hcju/proofwikips huggingface dataset:
Mind the Gap
Let $\left({S, \preccurlyeq}\right)$ be a totally ordered set. Let $a, b \in S$ with $a \prec b$. Suppose that: :$\left\{ {x \in S: a \prec x \prec b}\right\} = \varnothing$ Then: : $a^\succ = b^\succcurlyeq$ and : $b^\prec = a^\preccurlyeq$ where: : $a^\succ$ is the strict upper closure of $a$ : $b^\succcurlyeq$ is the weak upper closure of $b$ : $b^\prec$ is the strict lower closure of $b$ : $a^\preccurlyeq$ is the weak lower closure of $a$.
Proof by Contradiction/Sequent Form
:$\paren {p \vdash \bot} \vdash \neg p$
Left or Right Inverse of Matrix is Inverse
Let $\mathbf A, \mathbf B$ be square matrices of order $n$ over a commutative ring with unity $\left({R, +, \circ}\right)$. Suppose that: : $\mathbf A \mathbf B = \mathbf I_n$ where $\mathbf I_n$ is the unit matrix of order $n$. Then $\mathbf A$ and $\mathbf B$ are invertible matrices, and furthermore: :$\mathbf B = \mathbf A^{-1}$ where $\mathbf A^{-1}$ is the inverse of $\mathbf A$.
Banach-Alaoglu Theorem
Let $X$ be a separable normed space. Then the closed unit sphere in its dual $X^*$ is weak* sequentially compact.
Empty Product is Terminal Object
Let $\mathbf C$ be a metacategory. Suppose $\mathbf C$ admits a product $\displaystyle \prod \O$ for the empty set. Then $\displaystyle \prod \O$ is a terminal object of $\mathbf C$.
Unary Product for Object is Itself
Let $\mathbf C$ be a metacategory. Let $C$ be an object of $\mathbf C$. Then $\displaystyle \prod \set C = C$, where $\displaystyle \prod$ denotes product.
Vector Cross Product is Orthogonal to Factors
Let $\mathbf a$ and $\mathbf b$ be vectors in the Euclidean space $\R^3$. Let $\mathbf a \times \mathbf b$ denote the vector cross product. Then: :$(1): \quad$ $\mathbf a$ and $\mathbf a \times \mathbf b$ are orthogonal. :$(2): \quad$ $\mathbf b$ and $\mathbf a \times \mathbf b$ are orthogonal.
Squeeze Theorem/Sequences/Metric Spaces
Let $M = \left({S, d}\right)$ be a metric space or pseudometric space. Let $\left\langle{x_n}\right\rangle$ be a sequence in $S$. Let $p \in S$.  Let $\left\langle{r_n}\right\rangle$ be a sequence in $\R_{\ge 0}$. Let $\left\langle{r_n}\right\rangle$ converge to $0$. For each $n$, let $d \left({p, x_n}\right) \le r_n$. Then $\left\langle{x_n}\right\rangle$ converges to $p$.
Squeeze Theorem/Sequences/Linearly Ordered Space
Let $\left({S, \le, \tau}\right)$ be a linearly ordered space. Let $\left\langle{x_n}\right\rangle$, $\left\langle{y_n}\right\rangle$, and $\left\langle{z_n}\right\rangle$ be sequences in $S$. Let $p \in S$. Let $\left\langle{x_n}\right\rangle$ and $\left\langle{z_n}\right\rangle$ both converge to $p$ For each $n$, let $x_n \le y_n \le z_n$. Then $\left\langle{y_n}\right\rangle$ converges to $p$.
Non-Zero Vectors Orthogonal iff Perpendicular
Let $\mathbf u$, $\mathbf v$ be non-zero vectors in the Euclidean space $\R^n$. Then $\mathbf u$ and $\mathbf v$ are orthogonal  they are perpendicular.
Exponential Function is Continuous/Complex
:$\forall z_0 \in \C: \displaystyle \lim_{z \mathop \to z_0} \exp z = \exp z_0$
Order Topology on Convex Subset is Subspace Topology
Let $\left({S, \preceq,\tau}\right)$ be a linearly ordered space. Let $A \subseteq S$ be a convex set in $S$. Let $\upsilon$ be the order topology on $A$. Let $\tau'$ be the $\tau$-relative subspace topology on $A$. Then $\upsilon = \tau'$.
