avewright/proofwiki-math huggingface dataset:
Theorem: Jensen's Inequality (Measure Theory)/Concave Functions

Let $\Lambda: \hointr 0 \infty \to \hointr 0 \infty$ be a concave function.
Then for all positive measurable functions $g: X \to \R$, $g \in \map {\MM^+} \Sigma$:
:$\dfrac {\int \paren {\Lambda \circ g} \cdot f \rd \mu} {\int f \rd \mu} \le \map \Lambda {\dfrac {\int g \cdot f \rd \mu} {\int f \rd \mu} }$
where $\circ$ denotes composition, and $\cdot$ denotes pointwise multiplication.

Proof:
{{proof wanted}}
Category:Jensen's Inequality
Category:Measure Theory
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Theorem: Correspondence Theorem (Set Theory)

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $\mathscr A$ be the set of partitions of $S$ associated with equivalence relations $\RR'$ on $S$ such that:
:$\tuple {x, y} \in \RR \iff \tuple {x, y} \in \RR'$
Then there exists a bijection $\phi$ from $\mathscr A$ onto the set of partitions of the quotient set $S / \RR$.

Proof:
Denote the equivalence class of an element $x$ of $S$ by $\eqclass x {\RR}$ with respect to the relation $\RR$. 
Consider the relation on $S$:
:$\phi = \set {\tuple {\eqclass x {\RR'}, \eqclass x \RR}: x \in S}$
We prove that $\phi$ is a bijection. 
Let $\eqclass x \RR = \eqclass y \RR$.
Then:
:$\tuple {x, y} \in \RR$
Thus:
:$\tuple {x, y} \in \RR'$
Therefore:
:$\eqclass x {\RR'} = \eqclass y {\RR'}$
and so $\phi$ is one-to-many.
Let $\eqclass x {\RR'} = \eqclass y {\RR'}$.
Then:
:$\tuple {x, y} \in \RR'$
Thus:
:$\tuple {x, y} \in \RR$
Therefore $\eqclass x \RR = \eqclass y \RR$
Hence $\phi$ is many-to-one.
As $\phi$ is both one-to-many and many-to-one, it is by definition a one-to-one relation.
From the fact that $\Rng \phi = S / \RR$ we have that $\phi$ is both left-total and right-total.
We have demonstrated that $\phi$ is left-total and many-to-one, and so by definition is a mapping.
We have demonstrated that $\phi$ is right-total, and so by definition is a surjection.
We have demonstrated that $\phi$ is one-to-one, and so by definition is an injection.
Hence by definition $\phi$ is a bijection.
{{qed}}
Theorem: Power Set is Lattice

Let $S$ be a set.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the subset relation $\subseteq$.
Then $\struct {\powerset S, \subseteq}$ is a lattice.

Proof:
From Subset Relation on Power Set is Partial Ordering, we have that $\subseteq$ is a partial ordering.
Let $X, Y \in \powerset S$.
Then from Union is Smallest Superset:
:$X \subseteq T, Y \subseteq T \iff X \cup Y \subseteq T$
and from Intersection is Largest Subset:
:$X \subseteq T, Y \subseteq T \iff T \subseteq X \cap Y$
So $X \cap Y$ is the infimum and $X \cup Y$ is the supremum of $\set {X, Y}$.
Hence by definition $\powerset S$ is a lattice.
{{qed}}
Theorem: Divisor Relation induces Lattice

Let $\struct {\Z_{> 0}, \divides}$ be the ordered set comprising:
:The set of positive integers $\Z_{> 0}$
:The divisor relation $\divides$ defined as:
::$a \divides b := \exists k \in \Z_{> 0}: b = ka$
Then $\struct {\Z_{> 0}, \divides}$ is a lattice.

Proof:
It follows from Divisor Relation on Positive Integers is Partial Ordering that $\struct {\Z_{> 0}, \divides}$ is indeed an ordered set.
Let $a, b \in \Z_{>0}$.
Let $d = \gcd \set {a, b}$ be the greatest common divisor of $a$ and $b$.
By definition, $d$ is the infimum of $\set {a, b}$.
Similarly, let $m = \lcm \set {a, b}$ be the lowest common multiple of $a$ and $b$.
By definition, $m$ is the supremum of $\set {a, b}$.
Hence, as $\set {a, b}$ has both an infimum and a supremum in $\Z_{>0}$, it follows that $\struct {\Z_{> 0}, \divides}$ is a lattice.
{{qed}}
Theorem: Generator for Product Sigma-Algebra

Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be measurable spaces.
Let $\GG_1$ and $\GG_2$ be generators for $\Sigma_1$ and $\Sigma_2$, respectively.
Then $\GG_1 \times \GG_2$ is a generator for the product $\sigma$-algebra $\Sigma_1 \otimes \Sigma_2$.

