hcju/proofwikips huggingface dataset:
Closed Set in Linearly Ordered Space
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Let $C$ be a subset of $X$. Then $C$ is closed in $X$  for all non-empty subsets $S$ of $C$: :If $s \in X$ is a supremum or infimum of $S$ in $X$, then $s \in C$.
Factor Principles/Disjunction on Left/Formulation 2
: $\vdash \left({p \implies q}\right) \implies \left({\left({r \lor p}\right) \implies \left ({r \lor q}\right)}\right)$
Factor Principles/Disjunction on Right/Formulation 2
: $\vdash \left({p \implies q}\right) \implies \left({\left({p \lor r}\right) \implies \left ({q \lor r}\right)}\right)$
Factor Principles/Disjunction on Right/Formulation 1
:$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$
Factor Principles/Disjunction on Left/Formulation 1
: $p \implies q \vdash \left({r \lor p}\right) \implies \left ({r \lor q}\right)$
Disjunction of Implications
:$\vdash \paren {p \implies q} \lor \paren {q \implies r}$
Principle of Composition/Formulation 1/Forward Implication
:$\paren {p \implies r} \lor \paren {q \implies r} \vdash \paren {p \land q} \implies r$
Principle of Composition/Formulation 1/Reverse Implication
:$\paren {p \land q} \implies r \vdash \paren {p \implies r} \lor \paren {q \implies r}$
Principle of Composition/Formulation 2
:$\left({\left({p \implies r}\right) \lor  \left({q \implies r}\right)}\right) \iff \left({\left({p \land q}\right) \implies r}\right)$
Inversion Mapping on Topological Group is Homeomorphism
Let $T = \struct {G, \circ, \tau}$ be a topological group. Let $\phi: G \to G$ be the inversion mapping of $T$. Then $\phi$ is a homeomorphism.
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Forward Implication
: $\left({\neg p \land \neg q}\right) \implies \left({\neg \left({p \lor q}\right)}\right)$
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Reverse Implication
: $\left({\neg \left({p \lor q}\right)}\right) \implies \left({\neg p \land \neg q}\right)$
Equivalence of Definitions of Generalized Ordered Space/Definition 2 implies Definition 1
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 2: {{:Definition:Generalized Ordered Space/Definition 2}} Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by Definition 1: {{:Definition:Generalized Ordered Space/Definition 1}}
Inverse Image of Convex Set under Monotone Mapping is Convex
Let $\left({X, \le}\right)$ and $\left({Y, \preceq}\right)$ be ordered sets. Let $f: X \to Y$ be a monotone mapping. Let $C$ be a convex subset of $Y$. Then $f^{-1} \left[{C}\right]$ is convex in $X$.
Conjunction with Law of Excluded Middle
:$\vdash p \iff \paren {p \land q} \lor \paren {p \land \neg q}$
Proof by Cases with Contradiction
:$\vdash p \iff \left({p \lor q}\right) \land \left({p \lor \neg q}\right)$
Ray is Convex
Let $\left({S, \preceq}\right)$ be an ordered set. Let $I$ be a ray, either open or closed. Then $I$ is convex in $S$.
Intersection of Convex Sets is Convex Set (Order Theory)
Let $\left({S, \preceq}\right)$ be an ordered set. Let $\mathcal C$ be a set of convex sets in $S$. Then $\displaystyle \bigcap \mathcal C$ is convex.
Interval of Ordered Set is Convex
Let $\struct {S, \preceq}$ be a totally ordered set. Let $I$ be an interval: be it open, closed, or half-open in $S$. Then $I$ is convex in $S$.
Upper and Lower Closures are Convex
Let $\left({S, \preceq}\right)$ be an ordered set. Let $a \in S$. Then $a^\succeq$, $a^\succ$, $a^\preceq$, and $a^\prec$ are convex in $S$.
Transitive Closure of Symmetric Relation is Symmetric
Let $S$ be a set. Let $\mathcal R$ be a symmetric relation on $S$. Let $\mathcal T$ be the transitive closure of $\mathcal R$. The $\mathcal T$ is symmetric.
Transitive Closure of Reflexive Relation is Reflexive
Let $S$ be a set. Let $\mathcal R$ be a reflexive relation on $S$. Let $\mathcal T$ be the transitive closure of $\mathcal R$. Then $\mathcal T$ is reflexive.