Exponential Function is Continuous/Real Numbers
The real exponential function is continuous. That is: :$\forall x_0 \in \R: \displaystyle \lim_{x \mathop \to x_0} \ \exp x = \exp x_0$
Edge of Tree is Bridge
Let $T$ be a tree. Let $e$ be an edge of $T$. Then $e$ is a bridge of $T$.
Sub-Basis for Topological Subspace
Let $\left({X,\tau}\right)$ be a topological space. Let $K$ be a sub-basis for $\tau$. Let $\left({S, \tau'}\right)$ be a subspace of $\left({X, \tau}\right)$. Let $K' = \left\{{ U \cap S: U \in K }\right\}$. That is, $K'$ consists of the $\tau'$-open sets in $S$ corresponding to elements of $K$. Then $K'$ is a sub-basis for $\tau'$.
Modus Ponendo Ponens/Variant 1
:  $p \vdash \left({p \implies q}\right) \implies q$
Modus Ponendo Ponens/Variant 1/Proof 2
:$p \vdash \left({p \implies q}\right) \implies q$
Modus Ponendo Ponens/Variant 2
:  $\vdash p \implies \left({\left({p \implies q}\right) \implies q}\right)$
Modus Ponendo Ponens/Variant 2/Proof 2
:$\vdash p \implies \paren {\paren {p \implies q} \implies q}$
Rule of Commutation/Conjunction
Conjunction is commutative: === Formulation 1 === {{:Rule of Commutation/Conjunction/Formulation 1}} === Formulation 2 === {{:Rule of Commutation/Conjunction/Formulation 2}}
Rule of Commutation/Disjunction
Disjunction is commutative: === Formulation 1 === {{:Rule of Commutation/Disjunction/Formulation 1}} === Formulation 2 === {{:Rule of Commutation/Disjunction/Formulation 2}}
Factor Principles/Conjunction on Right/Formulation 1
: $p \implies q \vdash \left({p \land r}\right) \implies \left ({q \land r}\right)$
Factor Principles/Conjunction on Right/Formulation 2
: $\vdash \left({p \implies q}\right) \implies \left({\left({p \land r}\right) \implies \left ({q \land r}\right)}\right)$
Interior of Convex Angle is Convex Set
Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^2$, and let $p$ be a point in $\R^2$. Suppose that the angle between $\mathbf v$ and $\mathbf w$ is a convex angle. Then the set  :$U = \left\{ {p + st \mathbf v + \left({1-s}\right) t \mathbf w : s \in \left({0\,.\,.\,1}\right) , t \in \R_{>0} }\right\}$ is a convex set. {{expand|It'd be really nice to have a picture of $U$ to support intuition and connect with the page title}}
Factor Principles/Conjunction on Right
==== Formulation 1 ==== {{:Factor Principles/Conjunction on Right/Formulation 1}} ==== Formulation 2 ==== {{:Factor Principles/Conjunction on Right/Formulation 2}}
Factor Principles/Conjunction on Left
==== Formulation 1 ==== {{:Factor Principles/Conjunction on Left/Formulation 1}} ==== Formulation 2 ==== {{:Factor Principles/Conjunction on Left/Formulation 2}}
Factor Principles/Conjunction on Left/Formulation 1
: $p \implies q \vdash \left({r \land p}\right) \implies \left ({r \land q}\right)$
Factor Principles/Conjunction on Left/Formulation 2
:$\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} }$
Triangle is Convex Set
The interior of a triangle embedded in $\R^2$ is a convex set.
Subset Relation is Compatible with Subset Product
Let $\left({S,\circ}\right)$ be a magma. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $\circ_\mathcal P$ be the operation induced on $\mathcal P \left({S}\right)$ by $\circ$. Then the subset relation $\subseteq$ is compatible with $\circ_\mathcal P$.