Proof:
{{MissingLinks|throughout}}
{{handwaving|Much use of "clearly" in the below.}}
We begin by stating that this result is in fact incorrect.
For a simple counter-example, consider $X = Y = \set {1, 2}$ both equipped with the power set sigma algebra, that is, $\Sigma_1 = \Sigma_2 = \set {\O, \set 1, \set 2, \set {1, 2} }$.
Then clearly, we can take $\GG_1 = \GG_2 = \set 1$.
{{explain|How is $\set 1$ a generator for $\set {1, 2}$?}}
However, note that $\GG_1 \times \GG_2 = \set {\tuple {1, 1} }$ and hence the sigma algebra generated by this set will have exactly four elements.
Clearly, $\Sigma_1 \times \Sigma_2$ contains more than four elements and hence the product sigma algebra, i.e., the sigma algebra generated by the product $\Sigma_1\times\Sigma_2$ will contain more than four elements.
We also state and prove a minor modification of the above stated result in this light.
Theorem
Let $X, Y, \Sigma_1, \Sigma_2, \GG_1, \GG_2$ be as before and let $X \in \GG_1$ and $Y \in \GG_2$.
Then $\Sigma_1 \otimes \Sigma_2$ is generated by $\GG_1 \times \GG_2$.
Proof:
We begin by defining $\Sigma$ as the sigma algebra generated by $G_1 \times G_2$.
It is clear that $\Sigma \subset \Sigma_1 \otimes \Sigma_2$.
For the reverse inclusion, set:
:$M_1 := \set {A \subset X: A \times Y \in \Sigma}$
Clearly, $\GG_1 \subset M_1$.
Therefore, $M_1$ is non-empty.
We claim that $M_1$ is a sigma algebra.
Note that $X \in M_1$ obviously.
Also, $A \in M_1$ suffices $\paren {X \setminus A} \times Y = \paren {X \times Y} \setminus \paren {A \times Y} \in \Sigma$, proving closure of $M_1$ under complements.
Also, if $A_n \in M_1$ for $n \in \N$, then:
:$\ds \paren {\bigcup_{n \mathop = 1}^\infty A_n} \times Y = \bigcup_{n \mathop = 1}^\infty \paren {A_n \times Y} \in M_1$
proving closure under countable unions and establishing our claim.
Now since $M_1$ is a sigma algebra containing $\GG_1$, $M_1 \supset \Sigma_1$.
Similarly, setting $M_2 := \set {B \subset Y: X \times B \in \Sigma}$ gives us that $M_2 \supset \Sigma_2$.
Now, if $C \in \Sigma_1$ and $D \in \Sigma_2$, then clearly $C \in M_1$ and $D \in M_2$ giving $C \times Y, X \times D \in \Sigma$
Hence:
:$C \times D = \paren {C \times Y} \cap \paren {X \times D} \in \Sigma$
proving:
:$\Sigma_1 \times \Sigma_2 \subset \Sigma$
and hence giving:
:$\Sigma_1 \otimes \Sigma_2 \subset \Sigma$
establishing the result.
{{qed}}
Theorem: Tonelli's Theorem

Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be $\sigma$-finite measure spaces.
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$.
Let $f: X \times Y \to \overline \R_{\ge 0}$ be a positive $\Sigma_X \otimes \Sigma_Y$-measurable function.
Then:
:$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \int_X \map f {x, y} \map {\d \mu} x \map {\d \nu} y = \int_X \int_Y \map f {x, y} \map {\d \nu} y \map {\d \mu} x$