Transitive Closure of Reflexive Symmetric Relation is Equivalence
Let $S$ be a set. Let $\mathcal R$ be a symmetric and reflexive relation on $S$. Then the transitive closure of $\mathcal R$ is an equivalence relation.
Union of Overlapping Convex Sets in Toset is Convex
Let $\left({ S, \preceq }\right)$ be a totally ordered set. Let $U$ and $V$ be convex sets in $S$. Let $U \cap V \ne \varnothing$. Then $U \cup V$ is also convex.
Rule of Material Equivalence
==== Formulation 1 ==== {{:Rule of Material Equivalence/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Material Equivalence/Formulation 2}}
Union of Overlapping Convex Sets in Toset is Convex/Infinite Union
Let $\left({S, \preceq}\right)$ be a totally ordered set. Let $\mathcal A$ be a set of convex subsets of $S$. For any $P, Q \in \mathcal A$, let there be elements $C_0, \dotsc, C_n \in \mathcal A$ such that: :$C_0 = P$ :$C_n = Q$ :For $k = 0, \dots, n-1: C_k \cap C_{k+1} \ne \varnothing$ Then $\bigcup \mathcal A$ is convex in $S$.
Rule of Material Equivalence/Formulation 2
:$\vdash \left({p \iff q}\right) \iff \left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)$
Non-Equivalence as Disjunction of Conjunctions
==== Formulation 1 ==== {{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1}} ==== Formulation 2 ==== {{:Non-Equivalence as Disjunction of Conjunctions/Formulation 2}}
Continuous Involution is Homeomorphism
Let $\struct {S, \tau}$ be a topological space. Let $f: S \to S$ be a continuous involution. Then $f$ is a homeomorphism.
Involution is Permutation
Let $S$ be a set. Let $f: S \to S$ be an involution. Then $f$ is a permutation.
Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof
: $\neg \left ({p \iff q}\right) \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof
: $\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right) \vdash \neg \left ({p \iff q}\right)$
Non-Equivalence as Disjunction of Conjunctions/Formulation 2
:$\vdash \paren {\neg \paren {p \iff q} } \iff \paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} }$
Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication
:$\neg \paren {p \iff q} \vdash \paren {p \iff \neg q}$
Non-Equivalence as Equivalence with Negation/Formulation 1
:$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$
Sign of Function Matches Sign of Definite Integral
Let $f$ be a real function continuous on some closed interval $\closedint a b$, where $a < b$. Then: :If $\forall x \in \closedint a b: \map f x \ge 0$ then $\displaystyle \int_a^b \map f x \rd x \ge 0$ :If $\forall x \in \closedint a b: \map f x > 0$ then $\displaystyle \int_a^b \map f x \rd x > 0$ :If $\forall x \in \closedint a b: \map f x \le 0$ then $\displaystyle \int_a^b \map f x \rd x \le 0$ :If $\forall x \in \closedint a b: \map f x < 0$ then $\displaystyle \int_a^b \map f x \rd x < 0$
Non-Equivalence as Equivalence with Negation/Formulation 2
: $\vdash \neg \left ({p \iff q}\right) \iff \left({p \iff \neg q}\right)$
Symmetric Closure of Relation Compatible with Operation is Compatible
Let $\struct {S, \circ}$ be a magma. Let $\RR$ be a relation compatible with $\circ$. Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$. Then $\RR^\leftrightarrow$ is compatible with $\circ$.
Singleton is Convex Set (Order Theory)
Let $\left({S, \preceq}\right)$ be an ordered set. Let $x \in S$. Then the singleton $\left\{{x}\right\}$ is convex.
Rule of Transposition/Variant 1/Formulation 2
:$\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p}$
Rule of Transposition/Variant 1
==== Formulation 1 ==== {{:Rule of Transposition/Variant 1/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Transposition/Variant 1/Formulation 2}}
Rule of Transposition/Variant 1/Formulation 2/Reverse Implication
: $\vdash \left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)$
Rule of Transposition/Variant 1/Formulation 1
:$p \implies \neg q \dashv \vdash q \implies \neg p$
Rule of Transposition/Variant 2
==== Formulation 1 ==== {{:Rule of Transposition/Variant 2/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Transposition/Variant 2/Formulation 2}}
Rule of Transposition/Variant 2/Formulation 1
: $\neg p \implies q \dashv \vdash \neg q \implies p$
Rule of Transposition/Variant 2/Formulation 1/Proof 2
:$\neg p \implies q \dashv \vdash \neg q \implies p$
Rule of Transposition/Variant 2/Formulation 2
: $\vdash \left({\neg p \implies q}\right) \iff \left({\neg q \implies p}\right)$
Boundary of Polygon is Topological Boundary
Let $P$ be a polygon embedded in $\R^2$. Denote the boundary of $P$ as $\partial P$. Let $\Int P$ and $\Ext P$ denote the interior and exterior of $\partial P$, where $\partial P$ is considered as a Jordan curve. Then the topological boundary of $\Int P$ is equal to $\partial P$, and the topological boundary of $\Ext P$ is equal to $\partial P$.