Subset Relation is Compatible with Subset Product/Corollary 1
Let $A, B, C, D \in \powerset S$. Let $A \subseteq B$ and $C \subseteq D$. Then: :$A \circ_\PP C \subseteq B \circ_\PP D$
Subset Relation is Compatible with Subset Product/Corollary 2
Let $A,B \in \mathcal P \left({S}\right)$, the power set of $S$. {{improve|No need to use power set. $A \subseteq B \subseteq S$ sufficient.}} Let $A \subseteq B$. Let $x \in S$. Then: :$x \circ A \subseteq x \circ B$ :$A \circ x \subseteq B \circ x$
Equivalence of Definitions of Normal Subset/1 iff 2
Let $\left({G, \circ}\right)$ be a group. Let $S$ be a subset of $G$. Then Normal Subset/Definition 1 is equivalent to Normal Subset/Definition 2. That is, the following three statements are equivalent: :$(1)\quad \forall g \in G: g \circ S = S \circ g$ :$(2)\quad \forall g \in G: g \circ S \circ g^{-1} = S$ :$(3)\quad \forall g \in G: g^{-1} \circ S \circ g = S$
Subset Product with Identity
Let $\left({S, \circ}\right)$ be a magma. Let $\left({S, \circ}\right)$ have an identity element $e$. Then $e \circ S = S \circ e = S$, where $\circ$ is understood to be the subset product with singleton.
Equivalence of Definitions of Normal Subset/2 implies 3
Let $\left({G, \circ}\right)$ be a group. Let $S \subseteq G$. Let $S$ be a normal subset of $G$ by Definition 2. Then $S$ is a normal subset of $G$ by Definition 3. That is, if: :$\forall g \in G: g \circ S \circ g^{-1} = S$ or: :$\forall g \in G: g^{-1} \circ S \circ g = S$ then: :$\forall g \in G: g \circ S \circ g^{-1} \subseteq S$ and: :$\forall g \in G: g^{-1} \circ S \circ g \subseteq S$
Superset Relation is Compatible with Subset Product
Let $\left({S,\circ}\right)$ be a magma. Let $\circ_{\mathcal P}$ be the subset product on $\mathcal P \left({S}\right)$, the power set of $S$. Then the superset relation $\supseteq$ is compatible with $\circ_{\mathcal P}$.
Equivalence of Definitions of Normal Subset/3 iff 4
Let $\left({G,\circ}\right)$ be a group. Let $S \subseteq G$. Then: : $S$ is a normal subset of $G$ by Definition 3 iff: : $S$ is a normal subset of $G$ by Definition 4. That is, the following conditions are equivalent: : $(1)\quad \forall g \in G: g \circ S \circ g^{-1} \subseteq S$ : $(2)\quad \forall g \in G: g^{-1} \circ S \circ g \subseteq S$ : $(3)\quad \forall g \in G: S \subseteq g \circ S \circ g^{-1}$ : $(4)\quad \forall g \in G: S \subseteq g^{-1} \circ S \circ g$
Equivalence of Definitions of Normal Subset/3 and 4 imply 2
Let $\left({G,\circ}\right)$ be a group. Let $S \subseteq G$. Let $S$ be a normal subset of $G$ by Definition 3 and Definition 4. Then $S$ is a normal subset of $G$ by Definition 2.
Equivalence of Definitions of Normal Subset/3 iff 5
Let $\struct {G, \circ}$ be a group. Let $S \subseteq G$. Then: : $S$ is a normal subset of $G$ by Definition 3 : : $S$ is a normal subset of $G$ by Definition 5.
Equivalence of Definitions of Normal Subgroup
Let $G$ be a group. Let $N$ be a subgroup of $G$. {{TFAE|def = Normal Subgroup}}
Subgroup is Normal iff Normal Subset
Let $\left({G, \circ}\right)$ be a group. Let $N$ be a subgroup of $G$. Then $N$ is normal in $G$ (by definition 1)  it is a normal subset of $G$.
Rule of Explosion/Sequent Form
:$\bot \vdash \phi$
Law of Excluded Middle/Sequent Form
The Law of Excluded Middle can be symbolised by the sequent: :$\vdash p \lor \neg p$
Power Set of Monoid under Induced Operation is Monoid
Let $\left({G, \circ}\right)$ be a monoid with identity $e$. Let $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ be the algebraic structure consisting of the power set of $G$ and the operation induced on $\mathcal P \left({S}\right)$ by $\circ$. Then $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a monoid with identity $\left\{{e}\right\}$.