Proof:
We rewrite the demand as: 
:$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
where $f_x$ is the $x$-vertical section of $f$, and $f^y$ is the $y$-horizontal section of $f$.
From Horizontal Section of Measurable Function is Measurable, we have: 
:$f^y$ is $\Sigma_X$-measurable.
From Vertical Section of Measurable Function is Measurable, we have: 
:$f_x$ is $\Sigma_Y$-measurable.
From Integral of Horizontal Section of Measurable Function gives Measurable Function, we then have: 
:$\ds y \mapsto \int_X f^y \rd \mu$ is $\Sigma_Y$-measurable
and from Integral of Vertical Section of Measurable Function gives Measurable Function, we have: 
:$\ds x \mapsto \int_Y f_x \rd \nu$ is $\Sigma_X$-measurable.
So, both:
:$\ds \int_Y \paren {\int_X f^y \rd \mu} \rd \nu$
and:
:$\ds \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
are well-defined.
We first prove the case of: 
:$f = \chi_E$
where $E$ is a $\Sigma_X \otimes \Sigma_Y$-measurable set.
Theorem: Condition for Power Set to be Totally Ordered

Let $\powerset S$ be the power set of a set $S$.
Let $\struct {\powerset S, \subseteq}$ be the set $\powerset S$ ordered by $\subseteq$.
Then $\struct {\powerset S, \subseteq}$ is totally ordered {{iff}} $S$ is either the empty set or a singleton.

Proof:
From Subset Relation on Power Set is Partial Ordering we have that $\struct {\powerset S, \subseteq}$ is an ordered set.
We now need to show that $\struct {\powerset S, \subseteq}$ is a totally ordered set exactly when $S = \O$ or $S$ has exactly one element.
When $S = \O$ then $\powerset S = \set \O$ and it follows trivially that $\struct {\powerset S, \subseteq}$ is totally ordered.
{{qed|lemma}}
Let $S = \set x$.
Then $\powerset S = \set {\O, \set x}$.
From Empty Set is Subset of All Sets we have $\O \subseteq \set x$.
Again it follows by definition that $\struct {\powerset S, \subseteq}$ is totally ordered.
{{qed|lemma}}
Now suppose $S$ is neither the empty set nor a singleton.
Then:
:$\exists x, y \in S: x \ne y$
and so:
:$\exists \set x, \set y \in \powerset S: \set x \ne \set y$
But further, $\set x \nsubseteq \set y$ and $\set y \nsubseteq \set x$.
That is, $\set x$ and $\set y$ are non-comparable by $\subseteq$.
Thus, by definition, $\struct {\powerset S, \subseteq}$ is not totally ordered.
Hence the result.
{{qed}}
Theorem: Product Sigma-Algebra Generated by Projections

Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be measurable spaces.
Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X \times Y$.
Let $\pr_1: X \times Y \to X$ and $\pr_2: X \times Y \to Y$ be the first and second projections, respectively.
Then:
:$\Sigma_1 \otimes \Sigma_2 = \map \sigma {\pr_1, \pr_2}$
where $\sigma$ denotes generated $\sigma$-algebra.

Proof:
{{Proofread}}
{{tidy}}
Let $R = \set {A \times B: A \in \Sigma_1, B \in \Sigma_2}$ be a generator for the product $\sigma$-algebra $\Sigma_1 \otimes \Sigma_2$.
Note that by Preimage of Element under Projection:
:$\inv {\pr_1} A = A \times Y \in R$ 
and:
:$\inv {\pr_2} B = X \times B \in R$ 
for any $A \in \Sigma_1$ and $B \in \Sigma_2$.
Let:
:$G = \set {\inv {\pr_i} {C_i}: C_i \in \Sigma_i, i = 1, 2}$
Thus:
:$G \subseteq R$
However, for any $A \times B \in R$:
:$A \times B = \inv {\pr_1} A \cap \inv {\pr_2} B \in \map \sigma G$
So we have:
:$R \subseteq \map \sigma G$
But:
:$G \subseteq R$ 
and:
:$R \subseteq \map \sigma G$
we use Generated Sigma-Algebra Preserves Subset and $\paren {\map \sigma G = \map \sigma {\map \sigma G} }$.
Thus:
:$\map \sigma R = \map \sigma G$
It follows from the definition:
:$\Sigma_1 \otimes \Sigma_2 = \map \sigma R = \map \sigma G = \sigma (\pr_1, \pr_2)$
{{qed}}
Theorem: Cardinal Zero is Less than Cardinal One

The zero cardinal $0$ is less than one:
:$0 < 1$

Proof:
We have that the Cardinals are Totally Ordered.
Let $\RR \subseteq \O \times \set \O$ be any arbitrary relation between $\O$ and $\set \O$.
We have that $\RR$ is vacuously many-to-one.
Also vacuously, $\RR$ is left-total.
Thus by definition, $\RR$ is in fact a mapping.
From Empty Mapping is Unique, this relation $\RR$ is the unique mapping from $\O$ to $\set \O$.
Also vacuously, $\RR$ is an injection.
So, by definition, $\set \O$ dominates $\O$.
Since Empty Set is Subset of All Sets, we have $\O \subseteq \set \O$.
Hence the result.
{{qed}}
Theorem: Right Quasigroup if (1-3) Parastrophe of Magma is Magma

Let $\struct {S, \circ}$ be a magma.
Let the $(1-3)$ parastrophe of $\struct {S, \circ}$ be a magma.
Then $\struct {S, \circ}$ is a right quasigroup.

