nyuuzyou/wikis huggingface dataset:
Operation which is Right Distributive over Every Commutative Associative Operation is Left Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ have the property that:
:for every arbitrary operation $*$ on $S$ which is both commutative and associative, $\circ$ is right distributive over $*$.
Then $\circ$ is the left operation $\gets$:
:$\forall a, b \in S: a \gets b = a$
== Proof ==
First recall from Left Operation is Right Distributive over All Operations that the left operation is indeed right distributive over all operations, whether commutative or associative.
Let $*$ be an arbitrary operation on $S$ which is both commutative and associative.
As asserted, let $\circ$ be right distributive over $*$.
Let $c \in S$ be arbitrary.
Consider the constant operation $\sqbrk c$:
:$\forall a, b \in S: a \sqbrk c b := c$
We have from Constant Operation is Commutative and Constant Operation is Associative that $\sqbrk c$ is both commutative and associative.
Hence $\circ$ must be right distributive over $\sqbrk c$.
But then from Condition for Operation to be Right Distributive over Constant Operation:
:$c \circ x = c$
As $c$ is arbitrary:
:$\forall a, b \in S: a \circ b = a$
so $\circ$ has to be the left operation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.25$
Category:Left Operation
Category:Distributive Operations
Category:Commutativity
Category:Associativity
Operations of Boolean Algebra are Associative

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1.
Then:
That is, both $\vee$ and $\wedge$ are associative operations.
== Proof ==
Let $a, b, c \in S$.
Let:
:$x = a \wedge \paren {b \wedge c}$
:$y = \paren {a \wedge b} \wedge c$
Then:
Similarly:
Thus we see we have $a \vee x = a \vee y$.
Next:
Similarly:
Thus we see we have $\neg a \vee x = \neg a \vee y$.
In conclusion, we have:
:$a \vee x = a \vee y$
:$\neg a \vee x = \neg a \vee y$
Hence $x = y$ by Cancellation of Join in Boolean Algebra, that is:
:$\paren {a \wedge b} \wedge c = a \wedge \paren {b \wedge c}$
The result:
:$\paren {a \vee b} \vee c = a \vee \paren {b \vee c}$
follows from the Duality Principle.
== Sources ==
* : $\S 1.5$: Theorem $1.16$
* 
* : $\S 2$: Exercise $2$
* 
Category:Boolean Algebras
Operations of Boolean Algebra are Idempotent

== Definition ==
Let $\struct {S, \vee, \wedge}$ be a Boolean algebra.
Then:
:$\forall x \in S: x \wedge x = x = x \vee x$
That is, both $\vee$ and $\wedge$ are idempotent operations.
== Proof ==
Let $x \in S$.
Then:
So $x = x \vee x$.
The result $x = x \wedge x$ follows from Duality Principle (Boolean Algebras).
== Sources ==
* : $\S 1.5$: Theorem $1.13$
* 
* : $\S 2$: Exercise $2$
* 
Category:Boolean Algebras
Operations with Identities which Distribute over each other are Idempotent

== Theorem ==
Let $S$ be a set.
Let $\odot$ and $\otimes$ be operations on $S$ which have identity elements $e$ and $u$ respectively.
Let $\odot$ and $\otimes$ be distributive over each other.
Then $\odot$ and $\otimes$ are both idempotent.
== Proof ==
In the same way:
Then we have:
and:
Hence the result by definition of idempotent operation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.22$
Category:Idempotence
Category:Distributive Operations
Operator Commuting with Diagonalizable Operator

== Theorem ==
Let $H$ be a Hilbert space.
Let $A = \ds \sum_{i \mathop \in I} \alpha_i P_i$ be a diagonalizable operator on $H$.
Let $B \in \map B H$ be a bounded linear operator.
Then the following are equivalent:
:$(1): \quad A B = B A$
:$(2): \quad$ For all $i \in I$, $\Rng {P_i}$ is a reducing subspace for $B$
where $\Rng {P_i}$ denotes range.
== Proof ==
== Sources ==
* $II.7.5$
Category:Linear Transformations on Hilbert Spaces
Operator Diagonalizable iff Basis of Eigenvectors