Principle of Non-Contradiction/Sequent Form/Formulation 2
:$\vdash \neg \paren {p \land \neg p}$
Power Set of Semigroup under Induced Operation is Semigroup
Let $\left({S, \circ}\right)$ be a semigroup. 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 $\circ$. Then $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a semigroup.
Power Set of Group under Induced Operation is Monoid
Let $\left({G, \circ}\right)$ be a group with identity $e$. Let $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ be the algebraic structure consisting of the power set of $G$ and the operation induced on $\mathcal P \left({G}\right)$ by $\circ$. Then $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a monoid with identity $\left\{{e}\right\}$.
Power Set of Magma under Induced Operation is Magma
Let $\left({S, \circ}\right)$ be a magma. 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 $\circ$. Then $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a magma.
Element in Left Coset iff Product with Inverse in Subgroup
Let $y H$ denote the left coset of $H$ by $y$. Then: :$x \in y H \iff x^{-1} y \in H$
Element in Right Coset iff Product with Inverse in Subgroup
Let $H \circ y$ denote the right coset of $H$ by $y$. Then: :$x \in H y \iff x y^{-1} \in H$
Inverse of Inverse of Subset of Group
Let $\struct {G, \circ}$ be a group. Let $X \subseteq G$. Then: : $\paren {X^{-1} }^{-1} = X$. where $X^{-1}$ denotes the inverse of $X$.
Exclusive Or is Commutative
: $p \oplus q \dashv \vdash q \oplus p$
Exclusive Or is Associative
Exclusive or is associative: : $p \oplus \left({q \oplus r}\right) \dashv \vdash \left({p \oplus q}\right) \oplus r$
Exclusive Or with Itself
Exclusive or destroys copies of itself: : $p \oplus p \dashv \vdash \bot$
Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 1
: $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 2
: $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
Non-Equivalence as Disjunction of Conjunctions/Formulation 1
: $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
Exclusive Or is Negation of Biconditional
Exclusive or is equivalent to the negation of the biconditional: :$p \oplus q \dashv \vdash \neg \paren {p \iff q}$
Exclusive Or as Disjunction of Conjunctions
:$p \oplus q \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$
Non-Equivalence as Disjunction of Negated Implications/Proof 1
: $\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$
Non-Equivalence as Disjunction of Negated Implications/Proof 2
: $\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$
Non-Equivalence as Disjunction of Negated Implications
: $\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$
Conjunction of Disjunction with Negation is Conjunction with Negation
:$\left({p \lor q}\right) \land \neg q \dashv \vdash p \land \neg q$
Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof 1
: $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$
Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof 2
: $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$
Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction
: $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$ That is, negation of the biconditional means the same thing as '''either-or but not both''', that is, exclusive or.
Inversion Mapping is Involution
Let $G$ be a group, and let $\iota: G \to G$ be the inversion mapping. Then $\iota$ is an involution. That is: :$\forall g \in G: \map \iota {\map \iota g} = g$
Rule of Association/Conjunction
Conjunction is associative: === Formulation 1 === {{:Rule of Association/Conjunction/Formulation 1}} === Formulation 2 === {{:Rule of Association/Conjunction/Formulation 2}}
Rule of Association/Disjunction
Disjunction is associative: === Formulation 1 === {{:Rule of Association/Disjunction/Formulation 1}} === Formulation 2 === {{:Rule of Association/Disjunction/Formulation 2}}
Rule of Association/Conjunction/Formulation 1
:$p \land \left({q \land r}\right) \dashv \vdash \left({p \land q}\right) \land r$
Rule of Association/Disjunction/Formulation 1
:$p \lor \paren {q \lor r} \dashv \vdash \paren {p \lor q} \lor r$
Rule of Association/Disjunction/Formulation 1/Proof 2
:$p \lor \left({q \lor r}\right) \dashv \vdash \left({p \lor q}\right) \lor r$
Rule of Association/Conjunction/Formulation 2
:$\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$
Rule of Association/Disjunction/Formulation 2
:$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$
Principle of Commutation/Forward Implication/Formulation 1
: $p \implies \left({q \implies r}\right) \vdash q \implies \left({p \implies r}\right)$
Principle of Commutation/Reverse Implication/Formulation 1
:$q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$
Principle of Commutation/Reverse Implication/Formulation 1/Proof
: $q \implies \left({p \implies r}\right) \vdash p \implies \left({q \implies r}\right)$
Principle of Commutation/Formulation 1/Proof 1
: $p \implies \left({q \implies r}\right) \dashv \vdash q \implies \left({p \implies r}\right)$
Principle of Commutation/Formulation 1/Proof 2
: $p \implies \left({q \implies r}\right) \dashv \vdash q \implies \left({p \implies r}\right)$
Principle of Commutation/Formulation 2
:$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$
Linearly Ordered Space is Connected iff Linear Continuum
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Then $S$ is a connected space  it is a linear continuum.