Proof:
By the definition of a right quasigroup it must be shown that:
:$\forall a, b \in S: \exists ! x \in S: x \circ a = b$
{{AimForCont}} there exists $a, b \in S$ such that $x \circ a = b$ does not have a unique solution for $x$.
Then in the $(1-3)$ parastrophe of $\struct {S, \circ}$ we see that $\circ$ as a mapping either fails to be left-total or many-to-one for $b \circ a$.
So $\struct {S, \circ}$ is not a magma which contradicts our assumption.
{{qed}}
Category:Parastrophes
Category:Quasigroups
Category:Magmas
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Theorem: Idempotent Non-Trivial Quasigroup is Not a Loop

Let $\struct {S, \circ}$ be an idempotent quasigroup whose underlying set $S$ comprises more than one element.
Then $\struct {S, \circ}$ is not an algebra loop, that is, it has no identity element.

Proof:
{{AimForCont}} $\struct {S, \circ}$ has an identity element $e$.
Then by Identity Element is Idempotent:
:$e \circ e = e$
Consider $e' \in S$ where $e' \ne e$.
Since $e$ is an identity element:
:$e' \circ e = e'$
Also, by assumption, $\circ$ is idempotent, so:
:$e' \circ e' = e'$
Then by the definition of a quasigroup, the left regular representation:
:$\lambda_{e'}: S \to S, x \mapsto e' \circ x$
is a permutation, and in particular, an injection.
Hence $e' = e$, which contradicts our assumption that $e' \ne e$.
So $\struct {S, \circ}$ has no identity element and is not an algebra loop.
{{qed}}
Category:Quasigroups
Category:Algebra Loops
Category:Idempotence
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Theorem: B-Algebra is Right Cancellable

Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\circ$ is right-cancellable for $X$. That is:
:$\forall x, y, z \in X: x \circ z = y \circ z \implies x = y$

Proof:
Let $x, y \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ y} \circ \paren {0 \circ y}
      | r = x \circ \paren {\paren {0 \circ y} \circ \paren {0 \circ y} }
      | c = {{B-algebra-axiom|3}}
}}
{{eqn | r = x \circ 0
      | c = {{B-algebra-axiom|1}}
}}
{{eqn | r = x
      | c = {{B-algebra-axiom|2}}
}}
{{end-eqn}}
Now if for some $z \in X$ we have $x \circ z = y \circ z$, then also:
:$\paren {x \circ z} \circ \paren {0 \circ z} = \paren {y \circ z} \circ \paren {0 \circ z}$ 
which by the above implies $x = y$.
Hence the result.
{{qed}}
Category:B-Algebras
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Theorem: Right Regular Representation of 0 is Bijection in B-Algebra

Let $\struct {X, \circ}$ be a $B$-algebra.
Then the right regular representation of $\struct {X, \circ}$ {{WRT}} $0$ is a bijection.

Proof:
{{B-algebra-axiom|2}} states:
:$\forall x \in X: x \circ 0 = x$
and so, for all $x \in X$:
:$\map {\rho_0} x = x$
That is:
:$\rho_0 = I_X$
which is the identity mapping on $X$.
The result follows from Identity Mapping is Bijection.
{{qed}}
Category:B-Algebras
Category:Regular Representations
7qdxvqnkmbjcb8n4ss82iikpiufw3oo
Theorem: Product of Cardinals is Commutative

Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
: $\mathbf a \mathbf b = \mathbf b \mathbf a$
where $\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$.