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $A: \HH \to \HH$ be a linear operator on $\HH$.
Then $A$ is diagonalizable there exists a basis $E$ of $\HH$, consisting of eigenvectors for $A$.
== Proof ==
== Sources ==
* $\text {II}.7.4$
Category:Linear Transformations on Hilbert Spaces
Operator Generated by Closure System Preserves Directed Suprema iff Closure System Inherits Directed Suprema

== Theorem ==
Let $L = \struct {X, \vee, \wedge, \preceq}$ be a complete lattice.
Let $S = \struct {Y, \precsim}$ be a closure system on $L$.
Then $\map {\operatorname {operator} } S$ preserves directed suprema
 $S$ inherits directed suprema.
where $\map {\operatorname {operator} } S$ denotes the operator generated by $S$.
== Proof ==
=== Sufficient Condition ===
Assume that $\map {\operatorname {operator} } S$ preserves directed suprema.
Let $Z$ be directed subset of $Y$ such that
:$Z$ admits a supremum in $L$.
By Image of Operator Generated by Closure System is Set of Closure System:
:$\map {\operatorname {operator} } S \sqbrk X = Y$
By Operator Generated by Closure System is Closure Operator:
:$\map {\operatorname {operator} } S$ is closure operator.
By definition of closure operator/idempotent:
:$\map {\operatorname {operator} } S \sqbrk Z = Z$
By definitions of Definition:Mapping Preserves Supremum/Directed and Definition:Mapping Preserves Supremum/Subset:
:$\sup_L Z = \map {\map {\operatorname {operator} } S} {\sup_L Z}$
Thus by definition of image of mapping:
:$\sup_L Z \in Y$
=== Necessary Condition ===
Assume that
:$S$ inherits directed suprema.
Define $f := \map {\operatorname {operator} } S$.
By Image of Operator Generated by Closure System is Set of Closure System:
:$f \sqbrk X = Y$
Let $A$ be a directed subset of $X$ such that
:$A$ admits a supremum in $L$.
Thus by definition of complete lattice:
:$f \sqbrk A$ admits a supremum in $L$.
We will prove that
:$\map f {\sup_L A}$ is upper bound for $f \sqbrk A$
Let $x \in f \sqbrk A$.
By definition of image of subset:
:$\exists y \in A: x = \map f y$
By definitions of supremum and upper bound:
:$y \preceq \sup_L A$
Thus by definition of closure operator/increasing:
:$x \preceq \map f {\sup_L A}$
By definition of supremum:
:$\map {\sup_L} {f \sqbrk A} \preceq \map f {\sup_L A}$
We will prove that
:$\map {\sup_L} {f \sqbrk A}$ is upper bound for $A$
Let $x \in A$.
By definition of image of set:
:$\map f x \in f \sqbrk A$
By definitions of supremum and upper bound:
:$\map f x \preceq \map {\sup_L} {f \sqbrk A}$
By definition of closure operator/inflationary:
:$x \preceq \map f x$
Thus by definition of transitivity:
:$x \preceq \map {\sup_L} {f \sqbrk A}$
By definition of supremum:
:$\sup_L A \preceq \map {\sup_L} {f \sqbrk A}$
By Image of Directed Subset under Increasing Mapping is Directed:
:$f \sqbrk A$ is directed in $L$.
By definition of ordered subset:
:$f \sqbrk A$ is directed in $S$.
By definition of directed suprema inheriting:
:$\map {\sup_L} {f \sqbrk A} \in Y$
By definition of closure operator/idempotent:
:$\map f {\map {\sup_L} {f \sqbrk A} } = \map {\sup_L} {f \sqbrk A}$
By definition of closure operator/increasing:
:$\map f {\sup_L A} \preceq \map {\sup_L} {f \sqbrk A}$
Thus by definition of antisymmetry:
:$\map f {\sup_L A} = \map {\sup_L} {f \sqbrk A}$
== Sources ==
* 
* 
Category:Closure Operators
Operator Generated by Closure System is Closure Operator

