nyuuzyou/wikis huggingface dataset:
Condition for Existence of Expectation of Real-Valued Measurable Function composed with Absolutely Continuous Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be an absolutely continuous random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $P_X$ be the probability distribution of $X$.
Let $\map \BB \R$ be the Borel $\sigma$-algebra of $\R$.
Let $h : \R \to \R$ be a $\map \BB \R$-measurable function.
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$.
Then $\map h X$ is integrable $h$ is $\lambda$-integrable function.
In this case: 
:$\ds \expect {\map h X} = \int_\R \map h x \rd \map {P_X} x$
where $\expect {\map h X}$ denotes the expectation of $\map h X$.
== Proof ==
From Composition of Measurable Mappings is Measurable, we have: 
:$\map h X$ is $\Sigma$-measurable.
So:
:$\map h X$ is a real-valued random variable.
From the definition of the probability distribution, we have: 
:$P_X = X_\ast \Pr$
where $X_\ast \Pr$ denotes the pushforward measure of $\Pr$ under $X$.
Since $X$ is an absolutely continuous random variable, it is $\Sigma$-measurable.
So, from Integral with respect to Pushforward Measure: Corollary:
:$\map h X$ is integrable $h$ is $\lambda$-integrable function.
If $\map h X$ is integrable, Integral with respect to Pushforward Measure: Corollary gives: 
:$\ds \int \map h X \rd \Pr = \int_\R \map h x \rd \map {P_X} x$
that is: 
:$\ds \expect {\map h X} = \int_\R \map h x \rd \map {P_X} x$
== Sources ==
* : $6.12$: The 'elementary formula' for expectation
Category:Absolutely Continuous Random Variables
Category:Expectation
Condition for Existence of Third Number Proportional to Two Numbers

== Theorem ==
Let $a, b, c \in \Z$ be integers.
Let $\tuple {a, b, c}$ be a geometric sequence.
In order for this to be possible, both of these conditions must be true:
:$(1): \quad a$ and $b$ cannot be coprime
:$(2): \quad a \divides b^2$
where $\divides$ denotes divisibility.
== Proof ==
Let $P = \tuple {a, b, c}$ be a geometric sequence.
Then by definition their common ratio is:
:$\dfrac b a = \dfrac c b$
From Two Coprime Integers have no Third Integer Proportional it cannot be the case that $a$ and $b$ are coprime.
Thus condition $(1)$ is satisfied.
From Form of Geometric Sequence of Integers, $P$ is in the form:
:$\tuple {k p^2, k p q, k q^2}$
from which it can be seen that:
:$k p^2 \divides k^2 p^2 q^2$
demonstrating that condition $(2)$ is satisfied.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Proportion
Condition for Factoring of Quotient Mapping between Modulo Addition Groups

== Theorem ==
Let $m, n \in \Z_{>0}$ be strictly positive integers.
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\struct {\Z_m, +_m}$ and $\struct {\Z_n, +_n}$ denote the additive groups of integers modulo $m$ and $n$ respectively.
Let $f: \Z \to \Z_n$ be the quotient epimorphism from $\struct {\Z, +}$ to $\struct {\Z_n, +_n}$.
Let $q: \Z \to \Z_m$ be the quotient epimorphism from $\struct {\Z, +}$ to $\struct {\Z_m, +_m}$.
Then:
:there exists a group homomorphism $\psi: \struct {\Z_m, +_m} \to \struct {\Z_n, +_n}$
 
:$m \divides n$
where $\divides$ denotes divisibility.
== Proof ==
An example of the use of Third Isomorphism Theorem/Groups/Corollary.
== Sources ==
* : Chapter $\text{II}$: Groups: Factoring Morphisms
Category:Additive Groups of Integer Multiples
Condition for Group Endomorphism to Commute with All Inner Automorphisms

== Theorem ==
Let $G$ be a group.
Let $\phi: G \to G$ be an endomorphism on $G$.
Let $\phi$ be such that:
:$\forall a \in G: \kappa_a \circ \phi = \phi \circ \kappa_a$
where:
:$\kappa_a$ denotes the inner automorphism of $G$ given by $a$
:$\circ$ denotes composition of mappings.
Then:
:$H = \set {x \in G: \map \phi {\map \phi x} = \map \phi x}$
is a normal subgroup of $G$.
Also, the quotient group $G / H$ is an abelian group.
== Proof ==
We have for all $a, g \in G$:
Let $x \in H$. 
Then by the above:
Hence $a x a^{-1} \in H$, and thus $a H a^{-1} \subseteq H$.
By definition of a normal subgroup, $H$ is a normal subgroup of $G$.
It remains to show that the quotient group $G / H$ is an abelian group.
We need to show that for any $a, b \in G$, we have:
:$a H b H = b H a H$
that is,
:$a b H = b a H$
By Cosets are Equal iff Product with Inverse in Subgroup, this is equivalent to:
:$\paren {b a}^{-1} a b = a^{-1} b^{-1} a b \in H$
We have:
and thus $a^{-1} b^{-1} a b \in H$.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.21$
Category:Group Endomorphisms
Category:Inner Automorphisms
Condition for Group given Semigroup with Idempotent Element

