hcju/proofwikips huggingface dataset:
Closed Ball in Euclidean Space is Compact
Let $x \in \R_n$ be a point in the Euclidean space $\R^n$. Let $\epsilon \in \R_{>0}$. Then the closed $\epsilon$-ball $\map {B_\epsilon^-} x$ is compact.
Isometric Image of Cauchy Sequence is Cauchy Sequence
Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be metric spaces. Let $f: S_1 \to S_2$ be an isometry. Let $\sequence {x_n}$ be a Cauchy sequence in $S_1$. Let $\sequence {y_n} = \sequence {\map f {x_n} }$ be the image of $\sequence {x_n}$ under $f$. Then $\sequence {y_n}$ is a Cauchy sequence.
Union of Subsets is Subset
Let $S_1$, $S_2$, and $T$ be sets. Let $S_1$ and $S_2$ both be subsets of $T$. Then: :$S_1 \cup S_2 \subseteq T$ That is: :$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$
Proof by Cases/Formulation 1
: $\left({p \implies r}\right) \land \left({q \implies r}\right) \dashv \vdash \left({p \lor q}\right) \implies r$
Distance between Closed Sets in Euclidean Space
Let $S, T \subseteq \R^n$ be closed, non-empty subsets of the real Euclidean space $R^n$. Suppose that $S$ is bounded, and $S$ and $T$ are disjoint. Then there exists $x \in S$ and $y \in T$ such that: :$\map d {x, y} = \map d {S, T} > 0$ where: :$d$ denotes the Euclidean metric :$\map d {S, T}$ is the distance between $S$ and $T$.
Complement in Distributive Lattice is Unique
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a bounded distributive lattice. Then every $a \in S$ admits at most one complement.
Equivalence of Definitions of Top
{{TFAE|def = Top (Lattice Theory)|view = $\top$ (top)|context = Lattice Theory}} Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice.
Equivalence of Definitions of Bottom
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice. Let $\bot$ be a bottom of $\left({S, \vee, \wedge, \preceq}\right)$. {{TFAE|def = Bottom (Lattice Theory)|view = Bottom|context = Lattice Theory}}
Top is Unique
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice. Then $S$ has at most one top.
Isometry is Homeomorphism of Induced Topologies
Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be metric spaces or pseudometric spaces. Let $f: S_1 \to S_2$ be an isometry from $\struct {S_1, d_1}$ to $\struct {S_2, d_2}$. Let $\tau_1$ and $\tau_2$ be the topologies induced on $S_1$ and $S_2$ by the metrics $d_1$ and $d_2$, respectively. Then $f$ is a homeomorphism from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.
Equivalence of Definitions of Distributive Lattice
{{TFAE|def = Distributive Lattice}} Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice. {{begin-axiom}} {{axiom | n = 1         | q = \forall x, y, z \in S         | m = x \wedge \left({y \vee z}\right) = \left({x \wedge y}\right) \vee \left({x \wedge z}\right) }} {{axiom | n = 1'         | q = \forall x, y, z \in S         | m = \left({x \vee y}\right) \wedge z = \left({x \wedge z}\right) \vee \left({y \wedge z}\right) }} {{axiom | n = 2         | q = \forall x, y, z \in S         | m = x \vee \left({y \wedge z}\right) = \left({x \vee y}\right) \wedge \left({x \vee z}\right) }} {{axiom | n = 2'         | q = \forall x, y, z \in S         | m = \left({x \wedge y}\right) \vee z = \left({x \vee z}\right) \wedge \left({x \vee y}\right) }} {{end-axiom}}
Distance-Preserving Surjection is Isometry of Metric Spaces
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces. Let $\phi: M_1 \to M_2$ be a surjective distance-preserving mapping. That is: :$\forall a, b \in M_1: d_1 \tuple {a, b} = d_2 \tuple {\map \phi a, \map \phi b}$ Then $\phi$ is an isometry.
Existence of Positive Root of Positive Real Number
Let $x \in \R$ be a real number such that $x > 0$. Let $n \in \Z$ be an integer such that $n \ne 0$. Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.
