nyuuzyou/wikis huggingface dataset:
Reflection Rule for Gaussian Binomial Coefficients

== Theorem ==
Let $q \in \R_{\ne 1}, n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k_q = q^{k \paren {n - k} } \dbinom n k_{q^{-1} }$
where $\dbinom n k_q$ is a Gaussian binomial coefficient.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $58$
Category:Gaussian Binomial Coefficients
Reflection is Isometry

== Theorem ==
Let $\Gamma = \R^n$ denote the real Euclidean space of $n$ dimensions, wher $n = 2$ or $n = 3$.
Let $\phi$ be a reflection in $\Gamma$.
Then $\phi$ is an isometry.
== Proof ==
== Sources ==
* 
* 
Category:Geometric Reflections
Category:Isometries (Euclidean Geometry)
Reflection of Plane in Line through Origin is Linear Operator

== Theorem ==
Let $M$ be a straight line in the plane $\R^2$ passing through the origin.
Let $s_M$ be the reflection of $\R^2$ in $M$.
Then $s_M$ is a linear operator for every straight line $M$ through the origin.
== Proof ==
Let the angle between $M$ and the $x$-axis be $\alpha$.
To prove that $s_M$ is a linear operator it is sufficient to demonstrate that:
:$(1): \quad \forall P_1, P_2 \in \R^2: \map {s_M} {P_1 + P_2} = \map {s_M} {P_1} + \map {s_M} {P_2}$
:$(2): \quad \forall \lambda \in \R: \map {s_M} {\lambda P_1} = \lambda \map {s_M} {P_1}$
So, let $P_1 = \tuple {x_1, y_1}$ and $P_2 = \tuple {x_2, y_2}$ be arbitrary points in the plane.
and:
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.4$
Category:Geometric Reflections
Category:Linear Operators
Reflexive Bilinear Form is Symmetric or Alternating

== Theorem ==
Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$.
Let $f$ be a bilinear form on $V$.
Let $f$ be reflexive.
Then $f$ is symmetric or alternating.
== Proof ==
Let $x, y, z \in V$.
Because $f$ is bilinear:
:$\map f {x, \map f {x, y} z - \map f {x, z} y} = \map f {x, y} \, \map f {x, z} - \map f {x, z} \, \map f {x, y} = 0$
Because $f$ is reflexive:
:$\map f {\map f {x, y} z - \map f {x, z} y, x} = 0$
Because $f$ is bilinear:
:$(1): \quad \map f {x, y} \, \map f {z, x} = \map f {x, z} \, \map f {y, x}$
Letting $z = x$, we obtain:
:$\map f {x, x} \, \map f {x, y} = \map f {x, x} \, \map f {y, x}$
Thus:
:$(2): \quad$ if $\map f {x, x} \ne 0$, then $\map f {x, y} = \map f {y, x}$ for all $y \in V$.
Suppose $f$ is not symmetric.
We show that $\map f {u, u} = 0$ for all $u \in V$.
Let $v, w \in V$ with $\map f {v, w} \ne \map f {w, v}$.
By $(2)$, $\map f {v, v} = \map f {w, w} = 0$.
Suppose $u \in V$ with $\map f {u, u} \ne 0$.
By $(2)$, $\map f {u, v} = \map f {v, u}$ and $\map f {u, w} = \map f {w, u}$.
By $(1)$, we have:
:$\begin{cases} \map f {v, w} \, \map f {u, v} = \map f {v, u} \, \map f {w, v} \\ \map f {w, v} \, \map f {u, w} = \map f {w, u} \, \map f {v, w} \end{cases}$.
Because $\map f {v, w} \ne \map f {w, v}$:
:$\begin{cases} \map f {u, v} = \map f {v, u} = 0 \\ \map f {u, w} = \map f {w, u} = 0 \end{cases}$.
Because $f$ is bilinear, $\map f {u + v, w} = \map f {v, w} \ne \map f {v, w} = \map f {u + v, w}$.
By $(2)$, $\map f {u + v, u + v} = 0$.
Because $f$ is bilinear and $\map f {u, v} = \map f {v, u} = \map f {v, v} = 0$, $\map f {u, u} = 0$.
This contradicts the assumption $\map f {u, u} \ne 0$.
Thus $f$ is alternating.
== Also see ==
* Bilinear Form is Reflexive iff Symmetric or Alternating
Category:Bilinear Forms (Linear Algebra)
Reflexive Circular Relation is Equivalence