== Theorem ==
Let $L = \left({X, \vee, \wedge, \preceq}\right)$ be a complete lattice.
Let $S = \left({T, \precsim}\right)$ be a closure system of $L$.
Then $\operatorname{operator}\left({S}\right)$ is closure operator,
where $\operatorname{operator}\left({S}\right)$ denotes the operator generated by $S$.
== Proof ==
Define $f = \operatorname{operator}\left({S}\right)$.
=== Inflationary ===
By Intersection is Subset:
:$x^\succeq \cap T \subseteq x^\succeq$
By Infimum of Subset:
:$\inf x^\succeq \preceq \inf\left({x^\succeq \cap T}\right)$
By Infimum of Upper Closure of Element:
:$x \preceq \inf\left({x^\succeq \cap T}\right)$
Thus by definition of operator generated by $S$:
:$x \preceq f\left({x}\right)$
=== Increasing ===
Let $x, y \in X$ such that
:$x \preceq y$
By Upper Closure is Decreasing:
:$y^\succeq \subseteq x^\succeq$
By Set Intersection Preserves Subsets/Corollary:
:$y^\succeq \cap T \subseteq x^\succeq \cap T$
By Infimum of Subset:
:$\inf\left({x^\succeq \cap T}\right) \preceq \inf\left({y^\succeq \cap T}\right)$
Thus by definition of operator generated by $S$:
:$f\left({x}\right) \preceq f\left({y}\right)$
=== Idempotent ===
Let $x \in X$.
By Intersection is Subset:
:$x^\succeq \cap T \subseteq T$
By definition of complete lattice:
:$x^\succeq \cap T$ admits an infimum in $L$.
By definition of infima inheriting:
:$x^\succeq \cap T$ admits an infimum in $S$ and $\inf_S\left({x^\succeq \cap T}\right) = \inf_L \left({x^\succeq \cap T}\right)$
Then by definition of operator generated by $S$:
:$f\left({x}\right) \in T$
and
:$f\left({f\left({x}\right)}\right) = \inf_L \left({f\left({x}\right)^\succeq \cap T}\right)$
By definitions of upper closure of element and reflexivity:
:$f\left({x}\right) \in f\left({x}\right)^\succeq$
By definition of intersection:
:$f\left({x}\right) \in f\left({x}\right)^\succeq \cap T$
By definitions of infimum and lower bound:
:$\inf_L \left({f\left({x}\right)^\succeq \cap T}\right) \preceq f\left({x}\right)$
By Intersection is Subset:
:$f\left({x}\right)^\succeq \cap T \subseteq f\left({x}\right)^\succeq$
By Infimum of Subset:
:$\inf_L \left({f\left({x}\right)^\succeq}\right) \preceq \inf_L \left({f\left({x}\right)^\succeq \cap T}\right)$
By Infimum of Upper Closure of Element:
:$f\left({x}\right) \preceq \inf_L \left({f\left({x}\right)^\succeq \cap T}\right)$
Thus by definition of antisymmetry:
:$f\left({f\left({x}\right)}\right) = f\left({x}\right)$
Hence $f$ is closure operator.
== Sources ==
* 
* 
Category:Closure Operators
Operator Generated by Image of Closure Operator is Closure Operator