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let there exist an idempotent element $e$ of $S$ such that for all $a \in S$:
:there exists at least one element $x$ of $S$ satisfying $x \circ a = e$
:there exists at most one element $y$ of $S$ satisfying $a \circ y = e$.
Then $\struct {S, \circ}$ is a group.
== Proof ==
Let $a$ be arbitrary.
We have:
So $e$ is a right identity.
Again, let $a$ be arbitrary.
Let $x \in S$ be such that $x \circ a = e$.
We have:
So $e$ is a left identity.
We note the meaning of the criterion:
:there exists at least one element $x$ of $S$ satisfying $x \circ a = e$
As we now know that $e$ is a left identity, the above means that $x$ is a left inverse for $a$ in $S$.
To summarise, we have an algebraic structure $\struct {S, \circ}$:
:$(1): \quad$ which is closed, from 
:$(2): \quad$ which is associative, from 
:$(3): \quad$ which has a left identity
:$(4): \quad$ for which every element has a left inverse.
That is, $\struct {S, \circ}$ fulfils all the left group axioms.
Hence the result.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.15$
* 
Category:Semigroups
Category:Group Theory
Category:Idempotence
Condition for Group to Act Effectively on Left Coset Space

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Then $G$ acts effectively on the left coset space $G / H$ :
:$\ds \bigcap_{a \mathop \in G} H^a = \set e$
where $H^a$ denotes the conjugate of $H$ by $a$.
== Proof ==
$G$ acts effectively on the left coset space $G / H$ $a H \mapsto g a H$ is faithful, :
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 53 \delta$
Category:Group Action on Coset Space
Category:Cosets
Condition for Homogeneous Quadratic Equation to describe Perpendicular Straight Lines

== Theorem ==
Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a homogeneous quadratic equation $E$ in two variables.
Let $\LL_1$ and $\LL_2$ be perpendicular.
Then $E$ is of the form:
:$a x^2 + 2 h x y - a y^2$
== Proof ==
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, $E$ is of the form:
:$a x^2 + 2 h x y + b y^2$
From Angle Between Two Straight Lines described by Homogeneous Quadratic Equation, the angle $\psi$ between $\LL_1$ and $\LL_2$ is given by:
:$\tan \psi = \dfrac {2 \sqrt {h^2 - a b} } {a + b}$
When $\psi = \dfrac \pi 2$, $\tan \psi$ is undefined.
Hence:
:$a + b = 0$
and so $b = -a$.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $16$. Angle between the two straight lines represented by the homogeneous equation
Category:Straight Lines
Category:Perpendiculars
Condition for Ideal to be Total Ring

== Theorem ==
Let $\struct {A, +, \circ}$ be a commutative ring with unity.
Let $I$ be an ideal of $A$ such that the quotient ring $A / I$ is a field.
Let $J$ be an ideal of $A$ such that $I \subsetneq J$.
Then:
:$A = J$
== Proof ==
Let $A$ be a commutative ring with unity.
Let $I$ be an ideal of $A$ such that the quotient ring $A / I$ is a field.
Let $J$ be an ideal of $A$ such that $I \subsetneq J$.
From Ideal is Subring:
:$J \subseteq A$
It remains to be proved that that $A \subseteq J$.
Let $a \in A$.
As $I \subsetneq J$, it follows from definition of proper subset that:
:$\exists j \in J: j \notin I$
Consider the coset $j + I \in A / I$.
As $A / I$ is a field:
:$\exists C \in A / I: \paren {j + I} \circ C = 1 + I$
Let $j' \in A$ be such that $C = j' + I$.
Then:
By definition of subset:
:$A \subseteq J$
Thus $A = J$ follows by definition of set equality.
Category:Ideal Theory
Condition for Incommensurability of Roots of Quadratic Equation