== Theorem ==
Let $\RR \subseteq S \times S$ be a reflexive and circular relation in $S$.
Then $\RR$ is an equivalence relation.
== Proof ==
To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive.
So, checking in turn each of the criteria for equivalence:
=== Reflexive ===
We have that $\RR$ is reflexive.
=== Symmetric ===
By reflexivity:
:$\tuple {x, x} \in \RR$
If $\tuple {x, y} \in \RR$ then by the definition of circular relation $\tuple {y, x} \in \RR$.
Hence $\RR$ is symmetric.
=== Transitive ===
Let $\tuple {x, y}, \tuple {y, z} \in \RR$.
By definition of circular relation:
:$\tuple {z, x} \in \RR$
By $\RR$ being symmetric:
:$\tuple {x, z} \in \RR$
Hence $\RR$ is transitive.
Thus is $\RR$ is reflexive, symmetric and transitive, and therefore by definition an equivalence.
== Sources ==
* : $\S 1.3$ Exercise $2$
Category:Equivalence Relations
Category:Circular Relations
Category:Reflexive Relations
Reflexive Closure is Closure Operator

== Theorem ==
Let $S$ be a set.
Let $R$ be the set of all endorelations on $S$.
Then the reflexive closure operator on $R$ is a closure operator.
== Proof 1 ==
== Proof 2 ==
Category:Reflexive Closures
Category:Closure Operators
Category:Reflexive Closure is Closure Operator
Reflexive Closure is Closure Operator/Proof 1

== Theorem ==
== Proof ==
Let $\QQ$ be the set of reflexive relations on $S$.
By Intersection of Reflexive Relations is Reflexive, the intersection of any subset of $\QQ$ is in $Q$.
By the definition of reflexive closure as the intersection of reflexive supersets:
:The reflexive closure of a relation $\RR$ on $S$ is the intersection of elements of $\QQ$ that contain $S$.
From Closure Operator from Closed Sets we conclude that reflexive closure is a closure operator.
Category:Reflexive Closure is Closure Operator
Reflexive Closure is Closure Operator/Proof 2

== Theorem ==
== Proof ==
=== Reflexive Closure is Inflationary ===
=== Reflexive Closure is Order Preserving ===
=== Reflexive Closure is Idempotent ===
Thus by the definition of closure operator, reflexive closure is a closure operator.
Category:Reflexive Closure is Closure Operator
Reflexive Closure is Idempotent

== Theorem ==
Let $S$ be a set.
Let $R$ denote the set of all endorelations on $S$.
Then the reflexive closure operator is an idempotent mapping on $R$.
That is:
:$\forall \RR \in R: \RR^= = \paren {\RR^=}^=$
where $\RR^=$ denotes the reflexive closure of $\RR$.
== Proof ==
Let $\RR \in R$.
By the definition of reflexive closure:
:$\RR^= = \RR \cup \Delta_S$
:$\paren {\RR^=}^= = \paren {\RR \cup \Delta_S} \cup \Delta_S$
By Union is Associative:
:$\paren {\RR^=}^= = \RR \cup \paren {\Delta_S \cup \Delta_S}$
By Set Union is Idempotent:
:$\paren {\RR^=}^= = \RR \cup \Delta_S$
Hence:
:$\forall \RR \in R: \RR^= = \paren {\RR^=}^=$
Category:Reflexive Closures
Reflexive Closure is Inflationary