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $c: S \to S$ be a closure operator on $L$.
Then $\map {\operatorname {operator} } {\struct {c \sqbrk S, \precsim} } = c$
where
:$\mathord \precsim = \mathord \preceq \cap \paren {c \sqbrk S \times c \sqbrk S}$
:$\map {\operatorname {operator} } {\struct {c \sqbrk S, \precsim} }$ denotes the operator generated by $\struct {c \sqbrk S, \precsim}$
== Proof ==
Let $x \in S$.
By definition of closure operator/inflationary:
:$x \preceq \map c x$
By definition of upper closure of element:
:$\map c x \in x^\succeq$
By definition of image of mapping:
:$\map c x \in c \sqbrk S$
By definition of intersection:
:$\map c x \in x^\succeq \cap c \sqbrk S$
By definitions of infimum and lower bound:
:$\map {\inf_L} {x^\succeq \cap c \sqbrk S} \preceq \map c x$
We will prove that
:$\map c x$ is lower bound for $x^\succeq \cap c \sqbrk S$
Let $y \in x^\succeq \cap c \sqbrk S$
By definition of intersection:
:$y \in x^\succeq$ and $y \in c \sqbrk S$
By definition of image of mapping:
:$\exists z \in S: y = \map c z$
By definition of closure operator/idempotent:
:$y = \map c y$
By definition of upper closure of element:
:$x \preceq y$
Thus by definition of closure operator/increasing:
:$\map c x \preceq y$
By definition of infimum:
:$\map c x \preceq \map {\inf_L} {x^\succeq \cap c \sqbrk S}$
By definition of antisymmetry:
:$\map c x = \map {\inf_L} {x^\succeq \cap c \sqbrk S}$
Thus by definition of operator generated by $\struct {c \sqbrk S, \precsim}$:
:$\map {\map {\operatorname {operator} } {\struct {c \sqbrk S, \precsim} } } x = \map c x$
Hence by Equality of Mappings:
:$\map {\operatorname {operator} } {\struct {c \sqbrk S, \precsim} } = c$
== Sources ==
* 
* 
Category:Closure Operators
Operator Norm of Quotient Mapping in Quotient Normed Vector Space is 1

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a normed vector space over $\Bbb F$.
Let $N$ be a closed linear subspace of $X$. 
Let $\struct {X/N, \norm {\, \cdot \,}_{X/N} }$ be the normed quotient vector space associated with quotient vector space $X/N$.
Let $\pi : X \to X/N$ be the quotient mapping associated with $X/N$. 
Then:
:$\norm \pi_{\map B {X, X/N} } = 1$ 
where $\norm {\, \cdot \,}_{\map B {X, X/N} }$ denotes the norm on the space of bounded linear transformations.
== Proof ==
Let $B_X$ be the unit open ball in $\struct {X, \norm {\, \cdot \,} }$.
Let $B_{X/N}$ be the unit open ball in $\struct {X/N, \norm {\, \cdot \,} }$.
From Quotient Mapping is Bounded in Normed Quotient Vector Space, $\pi$ is bounded.
We have: 
Category:Quotient Mappings
Category:Normed Quotient Vector Spaces
Operator Zero iff Inner Product Zero

== Theorem ==
Let $\HH$ be a Hilbert space over $\C$.
Let $A: \HH \to \HH$ be a bounded linear operator.
Suppose that:
:$\forall h \in \HH: \innerprod {A h} h_\HH = 0$
Then $A$ is the zero operator.
== Proof ==
For each $x, y \in \HH$ we have: 
So:
:$\innerprod {A x} y = 0$
for each $x, y \in \HH$. 
Setting $y = A x$, we have: 
:$\norm {A x}^2 = 0$
by the definition of the inner product norm.
So from , we have $A x = 0$ for each $x \in \HH$.
== Also see ==
* Norm of Hermitian Operator/Corollary
== Sources ==
* $\text {II}.2.15$
Category:Linear Transformations on Hilbert Spaces
Operator is Hermitian iff Numerical Range is Real