Proof by Cases/Formulation 2
:$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$
Proof by Cases/Formulation 2/Forward Implication
: $\vdash \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies \left({\left({p \lor q}\right) \implies r}\right)$
Proof by Cases/Formulation 2/Reverse Implication
: $\vdash \left({\left({p \lor q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right)$
Set is Subset of Union/Set of Sets
Let $\mathbb S$ be a set of sets. Then: : $\displaystyle \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
Proof by Cases/Formulation 3
: $\vdash \left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r$
Properties of Relation Compatible with Group Operation
Let $\left({G,\circ}\right)$ be a group with identity element $e$. Let $\mathcal R$ be a endorelation on $G$ which is compatible with $\circ$. Let $x,y,z \in G$. Then the following hold:
Existence of Positive Root of Positive Real Number/Positive Exponent
Let $x \in \R$ be a real number such that $x > 0$. Let $n \in \Z$ be an integer such that $n > 0$. Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.
Existence of Positive Root of Positive Real Number/Negative Exponent
Let $x \in \R$ be a real number such that $x > 0$. Let $n \in \Z$ be an integer such that $n < 0$. Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.
Power of Ring Negative
Let $\struct {R, +, \circ}$ be a ring. Let $n \in \N_{>0}$ be a strictly positive integer. Let $x \in R$. Then: :If $n$ is even: :::$\map {\circ^n} {-x} = \map {\circ^n} x$ :If $n$ is odd: :::$\map {\circ^n} {-x} = -\map {\circ^n} x$
Paving Lemma
Let $S$ be an open subset of the Euclidean space $\R^m$ or the set of complex numbers $\C$. Let $\gamma: \closedint a b \to S$ be a path in $S$. Then there exists $K \in \R_{>0}$ such that: :For all $\epsilon \in \openint 0 K$, there exists a normal subdivision $\set {x_0, x_1, \ldots, x_{n-1}, x_n}$ of the closed interval $\closedint a b$ such that: ::$\displaystyle \bigcup_{i \mathop = 0}^n \map {B_\epsilon} {\map \gamma {x_i} } \subseteq S$ :and for all $i \in \set {0, 1, \ldots, n - 1}$: ::$\map \gamma {\closedint {x_i} {x_{i + 1} } } \subseteq \map {B_\epsilon} {\map \gamma {x_i} }$ Here, $\map {B_\epsilon} {\map \gamma {x_i} }$ denotes the open ball of $\map \gamma {x_i}$ with radius $\epsilon$.
Proof by Cases/Formulation 1/Reverse Implication
: $\left({p \lor q}\right) \implies r \vdash \left({p \implies r}\right) \land \left({q \implies r}\right)$
Praeclarum Theorema/Formulation 1/Proof 2
:$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$
Praeclarum Theorema/Formulation 1
: $\left({p \implies q}\right) \land \left({r \implies s}\right) \vdash \left({p \land r}\right) \implies \left({q \land s}\right)$
Praeclarum Theorema/Formulation 2
: $\vdash \left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)$
Strictly Positive Integer Power Function Strictly Succeeds Each Element
Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity. Let $\struct {R, \le}$ be a directed set with no upper bound. Let $n \in \N_{>0}$. Let $f: R \to R$ be defined by: :$\forall x \in R: \map f x = \circ^n x$ Then the image of $f$ has elements strictly succeeding each elements of $R$.
Constructive Dilemma/Formulation 1/Proof 2
: $p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
Constructive Dilemma/Formulation 1
: $p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
Constructive Dilemma/Formulation 2
: $\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$
Clavius's Law/Formulation 1
:$\neg p \implies p \vdash p$
Equivalent Matrices may not be Similar
If two square matrices of order $n > 1$ over a ring with unity $R$ are equivalent, they are not necessarily similar.
Odd Power Function is Strictly Increasing/General Result
Let $\struct {R, +, \circ, \le}$ be a totally ordered ring. Let $n$ be an odd positive integer. Let $f: R \to R$ be the mapping defined by: :$\map f x = \map {\circ^n} x$ Then $f$ is strictly increasing on $R$.