Jordan Curve and Jordan Arc form Two Jordan Curves
Let $\closedint a b$ denote the closed real interval between $a \in \R, b \in \R: a \le b$. Let $\gamma: \closedint a b \to \R^2$ be a Jordan curve. Let the interior of $\gamma$ be denoted $\Int \gamma$. Let the image of $\gamma$ be denoted $\Img \gamma$. Let $\sigma: \closedint c d \to \R^2$ be a Jordan arc such that: :$\map \sigma c \ne \map \sigma d$ :$\map \sigma c, \map \sigma d \in \Img \gamma$ and: :$\forall t \in \openint c d: \map \sigma t \in \Int \gamma$ Let $t_1 = \map {\gamma^{-1} } {\map \sigma c}$. Let $t_2 = \map {\gamma^{-1} } {\map \sigma d}$. Let $t_1 < t_2$. Define: :$-\sigma: \closedint c d \to \Img \sigma$ by $-\map \sigma t = \map \sigma {c + d - t}$ Let $*$ denote concatenation of paths. Let $\gamma \restriction_{\closedint a {t_1} }$ denote the restriction of $\gamma$ to $\closedint a {t_1}$. Define: :$\gamma_1 = \gamma {\closedint a {t_1} } * \sigma * \gamma {\restriction_{\closedint {t_2} b} }$ Define: :$\gamma_2 = \gamma {\closedint a {t_1} } * \paren {-\sigma}$ Then $\gamma_1$ and $\gamma_2$ are Jordan curves such that: :$\Int {\gamma_1} \subseteq \Int \gamma$ and: :$\Int {\gamma_2} \subseteq \Int \gamma$
Restriction of Total Ordering is Total Ordering
Let $\struct {S, \preceq}$ be a total ordering. Let $T \subseteq S$. Let $\preceq \restriction_T$ be the restriction of $\preceq$ to $T$. Then $\preceq \restriction_T$ is a total ordering of $T$.
Ordering Cycle implies Equality/General Case
Let $\left({S,\preceq}\right)$ be an ordered set. Let $x_0, x_1, \dots, x_n \in S$. Suppose that for $k = 0, 1, \dots, n - 1: x_k \preceq x_{k+1}$. Suppose also that $x_n \preceq x_0$. Then $x_0 = x_1 = \dots = x_n$.
Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements
Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$ be a finite, non-empty subset of $S$. Then $T$ has a maximal element and a minimal element.
Star Convex Set is Path-Connected
Let $A$ be a star convex subset of a vector space $V$ over $\R$ or $\C$. Then $A$ is path-connected.
Minimal Element of Chain is Smallest Element
Let $\left({S, \preceq}\right)$ be an ordered set. Let $C$ be a chain in $S$. Let $m$ be a minimal element of $C$. Then $m$ is the smallest element of $C$.
Maximal Element of Chain is Greatest Element
Let $\left({S, \preceq}\right)$ be an ordered set. Let $C$ be a chain in $S$. Let $m$ be a maximal element of $C$. Then $m$ is the greatest element of $C$.