== Theorem ==
Consider the quadratic equation:
:$(1): \quad a x - x^2 = \dfrac {b^2} 4$
Then $x$ and $a - x$ are incommensurable $\sqrt {a^2 - b^2}$ and $a$ are incommensurable.
== Proof ==
We have that:
Let:
:$a \smile b$ denote that $a$ is incommensurable with $b$
:$a \frown b$ denote that $a$ is commensurable with $b$.
=== Necessary Condition ===
Let $\paren {a - x} \smile x$.
From Incommensurability of Sum of Incommensurable Magnitudes:
:$a \smile x$
From Magnitudes with Rational Ratio are Commensurable:
:$x \frown 2 x$
So:
=== Sufficient Condition ===
Let $a \smile \sqrt {a^2 - b^2}$.
Then:
From Magnitudes with Rational Ratio are Commensurable:
:$x \frown 2 x$
From Commensurable Magnitudes are Incommensurable with Same Magnitude:
:$a \smile x$
From Incommensurability of Sum of Incommensurable Magnitudes:
:$\paren {a - x} \smile x$
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Condition for Increasing Binomial Coefficients

== Theorem ==
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\dbinom n k$ denote a binomial coefficient for $k \in \N$.
Then:
:$\dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $1 \ \text{(a)}$
Category:Binomial Coefficients
Category:Condition for Increasing Binomial Coefficients
Condition for Increasing Binomial Coefficients/Proof 2

== Theorem ==
== Proof ==
The proof proceeds by induction on $n$.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$
First we investigate the edge case.
Let $n = 1$.
Then we have:
Thus we see:
:there are no $k$ such that $0 \le k < \dfrac {1 - 1} 2 = 0$
and:
:there are no $k$ such that $\dbinom 1 k < \dbinom 1 {k + 1}$
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
Let $n = 2$.
Then we have:
Thus we see:
:there is one $k$ such that $0 \le k < \dfrac {2 - 1} 2 = \dfrac 1 2$, and that is $k = 0$
and:
:$\dbinom 2 k < \dbinom 2 {k + 1}$ holds for exactly $k = 0$.
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\dbinom r k < \dbinom r {k + 1} \iff 0 \le k < \dfrac {r - 1} 2$
from which it is to be shown that:
:$\dbinom {r + 1} k < \dbinom {r + 1} {k + 1} \iff 0 \le k < \dfrac r 2$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{> 0}: \dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$
Category:Condition for Increasing Binomial Coefficients
Condition for Independence from Product of Expectations

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\expect X$ denote the expectation of $X$.
Then $X$ and $Y$ are independent :
:$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$
for all functions $g, h: \R \to \R$ for which the latter two expectations exist.
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
Suppose that $X$ and $Y$ were not independent.
That is:
:$\map \Pr {X = a, Y = b} \ne \map \Pr {X = a} \map \Pr {Y = b}$
for some $a, b \in \R$.
Now, suppose that
:$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$
for all functions $g, h: \R \to \R$ for which the latter two expectations exist.
Let us define $g$ and $h$ as examples of such functions as follows:
:$(1): \quad \map g x := \begin{cases}
1 & : x = a \\
0 & : x \ne a
\end{cases}$
:$(2): \quad \map h y := \begin{cases}
1 & : y = b \\
0 & : y \ne b
\end{cases}$
where $a \in \R$ and $b \in \R$ are arbitrary real numbers.
Then:
So by hypothesis:
:$\map \Pr {X = a, Y = b} = \map \Pr {X = a} \map \Pr {Y = b}$
But also by hypothesis:
:$\map \Pr {X = a, Y = b} \ne \map \Pr {X = a} \map \Pr {Y = b}$
This contradicts our supposition that:
:$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$
for all functions $g, h: \R \to \R$ for which the latter two expectations exist.
So, if the above supposition holds, then $X$ and $Y$ have to be independent.
=== Necessary Condition ===
Suppose $X$ and $Y$ are independent.
Let $g, h: \R \to \R$ be any real functions such that $\expect {\map g x}$ and $\expect {\map h y}$ exist.
Then:
thus proving that
:$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$
whatever $g$ and $h$ are.
== Sources ==
* : $\S 3.3$: Independence of discrete random variables: Theorem $3 \text{D}$
Category:Expectation
Category:Condition for Independence from Product of Expectations
Condition for Independence from Product of Expectations/Corollary