Number of Bijective Restrictions
Let $f: S \to T$ be a surjection. Let $B$ be the set of all bijective restrictions of $f$. Then the cardinality of $B$ is: :$\displaystyle \card {\prod_{i \mathop \in I} \family {S / \mathcal R_f}_i}$ where $S / \mathcal R_f$ denotes the quotient set of the induced equivalence of $f$ indexed by $I$.
Number of Injective Restrictions
Let $f: S \to T$ be a mapping. Let $Q$ be the set of all injective restrictions of $f$. Then the cardinality of $Q$ is: :$\displaystyle \card {\prod_{i \mathop \in I} \prod_{j \mathop \in J_i} \family {\family {\powerset {S / \mathcal R_f} }_i}_j}$ where: :$\mathcal P$ denotes power set :$S / \mathcal R_f$ denotes quotient set of the induced equivalence of $f$.
Box Topology on Finite Product Space is Tychonoff Topology
Let $n \in \N$. For all $k \in \set {1, \ldots, n}$, let $T_k = \struct {X_k, \tau_k}$ be topological spaces. Let $\displaystyle X = \prod_{k \mathop = 1}^n X_k$ be the cartesian product of $X_1, \ldots, X_n$. Then the box topology and the Tychonoff topology on $X$ are identical.
Strictly Positive Integer Power Function is Unbounded Above/General Case
Let $\struct {R, +, \circ, \le}$ be a totally ordered ring with unity. Suppose that $R$ has no upper bound. Let $n \in \N_{>0}$. Let $f: R \to R$ be defined by: :$\map f x = \circ^n x$ Then the image of $f$ is unbounded above in $R$.
Strictly Positive Integer Power Function is Unbounded Above
Let $\R$ be the real numbers with the usual ordering. Let $n \in \N_{>0}$. Let $f: \R \to \R$ be defined by: :$\map f x = x^n$ Then $f$ is unbounded above.
Ring Product preserves Inequalities on Positive Elements
Let $\struct {R, +, \circ, \le}$ be an ordered ring. Let $x, y, z, w \in R$. Let $0 < x < y$ and $0 < z < w$. Then: :$0 < z \circ x < w \circ y$
Power Function is Strictly Increasing on Positive Elements
Let $\struct {R, +, \circ, \le}$ be an ordered ring. Let $x, y \in R$. Let $n \in \N_{>0}$ be a strictly positive integer. Let $0 < x < y$. Then: :$0 < \map {\circ^n} x < \map {\circ^n} y$
Biconditional is Associative/Formulation 1
: $p \iff \paren {q \iff r} \dashv \vdash \paren {p \iff q} \iff r$
Biconditional is Associative/Formulation 2
:$\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r}$
Open Ball is Convex Set
Let $V$ be a normed vector space with norm $\left\Vert{\cdot}\right\Vert$ over $\R$ or $\C$. An open ball in the metric induced by $\left\Vert{\cdot}\right\Vert$ is a convex set.
Pseudometric Space is Metric Space iff Kolmogorov
Let $M = \struct {S, d}$ be a pseudometric space. Let $T = \struct {S, \tau}$ be the topological space over $S$ induced by $d$. Then $M$ is a metric space  $T$ is a $T_0$ (Kolmogorov} space.
Sequence Converges to Point Relative to Metric iff it Converges Relative to Induced Topology
Let $M = \struct {S, d}$ be a metric space or a pseudometric space. Let $T = \struct {S, \tau}$ be the topological space induced by $d$. Let $\sequence {x_n}$ be a infinite sequence in $S$. Let $l \in S$. Then $\sequence {x_n}$ converges to $l$ relative to $d$  $\sequence {x_n}$ converges to $l$ relative to $\tau$.
Isometry Preserves Sequence Convergence
Let $M_1 = \struct {S_1, d_1}$ and $M_2 = \struct {S_2, d_2}$ both be metric spaces or pseudometric spaces. Let $\phi: S_1 \to S_2$ be an isometry. Let $\sequence {x_n}$ be an infinite sequence in $S_1$. Suppose that $\sequence {x_n}$ converges to a point $p \in S_1$. Then $\sequence {\map \phi {x_n}}$ converges to $\map \phi p$.
Connected Domain is Connected by Staircase Contours
Let $D \subseteq \C$ be an open set. Then $D$ is a connected domain : :for all $z, w \in \C$, there exists a staircase contour in $D$ with start point $z$ and end point $w$.