== Theorem ==
Let $S$ be a set.
Let $R$ denote the set of all endorelations on $S$.
Then the reflexive closure operator is an inflationary mapping on $R$.
== Proof ==
Let $\RR \in R$.
The reflexive closure $\RR^=$ of $\RR$ is defined as:
:$\RR^= := \RR \cup \Delta_S$
From Set is Subset of Union:
:$\RR \subseteq \RR^=$
Hence the reflexive closure operator is an inflationary mapping.
Category:Reflexive Closures
Reflexive Closure is Order Preserving

== Theorem ==
Let $S$ be a set.
Let $R$ denote the set of all endorelations on $S$.
Then the reflexive closure operator is an order preserving mapping on $R$.
That is:
:$\forall \RR, \SS \in R: \RR \subseteq \SS \implies \RR^= \subseteq \SS^=$
where $\RR^=$ and $\SS^=$ denote the reflexive closure of $\RR$ and $\SS$ respectively.
== Proof ==
Let $\RR, \SS \in R$.
Suppose:
:$\RR \subseteq \SS$
Their respective reflexive closures $\RR^=$ and $\SS^=$ are defined as:
:$\RR^= := \RR \cup \Delta_S$
:$\SS^= := \SS \cup \Delta_S$
Hence by Corollary to Set Union Preserves Subsets:
:$\RR^= \subseteq \SS^=$
Category:Reflexive Closures
Reflexive Closure is Reflexive

== Theorem ==
Let $\RR$ be a relation on a set $S$.
Then $\RR^=$, the reflexive closure of $\RR$, is reflexive.
== Proof ==
Recall the definition of reflexive closure:
:$\RR^= := \RR \cup \Delta_S$
From Set is Subset of Union:
:$\Delta_S \subseteq \RR^=$
The result follows directly from Relation Contains Diagonal Relation iff Reflexive.
== Sources ==
* : $\S 1.19$: Some Important Properties of Relations: Exercise $2$
Category:Reflexive Closures
Reflexive Closure of Antisymmetric Relation is Antisymmetric

== Theorem ==
Let $S$ be a set.
Let $RR$ be an antisymmetric relation on $S$.
Let $\RR^=$ be the reflexive closure of $\RR$.
Then $\RR^=$ is also antisymmetric.
== Proof ==
Let $a, b \in S$.
Suppose that $a \mathrel {\RR^=} b$ and $b \mathrel {\RR^=} a$.
By definition of $\RR^=$, this means:
:$a \mathrel \RR b$ or $a = b$
:$b \mathrel \RR a$ or $b = a$
If $a = b$ or $b = a$ we are done, by definition of an antisymmetric relation.
So suppose $a \mathrel \RR b$ and $b \mathrel \RR a$.
Because $\RR$ is antisymmetric, it follows that $a = b$ in this case as well.
Hence $\RR^=$ is also antisymmetric.
Category:Reflexive Closures
Reflexive Closure of Relation Compatible with Operation is Compatible

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\RR$ be a relation on $S$ which is compatible with $\circ$.
Let $\RR^=$ be the reflexive closure of $\prec$.
That is, $\RR^=$ is defined as the union of $\RR$ with the diagonal relation for $S$.
Then $\RR^=$ is compatible with $\circ$.
== Proof ==
By Diagonal Relation is Universally Compatible, the diagonal relation is compatible with $\circ$.
Then by Union of Relations Compatible with Operation is Compatible, $\RR^=$ is compatible with $\circ$.
== Also see ==
* Reflexive Reduction of Relation Compatible with Group Operation is Compatible
Category:Compatible Relations
Category:Reflexive Closures
Reflexive Closure of Strict Ordering is Ordering

== Theorem ==
Let $S$ be a set.
Let $\prec$ be a strict ordering on $S$.
Let $\preceq$ be the reflexive closure of $\prec$.
Then $\preceq$ is an ordering.
== Proof ==
Since $\prec$ is a strict ordering, it is by definition transitive and asymmetric.
By Asymmetric Relation is Antisymmetric, $\prec$ is antisymmetric.
Thus by Reflexive Closure of Transitive Antisymmetric Relation is Ordering, $\preceq$ is an ordering.
== Also see ==
* Reflexive Reduction of Ordering is Strict Ordering
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.7$
* : $\S \text I.1$: Lemma 1
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $1 \ \text {(b)}$
Category:Strict Orderings
Category:Reflexive Closures
Reflexive Closure of Strict Total Ordering is Total Ordering