== Corollary to Condition for Independence from Product of Expectations ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be independent discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Then:
:$\expect {X Y} = \expect X \expect Y$
assuming the latter expectations exist.
=== General Result ===
== Proof ==
From Condition for Independence from Product of Expectations, setting both $g$ and $h$ to the identity functions:
:$\forall x \in \R: \map g x = x$
:$\forall y \in \R: \map h y = y$
It follows directly that if $X$ and $Y$ are independent, then:
:$\expect {X Y} = \expect X \expect Y$
assuming the latter expectations exist.
== Note on Converse ==
Note that the converse of the corollary does not necessarily hold.
== Sources ==
* : $\S 3.3$: Independence of discrete random variables: Theorem $3 \text{C}$
Category:Condition for Independence from Product of Expectations
Condition for Independence from Product of Expectations/Corollary/Converse

== Converse of Corollary to Condition for Independence from Product of Expectations ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$ such that:
:$\expect {X Y} = \expect X \expect Y$
Then it is not necessarily the case that $X$ and $Y$ are independent.
== Proof ==
Proof by Counterexample:
Let $X$ be a discrete random variable whose distribution is defined as:
:$\map {p_X} {-1} = \map {p_X} 0 = \map {p_X} 1 = \dfrac 1 3$
Let $Y$ be the discrete random variable defined as:
:$Y = \begin{cases}
0 & : X = 0 \\
1 & : X \ne 0
\end{cases}$
We have:
So $X$ and $Y$ are dependent.
But:
So $\expect {X, Y} = \expect X \expect Y$ but $X$ and $Y$ are not independent.
== Sources ==
* : $\S 3.3$: Independence of discrete random variables: Example $13$
Category:Condition for Independence from Product of Expectations
Condition for Independence from Product of Expectations/Corollary/General Result

== Corollary to Condition for Independence from Product of Expectations ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X_1, X_2, \ldots, X_n$ be independent discrete random variables.
Then:
:$\ds \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$
assuming the latter expectations exist.
== Proof ==
Proof by induction:
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\ds \expect {\prod_{k \mathop = 1}^1 {X_k} } = \expect {X_1} = \prod_{k \mathop = 1}^1 \expect {X_k}$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \expect {\prod_{k \mathop = 1}^r {X_k} } = \prod_{k \mathop = 1}^r \expect {X_k}$
from which it is to be shown that:
:$\ds \expect {\prod_{k \mathop = 1}^{r + 1} {X_k} } = \prod_{k \mathop = 1}^{r + 1} \expect {X_k}$
=== Induction Step ===
This is the induction step:
We have:
So $\map P r \implies \map P {r + 1}$ and thus it follows by the Principle of Mathematical Induction that:
:$\ds \forall n \in \Z_{> 0}: \expect {\prod_{k \mathop = 1}^n {X_k} } = \prod_{k \mathop = 1}^n \expect {X_k}$
Category:Condition for Independence from Product of Expectations
Condition for Independence of Discrete Random Variables

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Then $X$ and $Y$ are independent there exist functions $f, g: \R \to \R$ such that the joint mass function of $X$ and $Y$ satisfies:
:$\forall x, y \in \R: \map {p_{X, Y} } {x, y} = \map f x \map g y$
== Proof ==
We have by definition of joint mass function that:
:$x \notin \Omega_X \implies \map {p_{X, Y} } {x, y} = 0$
:$y \notin \Omega_Y \implies \map {p_{X, Y} } {x, y} = 0$
Hence we only need to worry about values of $x$ and $y$ in their appropriate $\Omega$ spaces.
=== Necessary Condition ===
Suppose there exist functions $f, g: \R \to \R$ such that:
:$\forall x, y \in \R: \map {p_{X, Y} } {x, y} = \map f x \map g y$
Then by definition of marginal probability mass function:
:$\ds \map {p_X} x = \map f x \sum_y \map g y$
:$\ds \map {p_Y} y = \map g y \sum_x \map f x$
Hence:
So it follows that:
Hence the result from the definition of independent random variables.
=== Sufficient Condition ===
Suppose that $X$ and $Y$ are independent.
Then we can take the variables:
:$\map f x = \map {p_X} x$
:$\map g y = \map {p_Y} y$
and the result follows by definition of independence.
== Sources ==
* : $\S 3.3$: Independence of discrete random variables: Theorem $3 \ \text{B}$
Category:Independent Random Variables
Category:Discrete Random Variables
Condition for Infimum of Subset to equal Infimum of Set