Dedekind Completeness is Self-Dual
Let $\struct {S, \preceq}$ be an ordered set. Then $\struct {S, \preceq}$ is Dedekind complete  every non-empty subset of $S$ that is bounded below admits an infimum in $S$. That is, an ordered set is Dedekind complete  its dual is Dedekind complete.
Biconditional is Transitive/Formulation 1
:$p \iff q, q \iff r \vdash p \iff r$
Biconditional is Transitive/Formulation 2
: $\vdash \left({\left({p \iff q}\right) \land \left({q \iff r}\right)}\right) \implies \left({p \iff r}\right)$
Biconditional as Disjunction of Conjunctions/Formulation 1/Reverse Implication
: $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q$
Biconditional as Disjunction of Conjunctions/Formulation 1
: $p \iff q \dashv \vdash \paren {p \land q} \lor \paren {\neg p \land \neg q}$
Biconditional as Disjunction of Conjunctions/Formulation 2
:$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$
Biconditional as Disjunction of Conjunctions
==== Formulation 1 ==== {{:Biconditional as Disjunction of Conjunctions/Formulation 1}} ==== Formulation 2 ==== {{:Biconditional as Disjunction of Conjunctions/Formulation 2}}
Biconditional Equivalent to Biconditional of Negations/Formulation 1/Reverse Implication
:$\neg p \iff \neg q \vdash p \iff q$
Biconditional Equivalent to Biconditional of Negations/Formulation 1
: $p \iff q \dashv \vdash \neg p \iff \neg q$
Biconditional Equivalent to Biconditional of Negations/Formulation 2
: $\vdash \left({p \iff q}\right) \iff \left({\neg p \iff \neg q}\right)$
Biconditional Equivalent to Biconditional of Negations
==== Formulation 1 ==== {{:Biconditional Equivalent to Biconditional of Negations/Formulation 1}} ==== Formulation 2 ==== {{:Biconditional Equivalent to Biconditional of Negations/Formulation 2}}
Biconditional iff Disjunction implies Conjunction/Formulation 1/Forward Implication
: $p \iff q \vdash \left({p \lor q}\right) \implies \left({p \land q}\right)$
Biconditional iff Disjunction implies Conjunction/Formulation 1/Reverse Implication
: $\left({p \lor q}\right) \implies \left({p \land q}\right) \vdash p \iff q$
Biconditional iff Disjunction implies Conjunction/Formulation 1
:$p \iff q \dashv \vdash \paren {p \lor q} \implies \paren {p \land q}$
Biconditional iff Disjunction implies Conjunction/Formulation 2
: $\vdash \left({p \iff q}\right) \iff \left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right)$
Biconditional iff Disjunction implies Conjunction
==== Formulation 1 ==== {{:Biconditional iff Disjunction implies Conjunction/Formulation 1}} ==== Formulation 2 ==== {{:Biconditional iff Disjunction implies Conjunction/Formulation 2}}
Biconditional is Transitive/Formulation 1/Proof 1
: $p \iff q, q \iff r \vdash p \iff r$
Order of Squares in Totally Ordered Ring without Proper Zero Divisors
Let $\struct {R, +, \circ, \le}$ be a totally ordered ring without proper zero divisors whose zero is $0_R$. Let $x, y \in R$ be positive, that is, $0_R \le x, y$. Then $x \le y \iff x \circ x \le y \circ y$. That is, the square mapping is an order embedding of $\struct {R_{\ge 0}, \le}$ into itself. When $R$ is one of the standard sets of numbers $\Z, \Q, \R$, then this translates into: :If $x, y$ are positive, then $x \le y \iff x^2 \le y^2$.
Order of Squares in Ordered Field
Let $\struct {R, +, \circ, \le}$ be an ordered field whose zero is $0_R$ and whose unity is $1_R$. Suppose that $\forall a \in R: 0 < a \implies 0 < a^{-1}$. Let $x, y \in \struct {R, +, \circ, \le}$ such that $0_R \le x, y$. Then $x \le y \iff x \circ x \le y \circ y$. That is, the square function is an order embedding of $\struct {R_{\ge 0}, \le}$ into itself. When $R$ is one of the standard fields of numbers $\Q$ and $\R$, then this translates into: :If $x, y$ are positive then $x \le y \iff x^2 \le y^2$.