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ be a Hilbert space over $\C$. 
Let $A : \HH \to \HH$ be a bounded linear operator.
Let $\map W A$ be the numerical range of $A$.
Then $A$ is Hermitian :
:$\map W A \subseteq \R$
That is:
:$\forall h \in \HH: \innerprod {A h} h_\HH \mathop \in \R$
== Proof ==
=== Necessary Condition ===
Suppose that $A$ is Hermitian. 
Then: 
:$A^* = A$
where $A^*$ denotes the adjoint of $A$.
Let $x \in \HH$. 
Then, by the definition of the adjoint, we have: 
:$\innerprod {A x} x_\HH = \innerprod x {A x}_\HH$
From the conjugate symmetry of the inner product, we have: 
:$\innerprod x {A x}_\HH = \overline {\innerprod {A x} x_\HH}$
So:
:$\innerprod {A x} x_\HH = \overline {\innerprod {A x} x_\HH}$
So, from Complex Number equals Conjugate iff Wholly Real:
:$\innerprod {A x} x_\HH$ is a real number.
=== Sufficient Condition ===
Suppose that: 
:$\innerprod {A x} x_\HH$ is a real number for each $x \in \HH$. 
Let $\alpha \in \C$. 
Let $x, y \in \HH$. 
We have: 
We have that: 
:$\innerprod {\map A {x + \alpha y} } {x + \alpha y}_\HH$
is a real number.
Note that both: 
:$\innerprod {A x} x_\HH$
and:
:$\size \alpha^2 \innerprod {A y} y_\HH$
are also real numbers.
So, we must have that:
:$\overline \alpha \innerprod {A x} y_\HH + \alpha \innerprod {A y} x_\HH$
is a real number.
We therefore have: 
Setting $\alpha = 1$, we have: 
:$\innerprod {A x} y_\HH + \innerprod {A y} x_\HH = \innerprod {A^* y} x_\HH + \innerprod {A^* x} y_\HH$
Setting $\alpha = i$, we have: 
:$-i \innerprod {A x} y_\HH + i \innerprod {A y} x_\HH = i \innerprod {A^* y} x_\HH - i \innerprod {A^* y} x_\HH$
Dividing by $i$ we have: 
:$-\innerprod {A x} y_\HH + \innerprod {A y} x_\HH = \innerprod {A^* y} x_\HH - \innerprod {A^* y} x_\HH$
So:
:$2 \innerprod {A y} x_\HH = 2 \innerprod {A^* y} x_\HH$
giving:
:$\innerprod {A y} x_\HH = \innerprod {A^* y} x_\HH$
From Inner Product is Sesquilinear:
:$\innerprod {A y - A^* y} x_\HH = 0$
Setting $x = A y - A^* y$ we have: 
:$\innerprod {A y - A^* y} {A y - A^* y}_\HH = 0$ 
From the positiveness of the inner product we have: 
:$A y - A^* y = 0$
so:
:$A y = A^* y$
Since $y \in \HH$ was arbitrary, we have: 
:$A = A^*$
That is:
:$A$ is Hermitian.
== Sources ==
* $\text {II}.2.12$
Category:Hermitian Operators
Operator of Integrated Weighted Derivatives is Linear Mapping

== Theorem ==
Let $n \in \N$.
Let $I := \closedint a b$ be a closed real interval.
Let $\map {a_i} x : I \to \R$ be Riemann integrable functions.
Let $f, g \in \map {C^n} I$ be Riemann integrable real-valued functions of differentiability class $k$.
Let $L : \map {C^n} I \to \R$ be the operator of integrated weighted derivatives.
Then $L$ is a linear mapping.
== Proof ==
=== Distributivity ===
=== Positive homogenity ===
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Linear Transformations
Category:Operator Theory
Operator with Zero Numerical Range is Zero Operator