== Lemma ==
Let $S$ be a real set.
Let $T$ be a subset of $S$.
Let $S$ and $T$ admit infima.
Then:
: $\inf S = \inf T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$
== Proof ==
=== Necessary Condition ===
Let $\inf S = \inf T$.
The aim is to establish that $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$.
We have:
=== Sufficient Condition ===
Let $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$.
The aim is to establish that $\inf S = \inf T$.
We have:
Category:Real Analysis
Condition for Injective Mapping on Ordinals

== Theorem ==
Let $F$ be a mapping satisfying the following properties:
:$(1): \quad$ The domain of $F$ is $\On$, the class of all ordinals
:$(2): \quad$ For all ordinals $x$, $\map F x = \map G {F \restriction x}$
:$(3): \quad$ For all ordinals $x$, $\map G {F \restriction x} \in \paren {A \setminus \Img x}$ where $\Img x$ is the image of $x$ under $F$.
Let $\Img F$ denote the image of $F$.
Then the following properties hold:
:$(1): \quad \Img F \subseteq A$
:$(2): \quad F$ is injective
:$(3): \quad A$ is a proper class.
Note that only the third property of $F$ is the most important.
For any function $G$, a function $F$ can be constructed satisfying the first two using the First Principle of Transfinite Recursion.
== Proof ==
Let $x$ be an ordinal.
Then $\map F x = \map G {F \restriction x}$ and $\map G {F \restriction x} \in A$ by hypothesis.
Therefore, $\map F x \in A$.
This satisfies the first statement.
Take two distinct ordinals $x$ and $y$.
, assume $x \in y$ (we are justified in this by Ordinal Membership is Trichotomy).
Then:
Thus for distinct ordinals $x$ and $y$, $\map F x \ne \map F y$.
Therefore, $F$ is injective.
$F$ is injective and $F: \On \to A$.
Therefore, if $A$ is a set, then $\On$ is a set.
But by the Burali-Forti Paradox, this is impossible, so $A$ is not a set.
Therefore, $A$ is a proper class.
== Also see ==
* Maximal Injective Mapping from Ordinals to a Set
* Transfinite Recursion Theorem
== Sources ==
* : $\S 7.46$
Category:Class of All Ordinals
Category:Class Mappings
Condition for Invertibility in Power Structure on Associative or Cancellable Operation

== Theorem ==
Let $\struct {S, \circ}$ be a unital magma.
Let $\struct {\powerset S, \circ_\PP}$ denote the power structure of $\struct {S, \circ}$.
Let identity element $e \in S$ be an identity element of $\struct {S, \circ}$.
Let $\circ$ be either:
:an associative operation
:a cancellable operation.
Let $X \subseteq S$ be a subset of $S$.
Then:
:$X$ is invertible for $\circ_PP$
:
:there exists an element $x \in S$ which is invertible for $\circ$ such that $X = \set x$.
== Proof ==
First we note that from Identity Element for Power Structure, the algebraic structure $\struct {\powerset S, \circ_\PP}$ has an identity element $J = \set e$.
=== Sufficient Condition ===
Let $X$ be invertible for $\circ_PP$.
Then:
Similarly:
That is, there exists $x \in X$ such that:
:$x \circ y = e = y \circ x$
That is:
:$x$ is invertible for $\circ$
:$y$ is the inverse of $x$.
Hence we have $Y \in \powerset S$ such that:
:$\forall y \in Y: x \circ y = e = y \circ x$
;$\circ$ is Associative
Let $\circ$ be an associative operation.
 $\exists z \in X$ such that $z \ne x$.
Then we have:
:$\forall y \in Y: z \circ y = e = y \circ z$
This contradicts our supposition that $z \ne x$.
Hence there can be no elements in $X$ apart from $x$.
That is:
:$X = \set x$
where $x$ is invertible for $\circ$.
;$\circ$ is Cancellable
Let $\circ$ be a cancellable operation.
 $\exists z \in X$ such that $z \ne x$.
Then we have:
:$\forall y \in Y: z \circ y = e = y \circ z$
Then:
This contradicts our supposition that $z \ne x$.
Hence there can be no elements in $X$ apart from $x$.
That is:
:$X = \set x$
where $x$ is invertible for $\circ$.
=== Necessary Condition ===
Let there exist an element $x \in S$ which is invertible for $\circ$.
Hence there exist $y \in S$ such that:
:$x \circ y = e = y \circ x$
Let $X = \set x$.
We have:
and:
That is, $X$ is invertible for $\circ_\PP$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.7 \ \text {(c)}$
Category:Power Structures
Category:Associativity
Category:Cancellability
Category:Inverse Elements
Category:Magmas