Law of Identity/Formulation 1/Proof 1
: $p \vdash p$
Law of Identity/Formulation 1/Proof 2
: $p \vdash p$
Law of Identity/Formulation 2
Every proposition entails itself: :$\vdash p \implies p$
Vandermonde Determinant/Alternative Formulations
The '''Vandermonde determinant of order $n$''' is the determinant defined as follows: :$V_n = \begin{vmatrix}      1 &    x_1 &  x_1^2 & \cdots & x_1^{n - 2} & x_1^{n - 1} \\      1 &    x_2 &  x_2^2 & \cdots & x_2^{n - 2} & x_2^{n - 1} \\ \vdots & \vdots & \vdots & \ddots &      \vdots &      \vdots \\      1 &    x_n &  x_n^2 & \cdots & x_n^{n - 2} & x_n^{n - 1} \end{vmatrix}$ Its value is given by: :$\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$
Neighborhood Condition for Coarser Topology
Let $S$ be a set. Let $\tau_1$ and $\tau_2$ be two topologies on $S$. Suppose that for all $z \in S$ and for all open neighborhoods $N_z$ of $z$ with respect to $\tau_1$, there exists $U \in \tau_2$ such that $U \subseteq N_z$. Then $\tau_1$ is coarser than $\tau_2$.
Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof 2
:$p \land \left({p \lor q}\right) \dashv \vdash p$
Absorption Laws (Logic)/Conjunction Absorbs Disjunction
:$p \land \paren {p \lor q} \dashv \vdash p$
Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Proof 2
:$p \lor \left ({p \land q}\right) \dashv \vdash p$
Absorption Laws (Logic)/Disjunction Absorbs Conjunction
:$p \lor \left ({p \land q}\right) \dashv \vdash p$
Extended Transitivity
Let $S$ be a set. Let $\RR$ be a transitive relation on $S$. Let $\RR^=$ be the reflexive closure of $\RR$. Let $a, b, c \in S$. Then: {{begin-eqn}} {{eqn | n = 1       | l = \paren {a \mathrel \RR b} \land \paren {b \mathrel \RR c}       | o = \implies       | r = a \mathrel \RR c }} {{eqn | n = 2       | l = \paren {a \mathrel \RR b} \land \paren {b \mathrel {\RR^=} c}       | o = \implies       | r = a \mathrel \RR c }} {{eqn | n = 3       | l = \paren {a \mathrel {\RR^=} b} \land \paren {b \mathrel \RR c}       | o = \implies       | r = a \mathrel \RR c }} {{eqn | n = 4       | l = \paren {a \mathrel {\RR^=} b} \land \paren {b \mathrel {\RR^=} c}       | o = \implies       | r = a \mathrel {\RR^=} c }} {{end-eqn}}
Rule of Idempotence/Conjunction
{{:Rule of Idempotence/Conjunction/Formulation 1}}
Rule of Idempotence/Disjunction
{{:Rule of Idempotence/Disjunction/Formulation 1}}
Rule of Idempotence/Disjunction/Formulation 1/Proof
: $p \dashv \vdash p \lor p$
Ordering Cycle implies Equality
Let $\left({S,\preceq}\right)$ be an ordered set. Let $x_1$, $x_2$, and $x_3$ be elements of $S$. Suppose that {{begin-eqn}} {{eqn|l = x_1      |o = \preceq      |r = x_2 }} {{eqn|l = x_2      |o = \preceq      |r = x_3 }} {{eqn|l = x_3      |o = \preceq      |r = x_1 }} {{end-eqn}} Then $x_1 = x_2 = x_3$.
Euclidean Topology is Tychonoff Topology
Let $T_1 = \left({\R, \tau_1}\right)$ be the topological space such that $\tau_1$ is the Euclidean topology on $\R$. Let $T_n = \left({\R^n, \tau_n}\right)$ be the topological space such that $\tau_n$ is the Tychonoff topology on the cartesian product $\displaystyle \R_n = \prod_{i \mathop = 1}^n \R$. Then the Euclidean topology on $\R^n$ and the Tychonoff topology on $\R^n$ are the same.
