \subseteq S$ and $T' \subseteq T$. Then $S'$ and $T'$ are disjoint.
Network with Positive Integer Mapping is Multigraph
Let $N = \struct {V, E, w}$ be a network whose weights are all strictly positive integers. Then $N$ can be represented as a multigraph. Conversely, any multigraph can be expressed as a network whose weights are all strictly positive integers.
Complete Graph is Regular
Let $K_p$ be the complete graph of order $p$. Then $K_p$ is $p-1$-regular.
No Simple Graph is Perfect
Let $G$ be a simple graph whose order is $2$ or greater. Then $G$ is not perfect.
Tukey-Teichmüller Lemma
Let $\FF$ be a non-empty set of finite character. Then $\FF$ has an element which is maximal with respect to the set inclusion relation.
Circuit has Three Edges or More
Let $G$ be a simple graph. Let $C$ be a circuit in $G$. Then $C$ has at least $3$ edges.
Dirac's Theorem/Proof 1
If a connected graph $G$ has $n \ge 3$ vertices and the degree of each vertex is at least $\dfrac n 2$, then $G$ is Hamiltonian.
Dirac's Theorem/Proof 2
If a connected graph $G$ has $n \ge 3$ vertices and the degree of each vertex is at least $\dfrac n 2$, then $G$ is Hamiltonian.
Graph of Cube is Hamiltonian
The graph of the cube is Hamiltonian.
Graph of Icosahedron is Hamiltonian
The graph of the icosahedron is Hamiltonian.
Path in Tree is Unique/Necessary Condition
Let $T$ be a tree. Then there is exactly one path between any two vertices.
Path in Tree is Unique/Sufficient Condition
Let $T$ be a graph. Let $T$ be such that between any two vertices there is exactly one path. Then $T$ is a tree.
Size of Tree is One Less than Order/Necessary Condition
Let $T$ be a tree of order $n$. Then the size of $T$ is $n-1$.
Size of Tree is One Less than Order/Sufficient Condition
Let $T$ be a connected simple graph of order $n$. Let the size of $T$ be $n-1$. Then $T$ is a tree.
Size of Tree is One Less than Order/Necessary Condition/Induction Step/Proof 1
Let the following hold: :A tree of order $k$ is of size $k - 1$. Then this holds: :A tree of order $k + 1$  is of size $k$.
Size of Tree is One Less than Order/Necessary Condition/Induction Step/Proof 2
Let the following hold: :For all $j \le k$, a tree of order $j$ is of size $j-1$. Then this holds: :A tree of order $k+1$  is of size $k$.
Kruskal's Algorithm produces Minimum Spanning Tree
Kruskal's Algorithm produces a minimum spanning tree.
Regular Graph is Tree iff Complete Graph of Order 2
Let $G$ be a non-edgeless regular graph. Then $G$ is a tree  $G$ is $K_2$, the complete graph of order $2$.
Perimeter of Circle
The perimeter $C$ of a circle with radius $r$ is given by: : $C = 2 \pi r$
Euler's Criterion
Let $p$ be an odd prime. Let $a \not \equiv 0 \pmod p$. Then: {{begin-eqn}} {{eqn | l = a^{\frac {p-1} 2}       | o = \equiv       | r = 1       | rr= \pmod p       | c =  $a$ is a quadratic residue of $p$ }} {{eqn | l=a ^{\frac {p-1} 2}       | o = \equiv       | r = -1       | rr= \pmod p       | c =  $a$ is a quadratic non-residue of $p$. }} {{end-eqn}}
Value of Degree in Radians
The value of a degree in radians is given by: :$1 \degrees = \dfrac {\pi} {180} \radians \approx 0.01745 \ 32925 \ 19943 \ 29576 \ 92 \ldots \radians$ {{OEIS|A019685}}
Difference of Two Squares/Geometric Proof
:$\forall x, y \in \R: x^2 - y^2 = \paren {x + y} \paren {x - y}$
Area of Parallelogram/Square
The area of a square equals the product of one of its bases and the associated altitude.
Area of Parallelogram/Rectangle
The area of a rectangle equals the product of one of its bases and the associated altitude.