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\norm {\, \cdot \,}$ be the inner product norm on $\struct {\HH, \innerprod \cdot \cdot}$.
Let $\struct {\map D T, T}$ be a densely-defined linear operator on $\HH$ such that:
:$\map W T = \set 0$
where $\map W T$ is the numerical range of $T$. 
Then $T = 0$. 
=== Corollary ===
== Proof ==
Let $x, y \in \map D T$. 
Since $\map W T = \set 0$, we have:
:$\innerprod {\map T {x + y} } {x + y} = \innerprod {T x} x = \innerprod {T y} y = 0$
From Inner Product is Sesquilinear, we have:
:$\innerprod {T x} x + \innerprod {T x} y + \innerprod {T y} x + \innerprod {T y} y = 0$
That is:
:$\innerprod {T x} y = -\innerprod {T y} x$
Similarly, replacing $y$ with $i y$ and applying Inner Product is Sesquilinear, we have:
:$-i \innerprod {T x} y = -\innerprod {\map T {i y} } x$
Since $T$ is linear, we have:
:$\innerprod {T x} y = \innerprod {T y} x$
Hence:
:$\innerprod {T x} y = -\innerprod {T x} y$
This gives:
:$\innerprod {T x} y = 0$ for all $x, y \in \map D T$.
That is:
:$T x \in \map D T^\bot$ for all $x \in \map D T$
where $\map D T^\bot$ is the orthocomplement of $\map D T$.
From Linear Subspace Dense iff Zero Orthocomplement, we have $\map D T^\bot = \set 0$.
Hence:
:$T x = 0$ for all $x \in \map D T$.
== Sources ==
* : $12.7$: Bounded Operators
Category:Numerical Range
Category:Operator with Zero Numerical Range is Zero Operator
Operator with Zero Numerical Range is Zero Operator/Corollary

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\norm {\, \cdot \,}$ be the inner product norm on $\struct {\HH, \innerprod \cdot \cdot}$.
Let $\DD$ be a dense linear subspace of $\HH$. 
Let $\struct {\DD, T}$ and $\struct {\DD, S}$ be densely-defined linear operators on $\HH$ such that:
:$\innerprod {T x} x = \innerprod {S x} x$ for each $x \in \DD$.
Then $T = S$.
== Proof ==
From Inner Product is Sesquilinear, we have:
:$\innerprod {\paren {T - S} x} x = 0$ for each $x \in \DD$.
Applying Operator with Zero Numerical Range is Zero Operator to the densely-defined linear operator $\struct {\DD, T - S}$, we have:
:$T - S = 0$
That is:
:$T = S$
== Sources ==
* : $12.7$: Bounded Operators
Category:Operator with Zero Numerical Range is Zero Operator
Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles

== Theorem ==
== Proof ==
:
Let $ABCD$ be a cyclic quadrilateral in a circle.
Join $AC$ and $BD$.
From Sum of Angles of Triangle equals Two Right Angles, we have that $\angle CAB + \angle ABC + \angle BCA$ equals two right angles.
But from Angles in Same Segment of Circle are Equal we have that $\angle CAB = \angle BDC$ and also $\angle ACB = \angle ADB$.
If we add $\angle ABC$ to each, we get that $\angle ABC + \angle BAC + \angle ACB = \angle ABC + \angle ADC$.
Therefore $\angle ABC + \angle ADC$ equals two right angles.
Similarly we can show that $\angle BAD + \angle DCB$ also equals two right angles.
== Sources ==
* : Book $\text{III}$. Propositions
* 
* 
Category:Cyclic Quadrilaterals
Opposite Group is Group

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\struct {G, *}$ be the opposite group to $G$.
Then $\struct {G, *}$ is a group.
== Proof 1==
=== ===
$\struct {G, *}$ is closed:
:$b \circ a \in G \implies a * b \in G$
=== ===
$*$ is associative on $G$:
=== ===
Let $e$ be the identity of $\struct {G, \circ}$:
Thus $e$ is the identity of $\struct {G, *}$.
=== ===
Let the inverse of $a \in \struct {G, \circ}$ be $a^{-1}$:
Thus $a^{-1}$ is the inverse of $a \in \struct {G, *}$
So all the group axioms are satisfied, and $\struct {G, *}$ is a group.
== Proof 2 ==
Let $e$ be the identity of $\struct {G, \circ}$:
So:
$e \in \struct {G, *}$
Hence:
:$\struct {G, *}$ is non-empty
As $\struct {G, \circ}$ is a group, it is closed and every element has an inverse.
Therefore:
:$\quad \forall a, b \in \struct {G, \circ}: b^{-1} \circ a \in \struct {G, \circ}$.
Hence, by definition of $*$:
:$\quad \forall a, b \in \struct {G, *}: a * b^{-1} \in \struct {G, *}$.
The result follows from the One-Step Subgroup Test and Group is Subgroup of Itself.
* : Chapter $2$: The Definition of Group Structure: $\S 26 \epsilon$
Category:Opposite Groups
Opposite Group of Opposite Group