Modus Ponendo Ponens/Variant 3
:$\vdash \paren {\paren {p \implies q} \land p} \implies q$
Rule of Material Equivalence/Formulation 1/Proof 2
:$p \iff q \dashv \vdash \left({p \implies q}\right) \land \left({q \implies p}\right)$
Rule of Material Equivalence/Formulation 2/Proof 1
:$\vdash \left({p \iff q}\right) \iff \left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)$
Rule of Material Equivalence/Formulation 2/Proof 2
:$\vdash \left({p \iff q}\right) \iff \left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)$
Modus Tollendo Ponens/Variant/Formulation 2/Proof 2
: $\vdash \left({p \lor q}\right) \iff \left({\neg p \implies q}\right)$
Rule of Transposition/Formulation 2/Proof 2
: $\vdash \left({p \implies q}\right) \iff \left({\neg q \implies \neg p}\right)$
Convex Set Characterization (Order Theory)
Let $\left({S, \preceq, \tau}\right)$ be an ordered set. Let $C \subseteq S$.  {{begin-axiom}} {{axiom | n = 1         | t = $C$ is convex. }} {{axiom | n = 2         | t = $C$ is the intersection of an upper set with a lower set. }} {{axiom | n = 3         | t = $C$ is the intersection of its upper closure with its lower closure. }} {{end-axiom}}
Constructive Dilemma/Formulation 3
:$\paren {p \implies q} \land \paren {r \implies s}, p \lor r \vdash q \lor s$
Upper Closure is Upper Set
Let $(S, \preceq, \tau)$ be an ordered set. Let $T$ be a subset of $S$. Let $U$ be the upper closure of $T$. Then $U$ is an upper set.
Destructive Dilemma/Formulation 1/Proof 3
:$p \implies q, r \implies s \vdash \neg q \lor \neg s \implies \neg p \lor \neg r$
Open Ray is Open in GO-Space/Definition 1
Let $\struct {S, \preceq, \tau}$ 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 $p$.
Upper and Lower Closures of Open Set in GO-Space are Open
Let $\left({X, \preceq, \tau}\right)$ be a Generalized Ordered Space/Definition 1. Let $A$ be open in $X$. Then the upper and lower closures of $A$ are open.
Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 1: {{:Definition:Generalized Ordered Space/Definition 1}} Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by Definition 3: {{:Definition:Generalized Ordered Space/Definition 3}}
Upper Set is Convex
Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$ be an upper set. Then $T$ is convex in $S$.
Lower Set is Convex
Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$ be a lower set. Then $T$ is convex in $S$.
GO-Space Embeds Densely into Linearly Ordered Space
Let $\left({Y, \preceq, \tau}\right)$ be a generalized ordered space (GO-space) by Definition 3. That is: : let $\left({Y, \tau}\right)$ be a Hausdorff space and: : let $\tau$ have a sub-basis consisting of upper sets and lower sets relative to $\preceq$. Then $\left({Y, \preceq, \tau}\right)$ is a GO-space by Definition 2. That is, there is a linearly ordered space $\left({X, \preceq', \tau'}\right)$ and a mapping from $Y$ to $X$ which is a order embedding and a topological embedding.
Upper Set with no Minimal Element
Let $\struct {S, \preceq}$ be an ordered set. Let $U \subseteq S$. Then: :$U$ is an upper set in $S$ with no minimal element : :$\displaystyle U = \bigcup \set {u^\succ: u \in U}$ where $u^\succ$ is the strict upper closure of $u$.
Lower Sets in Totally Ordered Set form Nest
Let $\left({S, \preceq}\right)$ be a totally ordered set. Let $\mathcal L$ be a set of lower sets in $S$. Then $\mathcal L$ is a nest. That is, $\mathcal L$ is totally ordered by $\subseteq$.
Exclusive Or is Self-Inverse
:$\paren {p \oplus q} \oplus q \dashv \vdash p$ where $\oplus$ denotes the exclusive or operator.
Conjunction has no Inverse
Let $\land$ denote the conjunction operation of propositional logic. Then there exists no binary logical connective $\circ$ such that: :$(1): \quad \forall p, q \in \left\{{T, F}\right\}: \left({p \land q}\right) \circ q = p$
Disjunction has no Inverse
Let $\lor$ denote the disjunction operation of propositional logic. Then there exists no binary logical connective $\circ$ such that: :$(1): \quad \forall p, q \in \left\{{T, F}\right\}: \left({p \lor q}\right) \circ q = p$
Binary Logical Connectives with Inverse
Let $\circ$ be a binary logical connective. Then there exists another binary logical connective $*$ such that: :$\forall p, q \in \set {\F, \T}: \paren {p \circ q} * q \dashv \vdash p \dashv \vdash q * \paren {p \circ q}$  $\circ$ is either: :$(1): \quad$ the exclusive or operator or: :$(2): \quad$ the biconditional operator. That is, the only truth functions that have an inverse operation are the exclusive or and the biconditional.