Proof:
Let $\mathbf a = \card A$ and $\mathbf b = \card B$ for some sets $A$ and $B$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a \mathbf b
      | r = \card {A \times B}
      | c = {{Defof|Product of Cardinals}}
}}
{{eqn | r = \card {B \times A}
      | c = Cardinality of Cartesian Product of Finite Sets/Corollary
}}
{{eqn | r = \mathbf b \mathbf a
      | c = {{Defof|Product of Cardinals}}
}}
{{end-eqn}}
{{qed}}
Theorem: Product of Cardinals is Associative

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
: $\mathbf a \paren {\mathbf {b c} } = \paren {\mathbf {a b} } \mathbf c$
where $\mathbf {a b}$ denotes the product of $\mathbf a$ and $\mathbf b$.

Proof:
Let $\mathbf a = \card A$, $\mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.
By definition of product of cardinals:
:$\mathbf a \paren {\mathbf {b c} }$ is the cardinal associated with $A \times \paren {B \times C}$.
Consider the mapping $f: A \times \paren {B \times C} \to \paren {A \times B} \times C$ defined as:
:$\forall a \in A, b \in B, c \in C: \map f {a, \tuple {b, c} } = \tuple {\tuple {a, b}, c}$
Let $a_1, a_2 \in A, b_1, b_2 \in B, c_1, c_2 \in C$ such that:
:$\map f {a_1, \tuple {b_1, c_1} } = \map f {a_2, \tuple {b_2, c_2} }$
Then:
{{begin-eqn}}
{{eqn | l = \map f {a_1, \tuple {b_1, c_1} }
      | r = \map f {a_2, \tuple {b_2, c_2} }
}}
{{eqn | ll= \leadsto
      | l = \tuple {\tuple {a_1, b_1}, c_1}
      | r = \tuple {\tuple {a_2, b_2}, c_2}
      | c = Definition of $f$
}}
{{eqn | ll= \leadsto
      | l = \tuple {a_1, b_1}
      | r = \tuple {a_2, b_2}
      | c = Equality of Ordered Tuples
}}
{{eqn | lo= \land
      | l = c_1
      | r = c_2
}}
{{eqn | ll= \leadsto
      | l = a_1
      | r = a_2
      | c = Equality of Ordered Tuples
}}
{{eqn | lo= \land
      | l = b_1
      | r = b_2
}}
{{eqn | lo= \land
      | l = c_1
      | r = c_2
}}
{{eqn | ll= \leadsto
      | l = a_1
      | r = a_2
      | c = Equality of Ordered Tuples
}}
{{eqn | lo= \land
      | l = \tuple {b_1, c_1}
      | r = \tuple {b_2, c_2}
}}
{{eqn | ll= \leadsto
      | l = \tuple {a_1, \tuple {b_1, c_1} }
      | r = \tuple {a_2, \tuple {b_2, c_2} }
      | c = Equality of Ordered Tuples
}}
{{end-eqn}}
Thus $f$ is an injection.
{{begin-eqn}}
{{eqn | q = \forall x \in \paren {A \times B} \times C: \exists a \in A, b \in B, c \in C
      | l = x
      | r = \tuple {\tuple {a, b}, c}
}}
{{eqn | r = \map f {a, \tuple {b, c} }
}}
{{end-eqn}}
Thus $f$ is a surjection.
So $f$ is both an injection and a surjection, and so by definition a bijection.
Thus a bijection has been established between $A \times \tuple {B \times C}$ and $\tuple {A \times B} \times C$.
It follows by definition that $A \times \tuple {B \times C}$ and $\tuple {A \times B} \times C$ are equivalent.
The result follows by definition of cardinal.
{{qed}}
Theorem: B-Algebra Identity: xy=x(0(0y))

Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x, y \in X: x \circ y = x \circ \paren {0 \circ \paren {0 \circ y} }$

Proof:
Let $x, y \in X$.
Then:
{{begin-eqn}}
{{eqn | l = x \circ y
      | r = \paren {x \circ y} \circ 0
      | c = {{B-algebra-axiom|2}}
}}
{{eqn | r = x \circ \paren {0 \circ \paren {0 \circ y} }
      | c = {{B-algebra-axiom|3}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:B-Algebras
pximfi3644a38aax08lelqdnv4cyajl
Theorem: B-Algebra Identity: x (y z) = (x (0 z)) y

Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x, y, z \in X: x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$

Proof:
Let $x, y, z \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ \paren {0 \circ z} } \circ y
      | r = x \circ \paren {y \circ \paren {0 \circ \paren {0 \circ z} } }
      | c = {{B-algebra-axiom|2}}
}}
{{eqn | r = x \circ \paren {y \circ z}
      | c = Identity: $y \circ \paren {0 \circ \paren {0 \circ z} } = y \circ z$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: B-Algebra Identity: xy = 0 iff x = y

Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x, y \in X: x \circ y = 0 \iff x = y$

Proof:
=== Sufficient Condition ===
Suppose that $x = y$.
Then by {{B-algebra-axiom|1}}:
:$x \circ y = x \circ x = 0$
{{qed|lemma}}
Theorem: First Power of Element in B-Algebra

Let $\left({X, \circ}\right)$ be a $B$-algebra.
Then:
:$\forall x \in X: x^1 = x$
where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.