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\struct {G, *}$ be the opposite group to $\struct {G, \circ}$.
Let $\struct {G, \circ'}$ be the opposite group to $\struct {G, *}$.
Then:
:$\struct {G, \circ} = \struct {G, \circ'}$
== Proof ==
We have, for all $a, b \in G$:
:$a \circ b = b * a = a \circ' b$
by definition of opposite group.
Hence the result.
Category:Opposite Groups
Opposite Planes of Solid contained by Parallel Planes are Equal Parallelograms

== Theorem ==
== Proof ==
:
Let the solid $CDHG$ be contained by the parallel planes $AC, GF, AH, DF, BF, AE$.
It is to be demonstrated that opposite planes are equal parallelograms.
We have that the two parallel planes $BG$ and $CE$ are cut by the plane $AC$.
From :
:the common sections of $BG$ and $CE$ are parallel lines.
Thus $AB \parallel DC$.
Again, we have that the two parallel planes $BF$ and $AE$ are cut by the plane $AC$.
Thus $BC \parallel AD$.
But $AB \parallel DC$.
Therefore $AC$ is by definition a parallelogram.
Similarly it can be shown that each of $GF, AH, DF, BF, AE$ are parallelograms.
Let $AH$ and $DF$ be joined.
We have that:
:$AB \parallel DC$
and:
:$BH \parallel CF$
Thus the two straight lines $AB$ and $BH$ which meet one another are parallel to the two straight lines $DC$ and $CF$ which also meet one another, but not in the same plane.
Therefore by :
:$\angle ABH = \angle DCF$
From :
:$AB$ and $BH$ are equal to $DC$ and $CF$.
and because:
:$\angle ABH = \angle DCF$
it follows that:
:$AH = DF$
So from :
:$\triangle ABH = \triangle DCF$
We have that the parallelogram $BG$ is double $\triangle ABH$.
From :
:the parallelogram $CE$ is double the $\triangle DCF$.
Therefore the parallelogram $BG$ equals the parallelogram $CE$.
Similarly it is shown that:
:the parallelogram $AC$ equals the parallelogram $GF$
and:
:the parallelogram $AE$ equals the parallelogram $BF$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallel Planes
Category:Parallelograms
Opposite Ring is Ring

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $\struct {R, +, *}$ be the opposite ring of $\struct {R, +, \times}$.
Then $\struct {R, +, *}$ is a ring.
== Proof ==
By definition of the opposite ring:
:$\forall x, y \in R: x * y = y \times x$.
By definition of the ring $R$, $\struct {R, +}$ is an abelian group.
To complete the proof, it remains to be shown that $\struct {R, *}$ is a semigroup.
That is, it remains to show that $\struct{R, *}$ is associative.
Let $a, b, c \in R$.
The result follows.
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Examples of Rings
Opposite Ring of Opposite Ring

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $\struct {R, +, *}$ be the opposite ring of $\struct {R, +, \times}$.
Let $\struct {R, +, \circ}$ be the opposite ring of $\struct {R, +, *}$.
Then $\struct {R, +, \circ} = \struct {R, +, \times}$.
== Proof ==
By definition of the opposite ring:
:$\forall x, y \in S: x * y = y \times x$
:$\forall x, y \in S: x \circ y = y * x$
Hence for all $x,y \in S$:
:$x \circ y = y * x = x \times y$
The result follows.
Category:Ring Theory
Opposite Sides Equal implies Parallelogram

== Theorem ==
Let $ABCD$ be a convex quadrilateral with $AB = CD$ and $BC = AD$.
Then $ABCD$ is a parallelogram.
== Proof ==
Join $AC$.
From $(7)$ and $(8)$, it follows by definition that $ABCD$ is a parallelogram.
== Sources ==
* 
Category:Parallelograms