== Theorem ==
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 $S$.
== Theorem ==
By the definition of strict total ordering, $\prec$ is a strict ordering which connects $S$.
By Reflexive Closure of Strict Ordering is Ordering, $\preceq$ is a ordering.
Since $\prec$ connects $S$, for each $a, b \in S$, either $a = b$, $a \prec b$ or $b \prec a$.
If $a = b$, then $a \preceq b$.
If $a \prec b$, then $a \preceq b$.
If $b \prec a$, then $b \preceq a$.
Thus in all cases either $a \preceq b$ or $b \preceq a$.
Thus $\preceq$ is a total ordering of $S$.
Category:Strict Orderings
Category:Total Orderings
Category:Reflexive Closures
Reflexive Closure of Transitive Antisymmetric Relation is Ordering

== Theorem ==
Let $S$ be a set.
Let $\prec$ be a transitive, antisymmetric relation on $S$.
Let $\preceq$ be the reflexive closure of $\prec$.
Then $\preceq$ is an ordering on $S$.
== Proof ==
=== Reflexive ===
Follows from Reflexive Closure is Reflexive.
=== Transitive ===
Follows from Reflexive Closure of Transitive Relation is Transitive.
=== Antisymmetric ===
Let $a, b \in S$.
Suppose that $a \ne b$.
Then by the definition of the diagonal relation:
:$\tuple {a, b} \notin \Delta_S$ and 
:$\tuple {b, a} \notin \Delta_S$
Since $\prec$ is antisymmetric, $\tuple {a, b}$ and $\tuple {b, a}$ are not both in $\prec$.
Thus $a \not \preceq b$ or $b \not \preceq a$.
That is, $\preceq$ is antisymmetric.
It has been demonstrated that $\preceq$ is reflexive, antisymmetric and transitive.
Hence by definition $\preceq$ is an ordering,
Category:Order Theory
Category:Reflexive Closures
Category:Transitive Relations
Category:Antisymmetric Relations
Reflexive Closure of Transitive Relation is Transitive

== Theorem ==
Let $S$ be a set.
Let $\RR$ be a transitive relation.
Let $\RR^=$ be the reflexive closure of $\RR$.
Then $\RR^=$ is also transitive.
== Proof ==
Let $a, b, c \in S$.
Suppose that $a \mathrel {\RR^=} b$ and $b \mathrel {\RR^=} c$.
If $a = b$, then since $b \mathrel {\RR^=} c$, also $a \mathrel {\RR^=} c$.
If $b = c$, then since $a \mathrel {\RR^=} b$, also $a \mathrel {\RR^=} c$.
The only case that remains is that $a \ne b$ and $b \ne c$.
Then by the definition of $\RR^=$, $a \mathrel \RR b$ and $b \mathrel \RR c$.
Since $\RR$ is transitive, it follows that:
:$a \mathrel \RR c$
and hence also $a \mathrel {\RR^=} c$.
Thus $\RR^=$ is transitive.
== Also see ==
* Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering
* Transitive Closure of Reflexive Relation is Reflexive
== Sources ==
* : $\S 1.19$: Some Important Properties of Relations: Exercise $2$
Category:Reflexive Closures
Reflexive Euclidean Relation is Equivalence