Proof:
{{begin-eqn}}
{{eqn | l = x^1
      | r = x^0 \circ \left({0 \circ x}\right)
      | c = {{Defof|B-Algebra Power of Element}}
}}
{{eqn | r = 0 \circ \left({0 \circ x}\right)
      | c = {{Defof|B-Algebra Power of Element}}
}}
{{eqn | r = 0 \circ x
      | c = $0$ in $B$-Algebra is Left Cancellable Element
}}
{{eqn | r = x
      | c = $0$ in $B$-Algebra is Left Cancellable Element
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:B-Algebras
nhb6c3mrjvjpzkymf4c88s6jes275y5
Theorem: B-Algebra Power Law

Let $\struct {X, \circ}$ be a $B$-algebra.
Let $n, m \in \N$ such that $n \ge m$.
Then:
:$\forall x \in X: x^n \circ x^m = x^{n - m}$
where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.

Proof:
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall m \in \N_{> 0}, m \le n: \forall x \in X: x^n \circ x^m = x^{n - m}$
Theorem: B-Algebra is Left Cancellable

Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\circ$ is a left cancellable operation.

Proof:
Let $x, y, z \in X$:
{{begin-eqn}}
{{eqn | l = x \circ y
      | r = x \circ z
}}
{{eqn | ll= \leadstoandfrom
      | l = 0 \circ \paren {x \circ y}
      | r = 0 \circ \paren {x \circ z}
      | c = $0$ in $B$-Algebra is Left Cancellable Element
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {0 \circ \paren {0 \circ y} } \circ x
      | r = \paren {0 \circ \paren {0 \circ z} } \circ x
      | c = Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {0 \circ y} \circ x
      | r = \paren {0 \circ z} \circ x
      | c = $0$ in $B$-Algebra is Left Cancellable Element
}}
{{eqn | ll= \leadstoandfrom
      | l = y \circ x
      | r = z \circ x
      | c = $0$ in $B$-Algebra is Left Cancellable Element
}}
{{eqn | ll= \leadstoandfrom
      | l = y
      | r = z
      | c = $B$-Algebra is Right Cancellable
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:B-Algebras
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Theorem: B-Algebra is Quasigroup

Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\struct {X, \circ}$ is a quasigroup.

Proof:
By the definition of a quasigroup it must be shown that $\forall x \in X$ the left and right regular representations $\lambda_x$ and $\rho_x$ are permutations on $X$.
As $\struct {X, \circ}$ is a magma:
:$\forall x \in X$ the codomain of $\lambda_x$ and $\rho_x$ is $X$.
Hence it is sufficient to prove that $\lambda_x$ and $\rho_x$ are bijections.
We have that:
:$B$-Algebras are left cancellable
:$B$-Algebras are right cancellable
:Cancellable elements have injective regular representations.
Therefore:
:$\forall x \in X$: $\lambda_x$ and $\rho_x$ are injective mappings.
We also have that regular representations of $B$-algebras are surjective.
Therefore: 
:$\forall x \in X$: $\lambda_x$ and $\rho_x$ are both injective and surjective mappings.
Hence by definition are $\lambda_x$ and $\rho_x$ bijections for all $x \in X$.
{{qed}}
Theorem: Quasigroup is not necessarily B-Algebra

Let $\struct {S, \circ}$ be a quasigroup.
Then $\struct {S, \circ}$ is not necessarily a $B$-algebra.

Proof:
;Proof by Counterexample
From Group is Quasigroup we take an arbitrary small group.
Consider the Cayley table of the group of order $3$:
:$\begin{array}{c|cccccc}
  & 0 & 1 & 2 \\
\hline
0 & 0 & 1 & 2 \\
1 & 1 & 2 & 0 \\
2 & 2 & 0 & 1 \\
\end{array}$
By inspection we see that {{B-algebra-axiom|2}} does not hold as $1 \circ 1 \ne 0$.
{{qed}}