Rule of Transposition/Variant 1/Formulation 2/Forward Implication
: $\vdash \left({p \implies \neg q}\right) \implies \left({q \implies \neg p}\right)$
Boundary of Polygon is Jordan Curve
Let $P$ be a polygon embedded in $\R^2$. Then there exists a Jordan curve $\gamma: \closedint 0 1 \to \R^2$ such that the image of $\gamma$ is equal to the boundary $\partial P$ of $P$.
Rule of Explosion/Variant 1
:$\vdash p \implies \paren {\neg p \implies q}$
Negation of Conditional implies Antecedent
:$\vdash \neg \paren {p \implies q} \implies p$
Negation of Conditional implies Negation of Consequent
:$\vdash \neg \left({p \implies q}\right) \implies \neg q$
Compactness from Basis
Let $\struct {X, \tau}$ be a topological space. Let $B$ be a basis for $\tau$. Then the following propositions are equivalent: {{begin-axiom}} {{axiom | n = 1         | t = $\struct {X, \tau}$ is compact. }} {{axiom | n = 2         | t = Every open cover of $X$ by elements of $B$ has a finite subcover. }} {{end-axiom}}
Implication Equivalent to Negation of Conjunction with Negative/Formulation 2/Reverse Implication
:$\vdash \left({\neg \left({p \land \neg q}\right)}\right) \implies \left({p \implies q}\right)$
Biconditional Elimination
The rule of '''biconditional  elimination''' is a valid deduction sequent in propositional logic.
Biconditional Introduction/Sequent Form
:$p \implies q, q \implies p \vdash p \iff q$
Biconditional Elimination/Sequent Form/Proof 1/Form 1
:$p \iff q \vdash p \implies q$
Biconditional Elimination/Sequent Form/Proof 1/Form 2
:$p \iff q \vdash q \implies p$
Biconditional Elimination/Sequent Form/Proof 1
:$(1): \quad p \iff q \vdash p \implies q$ :$(2): \quad p \iff q \vdash q \implies p$
Biconditional Elimination/Sequent Form/Proof 2
:$(1): \quad p \iff q \vdash p \implies q$ :$(2): \quad p \iff q \vdash q \implies p$
Biconditional is Commutative/Formulation 1/Proof 1
: $p \iff q \dashv \vdash q \iff p$
Rule of Material Equivalence/Formulation 1
:$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$
Equivalences are Interderivable/Proof 1
If two propositional formulas are interderivable, they are equivalent: : $\left ({p \dashv \vdash q}\right) \dashv \vdash \left ({p \iff q}\right)$
Equivalences are Interderivable/Forward Implication
: $\left ({p \dashv \vdash q}\right) \vdash \left ({p \iff q}\right)$
Compact Subspace of Linearly Ordered Space/Reverse Implication
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Let $Y \subseteq X$ be a non-empty subset of $X$. Let the following hold: :$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$. :$(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$. Then $Y$ is a compact subspace of $\struct {X, \tau}$.
Modus Ponendo Tollens/Sequent Form
==== Case 1 ==== {{:Modus Ponendo Tollens/Sequent Form/Case 1}} ==== Case 2 ==== {{:Modus Ponendo Tollens/Sequent Form/Case 2}}
Modus Ponendo Tollens/Sequent Form/Case 1
:$\neg \paren {p \land q}, p \vdash \neg q$
Modus Ponendo Tollens/Sequent Form/Case 2
:$\neg \left({p \land q}\right), q \vdash \neg p$
Union of Subsets is Subset/Subset of Power Set
Let $S$ and $T$ be sets. Let $\powerset S$ be the power set of $S$. Let $\mathbb S$ be a subset of $\powerset S$. Then: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
Conjunction with Negative Equivalent to Negation of Implication/Formulation 2/Forward Implication
:$\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$
Neighborhood Sub-Basis Criterion for Filter Convergence
Let $\left({X, \tau}\right)$ be a topological space. Let $\mathcal F$ be a filter on $X$. Let $p \in X$. Then $\mathcal F$ converges to $p$ iff $\mathcal F$ contains as a subset a neighborhood sub-basis at $x$.