== Theorem ==
A relation is an equivalence it is either left-Euclidean or right-Euclidean, and also reflexive.
== Proof ==
=== Sufficient Condition ===
Let $\RR$ be an equivalence relation.
Then by definition it is reflexive.
It is also transitive and symmetric.
So, let $x, y, z \in \RR$ such that $x \mathrel \RR y$ and $x \mathrel \RR z$.
From symmetry, we have $y \mathrel \RR x$, and by transitivity it follows that $y \mathrel \RR z$.
Hence $\RR$ is right-Euclidean.
Now let $x, y, z \in \RR$ such that $x \mathrel \RR z$ and $y \mathrel \RR z$.
From symmetry, we have $z \mathrel \RR y$, and by transitivity it follows that $x \mathrel \RR y$.
Hence $\mathrel \RR$ is left-Euclidean.
So $\mathrel \RR$ is both left-Euclidean and right-Euclidean.
We know by definition that it is reflexive.
Thus an equivalence relation is both left-Euclidean and right-Euclidean and also reflexive.
=== Necessary Condition ===
Let $\RR$ be a relation which is both left-Euclidean and reflexive.
Checking in turn each of the criteria for equivalence:
==== Reflexive ====
By definition, $\RR$ has been defined as a relation which is reflexive.
==== Symmetric ====
Suppose $x \mathrel \RR y$.
We have that $y \mathrel \RR y$ from the reflexivity of $\RR$.
So we have $y \mathrel \RR y$ and $x \mathrel \RR y$, and so $y \mathrel \RR x$ from the fact that $\RR$ is left-Euclidean.
Hence $\RR$ is symmetric.
==== Transitive ====
Suppose $x \mathrel \RR y$ and $y \mathrel \RR z$.
As $\RR$ is symmetric (from above), we have that $z \mathrel \RR y$.
So we have $x \mathrel \RR y$ and $z \mathrel \RR y$, and by the fact that $\RR$ is left-Euclidean, we have that $x \mathrel \RR z$.
So, if $\RR$ be a relation which is both left-Euclidean and reflexive, it is also an equivalence relation.
A similar argument shows that if $\RR$ be a relation which is both right-Euclidean and reflexive, it is also an equivalence relation.
Category:Reflexive Relations
Category:Euclidean Relations
Category:Equivalence Relations
Reflexive Reduction is Antireflexive

== Theorem ==
Let $\RR$ be a relation on a set $S$.
Let $\RR^\ne$ denote the reflexive reduction of $\RR$.
Then $\RR^\ne$ is antireflexive.
== Proof ==
By the definition of reflexive reduction:
:$\RR^\ne = \RR \setminus \Delta_S$
where $\Delta_S$ denotes the diagonal relation on $S$.
By Set Difference Intersection with Second Set is Empty Set:
:$\paren {\RR \setminus \Delta_S} \cap \Delta_S = \O$
Hence by Relation is Antireflexive iff Disjoint from Diagonal Relation, $\RR^\ne$ is antireflexive.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.7: 1^\circ$
Category:Reflexive Reductions
Reflexive Reduction of Antisymmetric Relation is Asymmetric

== Theorem ==
Let $S$ be a set.
Let $\RR$ be an antisymmetric relation on $S$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is asymmetric.
== Proof ==
 $\RR^\ne$ is not asymmetric.
That is:
:$\exists a, b \in S: a \mathrel {\RR^\ne} b$ and $b \mathrel {\RR^\ne} a$
Then by the definition of reflexive reduction:
:$a \mathrel \RR b$, $b \mathrel \RR a$
and $a \ne b$.
But this contradicts the antisymmetry of $\RR$.
Thus, by definition, $\RR^\ne$ is an asymmetric relation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.7: 2^\circ$
Category:Reflexive Reductions
Reflexive Reduction of Ordering is Strict Ordering

== Theorem ==
Let $\RR$ be an ordering on a set $S$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is a strict ordering on $S$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Reflexive Closure of Strict Ordering is Ordering
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings (passim)
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $1 \ \text {(b)}$
Category:Strict Orderings
Category:Reflexive Reductions
Category:Reflexive Reduction of Ordering is Strict Ordering
Reflexive Reduction of Ordering is Strict Ordering/Proof 1