Modus Ponendo Tollens/Variant/Formulation 1/Proof
:$\neg \paren {p \land q} \dashv \vdash p \implies \neg q$
Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication
: $\neg \left({p \land q}\right) \vdash p \implies \neg q$
Modus Ponendo Tollens/Variant/Formulation 1/Reverse Implication
:$p \implies \neg q \vdash \neg \paren {p \land q}$
Modus Ponendo Tollens/Variant/Formulation 1
: $\neg \left({p \land q}\right) \dashv \vdash p \implies \neg q$
Modus Ponendo Tollens/Variant/Formulation 2
: $\vdash \left({\neg \left({p \land q}\right)}\right) \iff \left({p \implies \neg q}\right)$
Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication
:$\neg \left({p \land \neg q}\right) \vdash p \implies q$
Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implication
:$p \implies q \vdash \neg \left({p \land \neg q}\right)$
Implication Equivalent to Negation of Conjunction with Negative/Formulation 1
:$p \implies q \dashv \vdash \neg \paren {p \land \neg q}$
Implication Equivalent to Negation of Conjunction with Negative
==== Formulation 1 ==== {{:Implication Equivalent to Negation of Conjunction with Negative/Formulation 1}} ==== Formulation 2 ==== {{:Implication Equivalent to Negation of Conjunction with Negative/Formulation 2}}
Zero Derivative implies Constant Complex Function
Let $D \subseteq \C$ be a connected domain of $\C$. Let $f: D \to \C$ be a complex-differentiable function. For all $z \in D$, let $\map {f'} z = 0$. Then $f$ is constant on $D$.
Relation between P-Product Metric and Chebyshev Distance on Real Vector Space
For $n \in \N$, let $\R^n$ be a Euclidean space. Let $p \in \R_{\ge 1}$. Let $d_p$ be the $p$-product metric on $\R^n$.  Let $d_\infty$ be the Chebyshev distance on $\R^n$. Then :$\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_p} {x, y} \le n^{1/p} \map {d_\infty} {x, y}$
P-Product Metrics on Real Vector Space are Topologically Equivalent/Inequality for General Case
For $n \in \N$, let $\R^n$ be a real vector Space. Let $r, t \in \R_{\ge 1}$. Let $d_r$ and $d_t$ be $p$-product metrics on $\R^n$. Then $d_r$ and $d_t$ are topologically equivalent.
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Forward Implication
: $\neg p \lor \neg q \vdash \neg \left({p \land q}\right)$
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Reverse Implication
: $\neg \left({p \land q}\right) \vdash \neg p \lor \neg q$
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1
:$\neg p \lor \neg q \dashv \vdash \neg \paren {p \land q}$
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 2
:$\vdash \paren {\neg p \lor \neg q} \iff \paren {\neg \paren {p \land q} }$
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Forward Implication
:$\neg p \land \neg q \vdash \neg \paren {p \lor q}$
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication
:$\neg \paren {p \lor q} \vdash \neg p \land \neg q$
De Morgan's Laws (Logic)/Conjunction of Negations
{{:De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1}}
De Morgan's Laws (Logic)/Conjunction
{{:De Morgan's Laws (Logic)/Conjunction/Formulation 1}}
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Forward Implication
: $p \land q \vdash \neg \left({\neg p \lor \neg q}\right)$
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Reverse Implication
: $\neg \left({\neg p \lor \neg q}\right) \vdash p \land q$
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Proof
: $p \land q \dashv \vdash \neg \left({\neg p \lor \neg q}\right)$
De Morgan's Laws (Logic)/Disjunction
{{:De Morgan's Laws (Logic)/Disjunction/Formulation 1}}
Rule of Distribution/Disjunction Distributes over Conjunction
=== Disjunction is Left Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === Disjunction is Right Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}}
Rule of Distribution/Conjunction Distributes over Disjunction
=== Conjunction is Left Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === Conjunction is Right Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive}}
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Proof
:$p \land \left({q \lor r}\right) \dashv \vdash \left({p \land q}\right) \lor \left({p \land r}\right)$
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2
:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2}}