Biconditional is Self-Inverse
:$\paren {p \iff q} \iff q \dashv \vdash p$ where $\iff$ denotes the biconditional operator.
Finite Chain is Order-Isomorphic to Finite Ordinal
Let $\left({S, \preceq}\right)$ be an ordered set. Let $C$ be a finite chain in $S$. Then for some finite ordinal $\mathbf n$: :$\left({C, {\preceq \restriction_C} }\right)$ is order-isomorphic to $\mathbf n$. That is: :$\left({C, {\preceq \restriction_C} }\right)$ is order-isomorphic to $\N_n$ where $\N_n$ is the initial segment of $\N$ determined by $n$: :$\N_n = \left\{ {k \in \N: k < n}\right\} = \left\{ {0, 1, \ldots, n - 1}\right\}$
Complete Linearly Ordered Space is Compact
Let $\left({X, \preceq, \tau}\right)$ be a linearly ordered space. Let $\left({X, \preceq}\right)$ be a complete lattice. Then $\left({X, \tau}\right)$ is compact.
Condition for Well-Foundedness/Reverse Implication
Let $\left({S, \preceq}\right)$ be an ordered set. Suppose that there is no infinite sequence $\left \langle {a_n}\right \rangle$ of elements of $S$ such that $\forall n \in \N: a_{n+1} \prec a_n$. Then $\left({S, \preceq}\right)$ is well-founded.
Condition for Well-Foundedness/Reverse Implication/Proof 1
Let $\struct {S, \preceq}$ be an ordered set. Suppose that there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that $\forall n \in \N: a_{n + 1} \prec a_n$. Then $\struct {S, \preceq}$ is well-founded.
Inversion Mapping on Ordered Group is Dual Order-Isomorphism
Let $\struct {G, \circ, \preceq}$ be an ordered group. Let $\iota: G \to G$ be the inversion mapping, defined by $\map \phi x = x^{-1}$. Then $\iota$ is a dual order-isomorphism.
Properties of Ordered Group/OG2/Proof 2
Let $\left({G, \circ, \preceq}\right)$ be an ordered group with identity $e$. Let $\prec$ be the reflexive reduction of $\preceq$. Let $x, y \in G$. Then the following equivalences hold: :$(\operatorname{OG}2.1):\quad x \preceq y \iff e \preceq y \circ x^{-1}$ :$(\operatorname{OG}2.2):\quad x \preceq y \iff e \preceq x^{-1} \circ y$ :$(\operatorname{OG}2.3):\quad x \preceq y \iff x \circ y^{-1} \preceq e$ :$(\operatorname{OG}2.4):\quad x \preceq y \iff y^{-1} \circ x \preceq e$ :$(\operatorname{OG}2.1'):\quad x \prec y \iff e \prec y \circ x^{-1}$ :$(\operatorname{OG}2.2'):\quad x \prec y \iff e \prec x^{-1} \circ y$ :$(\operatorname{OG}2.3'):\quad x \prec y \iff x \circ y^{-1} \prec e$ :$(\operatorname{OG}2.4'):\quad x \prec y \iff y^{-1} \circ x \prec e$
Properties of Ordered Group/OG4/Proof 1
Let $\struct {G, \circ, \preceq}$ be an ordered group with identity $e$. Let $x \in G$. Then the following equivalences hold: {{begin-axiom}} {{axiom | n = \operatorname {OG} 4.1         | m = x \preceq e \iff e \preceq x^{-1} }} {{axiom | n = \operatorname {OG} 4.2         | m = e \preceq x \iff x^{-1} \preceq e }} {{axiom | n = \operatorname {OG} 4.1'         | m = x \prec e \iff e \prec x^{-1} }} {{axiom | n = \operatorname {OG} 4.2'         | m = e \prec x \iff x^{-1} \prec e }} {{end-axiom}}
Properties of Ordered Group/OG4/Proof 2
Let $\left({G, \circ, \preceq}\right)$ be an ordered group with identity $e$. Let $x \in G$. Then the following equivalences hold: {{begin-axiom}} {{axiom|n = \operatorname{OG}4.1        |m = x \preceq e \iff e \preceq x^{-1} }} {{axiom|n = \operatorname{OG}4.2        |m = e \preceq x \iff x^{-1} \preceq e }} {{axiom|n = \operatorname{OG}4.1'        |m = x \prec e \iff e \prec x^{-1} }} {{axiom|n = \operatorname{OG}4.2'        |m = e \prec x \iff x^{-1} \prec e }} {{end-axiom}}
Properties of Ordered Group/OG5/Proof 1
Let $\left({G, \circ, \preceq}\right)$ be an ordered group with identity $e$. Let $\prec$ be the reflexive reduction of $\preceq$. Let $x \in G$. Let $n \in \N_{>0}$ be a strictly positive integer. Then the following hold: :$x \preceq e \implies x^n \preceq e$ :$e \preceq x \implies e \preceq x^n$ :$x \prec e \implies x^n \prec e$ :$e \prec x \implies e \prec x^n$