== Theorem ==
== Proof ==
=== Antireflexivity ===
Follows from Reflexive Reduction is Antireflexive.
=== Transitivity ===
Suppose $\tuple {x, y}, \tuple {y, z} \in \RR^\ne$.
By antireflexivity $x \ne y$ and $y \ne z$.
We consider the two remaining cases.
==== Case 1: $x = z$ ====
If $x = z$ then:
:$\tuple {x, y}, \tuple {y, x} \in \RR^\ne$
and so:
:$\tuple {x, y}, \tuple {y, x} \in \RR$
Then by the antisymmetry of $\RR$:
:$x = y$
and:
:$\tuple {x, x} \in \RR^\ne$
which contradicts that $\RR^\ne$ is antireflexive.
==== Case 2: $x \ne z$ ====
By the transitivity of $\RR$:
:$\tuple {x, z} \in \RR$
and by $x$ and $z$ being distinct:
:$\tuple {x, z} \notin \Delta_S$
It follows by the definition of reflexive reduction:
:$\tuple {x, z} \in \RR^\ne$
Hence $\RR^\ne$ is transitive.
Category:Reflexive Reduction of Ordering is Strict Ordering
Reflexive Reduction of Relation Compatible with Cancellable Operation is Compatible

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure such that $\circ$ is a cancellable operation.
Let $\RR$ be a relation on $S$ which is compatible with $\circ$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is compatible with $\circ$.
== Proof ==
 $\RR^\ne$ is not compatible with $\circ$.
Let $x, y \in S$ such that:
:$x \mathrel \RR y$
but:
:$x \mathrel {\RR^\ne} y$
Then by definition of reflexive reduction:
:$x \ne y$
Then as $\RR^\ne$ is not compatible with $\circ$:
:$\exists z \in S: \lnot \paren {z \circ x \mathrel {\RR^\ne} z \circ y}$
But as $\RR$ is compatible with $\circ$:
:$z \circ x \mathrel \RR z \circ y$
That means:
:$z \circ x = z \circ y$
As $\circ$ is cancellable this leads to:
:$x = y$
which is a contradiction.
Hence $\RR^\ne$ must be compatible with $\circ$.
Category:Cancellability
Category:Compatible Relations
Category:Reflexive Reductions
Reflexive Reduction of Relation Compatible with Group Operation is Compatible

== Theorem ==
Let $\struct {S, \circ}$ be a group.
Let $\RR$ be a relation on $S$ which is compatible with $\circ$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is compatible with $\circ$.
== Proof 1 ==
== Proof 2 ==
Category:Compatible Relations
Category:Reflexive Reductions
Category:Reflexive Reduction of Relation Compatible with Group Operation is Compatible
Reflexive Reduction of Relation Compatible with Group Operation is Compatible/Proof 1

== Theorem ==
== Proof ==
By definition of reflexive reduction, for all $a, b \in S$:
:$a \mathrel {\RR^\ne} b$ $a \mathrel \RR b$ but $a \ne b$.
By definition of the diagonal relation $\Delta_S$:
:$a \ne b$ $\tuple {a, b} \notin \Delta_S$
Thus, considered as subsets of $S \times S$, we have:
:$\RR^\ne = \RR \setminus \Delta_S$
By Diagonal Relation is Universally Compatible, $\Delta_S$ is compatible with $\circ$.
Thus by Set Difference of Relations Compatible with Group Operation is Compatible, $\RR^\ne$ is compatible with $\circ$.
Category:Reflexive Reduction of Relation Compatible with Group Operation is Compatible
Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering

== Theorem ==
Let $S$ be a set.
Let $\RR$ be a transitive, antisymmetric relation on $S$.
Let $\RR^\ne$ denote the reflexive reduction of $\RR$.
Then $\RR^\ne$ is a strict ordering.
== Proof ==
To show that $\RR^\ne$ is a strict ordering, it is sufficient to show that $\RR^\ne$ is antireflexive and transitive.
=== Antireflexive ===
Follows from Reflexive Reduction is Antireflexive.
=== Transitive ===
Let $a, b, c \in S$.
Let $a \mathrel {\RR^\ne} b$ and $b \mathrel {\RR^\ne} c$.
By the definition of reflexive reduction:
:$a \ne b$ and $a \mathrel \RR b$
:$b \ne c$ and $b \mathrel \RR c$
Since $\mathrel \RR$ is transitive:
:$a \mathrel \RR c$
 $a = c$.
Since $a \mathrel \RR b$ it follows that $c \mathrel \RR b$.
Since $c \mathrel \RR b$, $b \mathrel \RR c$, and $\mathrel \RR$ is antisymmetric, $b = c$.
But this contradicts $b \ne c$.
The conclusion is that $a \ne c$.
Recall that $a \mathrel \RR c$.
By the definition of reflexive reduction:
:$a \mathrel {\RR^\ne} c$
Category:Reflexive Reductions
Category:Strict Orderings
Reflexive Reduction of Transitive Relation is Transitive