Rule of Material Implication/Formulation 2/Forward Implication
: $\vdash \left({p \implies q}\right) \implies \left({\neg p \lor q}\right)$
Implication is Left Distributive over Conjunction/Formulation 1/Proof 2
:$p \implies \left({q \land r}\right) \dashv \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)$
Compact Subspace of Linearly Ordered Space
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Let $Y \subseteq X$ be a non-empty subset of $X$. Then $Y$ is a compact subspace of $\struct {X, \tau}$  both of the following hold: :$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$. :$(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.
Order Topology is Hausdorff
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Then $\struct {X, \tau}$ is a Hausdorff space.
Subset of Linearly Ordered Space which is Order-Complete and Closed but not Compact
Let $X = \left[{0 \,.\,.\, 1}\right) \cup \left({2 \,.\,.\, 3}\right) \cup \left\{{4}\right\}$. Let $\preceq$ be the ordering on $X$ induced by the usual ordering of the real numbers. Let $\tau$ be the $\preceq$ order topology on $X$. Let $Y = \left[{0 \,.\,.\, 1}\right) \cup \left\{{4}\right\}$. Let $\tau'$ be the $\tau$-relative subspace topology on $Y$. Then: :$\left({Y, \preceq}\right)$ is a complete lattice :$Y$ is closed in $X$ but: :$\left({Y, \tau'}\right)$ is not compact.
Rule of Explosion/Variant 2
:$\vdash \paren {p \land \neg p} \implies q$
Jordan Curve Theorem
Let $\gamma: \closedint 0 1 \to \R^2$ be a Jordan curve. Let $\Img \gamma$ denote the image of $\gamma$. Then $\R^2 \setminus \Img \gamma$ is a union of two disjoint connected components. Both components are open in $\R^2$, and both components have $\Img \gamma$ as their boundary. One component is bounded, and is called the interior of $\gamma$. The other component is unbounded, and is called the exterior of $\gamma$.
Heine-Borel Theorem/Dedekind Complete Space
Let $T = \left({X, \preceq, \tau}\right)$ be a Dedekind-complete linearly ordered space. Let $Y$ be a non-empty subset of $X$. Then $Y$ is compact  $Y$ is closed and bounded in $T$.
Heine-Borel iff Dedekind Complete
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space. Then $X$ is Dedekind complete  every closed, bounded subset of $X$ is compact.
Jordan Polygon Theorem
Let $P$ be a polygon embedded in $\R^2$. Denote the boundary of $P$ as $\partial P$. Then, $\R^2 \setminus \partial P$ is a union of two connected components. Both components are open in $\R^2$. One component is bounded, and is called the interior of $P$. The other component is unbounded, and is called the exterior of $P$.
Jordan Polygon Interior and Exterior Criterion
Let $P$ be a polygon embedded in $\R^2$. Let $q \in \R^2 \setminus \partial P$, where $\partial P$ denotes the boundary of $P$. Let $\mathbf v \in \R^2 \setminus \left\{ {\mathbf 0}\right\}$ be a non-zero vector, and let $\mathcal L = \left\{ {q + s \mathbf v: s \in \R_{\ge 0} }\right\}$ be a ray with start point $q$. Let $N \left({q}\right) \in \N$ be the number of crossings between $\mathcal L$ and $\partial P$. Then: :$(1): \quad$ $q \in \operatorname{Int} \left({P}\right)$, iff $N \left({q}\right) \equiv 1 \pmod 2$ :$(2): \quad$ $q \in \operatorname{Ext} \left({P}\right)$, iff $N \left({q}\right) \equiv 0 \pmod 2$ Here, $\operatorname{Int} \left({P}\right)$ and $\operatorname{Ext} \left({P}\right)$ denote the interior and exterior of $\partial P$, when $\partial P$ is considered as a Jordan curve.