Properties of Ordered Group/OG5/Proof 2
Let $\left({G, \circ, \preceq}\right)$ be an ordered group with identity $e$. Let $\prec$ be the reflexive reduction of $\preceq$. Let $x \in G$. Let $n \in \N_{>0}$ be a strictly positive integer. Then the following hold: :$x \preceq e \implies x^n \preceq e$ :$e \preceq x \implies e \preceq x^n$ :$x \prec e \implies x^n \prec e$ :$e \prec x \implies e \prec x^n$
Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing
Let $\left({S, \preceq_1}\right)$ be a totally ordered set. Let $\left({T, \preceq_2}\right)$ be an ordered set. Let $\phi: S \to T$ be a mapping. Then $\phi$ is a dual order embedding  $\phi$ is strictly decreasing. That is: :$\forall x, y \in S: x \preceq_1 y \iff \phi \left({y}\right) \preceq_2 \phi \left({x}\right)$  :$\forall x, y \in S: x \prec_1 y \implies \phi \left({y}\right) \prec_2 \phi \left({x}\right)$
Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication
Let $\struct {S, \preceq_1}$ be a totally ordered set and let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: S \to T$ be a strictly increasing mapping. Then $\phi$ is an order embedding. <section notitle="true" name="proofs"> == Proof 1 == {{:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 1}} == Proof 2 == {{:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 2}} </section>
Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 2
Let $\struct {S, \preceq_1}$ be a totally ordered set and let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: S \to T$ be a strictly increasing mapping. Then $\phi$ is an order embedding.
Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 1
Let $\struct {S, \preceq_1}$ be a totally ordered set. Let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: S \to T$ be a strictly increasing mapping. Then $\phi$ is an order embedding .
Foundational Relation is Antireflexive
Let $\mathcal R$ be a foundational relation on a set or class $A$. Then $\mathcal R$ is antireflexive.
Foundational Relation is Asymmetric
Let $\struct {S, \RR}$ be a relational structure, where $S$ is a set or a proper class. Let $\RR$ be a foundational relation. Then $\RR$ is asymmetric.
Upper Set with no Smallest Element is Open in GO-Space
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space. Let $U$ be an upper set in $S$ with no smallest element. Then $U$ is open in $\struct {S, \preceq, \tau}$.
Lower Set with no Maximal Element
Let $\left({S, \preceq}\right)$ be an ordered set. Let $L \subseteq S$. Then: : $L$ is a lower set in $S$ with no maximal element : : $\displaystyle L = \bigcup \left\{{l^\prec: l \in L }\right\}$ where $l^\prec$ is the strict lower closure of $l$.
Lower Set with no Greatest Element is Open in GO-Space
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space. Let $L$ be a lower set in $S$ with no greatest element. Then $L$ is open in $\struct {S, \preceq, \tau}$.
Lower Set is Dual to Upper Set
Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$. The following are dual statements: :$T$ is a lower set in $S$ :$T$ is an upper set in $S$
Open Ray is Dual to Open Ray
Let $\struct {S, \preceq}$ be a totally ordered set. Let $R$ be an open ray in $\struct {S, \preceq}$. Then $R$ is an open ray in $\struct {S, \succeq}$, where $\succeq$ is the dual ordering of $\preceq$.
Topologies on Set form Complete Lattice
Let $X$ be a non-empty set. Let $\mathcal L$ be the set of topologies on $X$. Then $\left({\mathcal L, \subseteq}\right)$ is a complete lattice.
Complement of Lower Set is Upper Set
Let $\left({S, \preceq}\right)$ be an ordered set. Let $L$ be a lower set. Then $S \setminus L$ is an upper set.