== Theorem ==
Let $S$ be a set.
Let $\RR$ be a transitive relation on $S$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is transitive.
== Proof ==
Let $a, b, c \in S$.
Let $a \mathrel {\RR^\ne} b$ and $b \mathrel {\RR^\ne} c$.
By the definition of reflexive reduction:
:$a \ne b$ and $a \mathrel \RR b$
:$b \ne c$ and $b \mathrel \RR c$
Since $\mathrel \RR$ is transitive:
:$a \mathrel \RR c$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.7: 3^\circ$
Category:Reflexive Reductions
Reflexive Reduction of Well-Founded Relation is Strictly Well-Founded Relation

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be a well-founded relation on $S$.
Let $\RR^{\ne}$ be the reflexive reduction of $\RR$.
Then $\RR^{\ne}$ is a strictly well-founded relation.
== Proof ==
Let $T$ be a non-empty subset of $S$.
Since $\RR$ is a well-founded relation, $T$ has a minimal element with respect to the relation $\RR$.
That is, there is an element $m \in T$ such that $\forall x \in T: \paren {\tuple {x, m} \notin \RR} \lor \paren {x = m}$.
Let $x \in T$.
Then $\tuple {x, m} \notin \RR$ or $x = m$.
By the definition of reflexive reduction, $\RR^{\ne}$ is a subset of $\RR$.
Thus:
:if $\tuple {x, m} \notin \RR$, it follows that $\tuple {x, m} \notin \RR^{\ne}$.
:if $x = m$ then by Reflexive Reduction is Antireflexive, $\tuple {x, m} \notin \RR^{\ne}$.
As $x$ is arbitrary, this holds for all $x \in T$.
Thus $m$ is a strictly minimal element under $\RR^{\ne}$ in $T$.
As $T$ is arbitrary, this condition holds for all non-empty subsets of $S$.
That is, every non-empty subset of $S$ has a strictly minimal element under $\RR^{\ne}$.
Hence, by definition, $\RR^{\ne}$ is a strictly well-founded relation on $S$.
Category:Reflexive Reductions
Category:Well-Founded Relations
Reflexive Relation/Examples/Arbitrary Reflexive Relation

== Examples of Reflexive Relations ==
Let $V_0 = \set {a, b, c}$.
A reflexive relation on $V_0$ must include the ordered pairs:
:$\tuple {a, a}, \tuple {b, b}, \tuple {c, c}$
== Sources ==
* : Appendix $\text{A}.2$: Cartesian Products and Relations
Category:Examples of Reflexive Relations
Reflexive Relation/Examples/Distance Less than 1

== Example of Reflexive Relation ==
Let $\sim$ be the relation on the set of real numbers $\R$ defined as:
:$x \sim y \iff \size {x - y} < 1$
Then $\sim$ is reflexive and symmetric, but not transitive.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 7$: Relations
Category:Examples of Reflexive Relations
Reflexive Relation/Examples/Divisibility on Natural Numbers

== Example of Reflexive Relation ==
Let $\sim$ be the relation on the set of natural numbers $\N$ defined as:
:$x \sim y \iff x \divides y$
where $\divides$ denotes divisibility.
Then $\sim$ is reflexive.
== Proof ==
This is demonstrated in Divides is Reflexive.
== Sources ==
* 
* 
Category:Examples of Reflexive Relations