hcju/proofwikiqa huggingface dataset:
{{MissingLinks}} Let $\d \map \nu x := \dfrac {\map f x} {\int \map f s \rd \map \mu s} \rd \map \mu x$ be a probability measure. {{explain|This proof invokes a probability measure. Needs to be for a measure space. Does the proof work for both?}} Let $x_0 := \displaystyle \int \map g s \rd \map \nu s$. Then by convexity there exists constants $a, b$ such that $\forall x \in \R_{\ge 0}$: :$\map V {x_0} = a x_0 + b$ :$\map V x \ge a x + b$ In other words, there is a tangent line at $\tuple {x_0, V_0}$ that falls below the graph of $V$. Therefore: {{begin-eqn}} {{eqn | l = \map V {\map g s}       | o = \ge       | r = a \map g s + b       | c =  }} {{eqn | ll= \leadsto       | l = \int \map V {\map g s} \rd \map \nu s       | o = \ge       | r = a \int \map g s \rd \map \nu s + b       | c = Integration {{WRT|Integration}} $\map \nu s$ }} {{eqn | r = \map V {x_0}       | c =  }} {{eqn | r = \map V {\int \map g s \rd \map \nu s}       | c =  }} {{end-eqn}} {{explain|why this does what it purports to}} Category:Measure Theory sbehg3o0qvufoy40zc8164qkmv325t3
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}}
From Subset Relation on Power Set is Partial Ordering, we have that $\subseteq$ is a partial ordering. Let $X, Y \in \mathcal P \left({S}\right)$. 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 $\left\{{X, Y}\right\}$. Hence by definition $\mathcal P \left({S}\right)$ is a lattice. {{qed}}
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}}
From Subset Relation on Power Set is Partial Ordering we have that $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is an ordered set. We now need to show that $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is a totally ordered set exactly when $S = \varnothing$ or $S$ has exactly one element. When $S = \varnothing$ then $\mathcal P \left({S}\right) = \left\{{\varnothing}\right\}$ and it follows trivially that $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is totally ordered. {{qed|lemma}} Let $S = \left\{{x}\right\}$. Then $\mathcal P \left({S}\right) = \left\{{\varnothing, \left\{{x}\right\}}\right\}$. From Empty Set is Subset of All Sets we have $\varnothing \subseteq \left\{{x}\right\}$. Again it follows by definition that $\left({\mathcal P \left({S}\right), \subseteq}\right)$ 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: :$\exists \left\{{x}\right\}, \left\{{y}\right\} \in \mathcal P \left({S}\right): \left\{{x}\right\} \ne \left\{{y}\right\}$ But further, $\left\{{x}\right\} \nsubseteq \left\{{y}\right\}$ and $\left\{{y}\right\} \nsubseteq \left\{{x}\right\}$. That is, $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are non-comparable by $\subseteq$. Thus, by definition, $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is not totally ordered. Hence the result. {{qed}}
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}}
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 oe120uxkbkidt7wc8f6rgnstq7zr8fj
By the definition of a left quasigroup it must be shown that: :$\forall a, b \in S: \exists ! x \in S: a \circ x = b$ {{AimForCont}} there exists $a, b \in S$ such that $a \circ x = b$ does not have a unique solution for $x$. Then in the $\paren {2 - 3}$ parastrophe of $\struct {S, \circ}$ we see that $\circ$ as a mapping either fails to be left-total or many-to-one for $a \circ b$. So $\struct {S, \circ}$ is not a magma. This contradicts our assumption. {{qed}} Category:Magmas Category:Quasigroups Category:Parastrophes feg90u6n3458v4t13yipjjk86ivjr7f
{{AimForCont}} $\left({S, \circ}\right)$ 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 $\left({S, \circ}\right)$ has no identity element and is not an algebra loop. {{qed}} Category:Quasigroups Category:Algebra Loops Category:Idempotence 5weuldhp0cckv748h7h804ubcb8unu9
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 = Axiom $(\text A 3)$ for $B$-algebras }} {{eqn | r = x \circ 0       | c = Axiom $(\text A 1)$ for $B$-algebras }} {{eqn | r = x       | c = Axiom $(\text A 2)$ for $B$-algebras }} {{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 cjiciyl94ahtk4egukvgk0nq7rn6brp
Axiom $(A2)$ for $B$-algebras states: :$\forall x \in X: x \circ 0 = x$ and so, for all $x \in X$: :$\rho_0 \left({x}\right) = x$ That is, $\rho_0 = I_X$, the identity mapping on $X$. The result follows from Identity Mapping is Bijection. {{qed}} Category:B-Algebras Category:Regular Representations 3m4o37pdjcnswwnk01d0lgd84i5xwfy
Let $\mathbf a = \operatorname{Card} \left({A}\right)$ and $\mathbf b = \operatorname{Card} \left({B}\right)$ for some sets $A$ and $B$. Then: {{begin-eqn}} {{eqn | l=\mathbf a \mathbf b       | r=\operatorname{Card} \left({A \times B}\right)       | c=Definition of product of cardinals }} {{eqn | r=\operatorname{Card} \left({B \times A}\right)       | c=Cardinality of Cartesian Product/Corollary }} {{eqn | r=\mathbf b \mathbf a       | c=Definition of product of cardinals }} {{end-eqn}} {{qed}}
Let $\mathbf a = \operatorname{Card} \left({A}\right)$, $\mathbf b = \operatorname{Card} \left({B}\right)$ and $\mathbf c = \operatorname{Card} \left({C}\right)$ for some sets $A$, $B$ and $C$. We have by definition of product of cardinals that $\mathbf a \left({\mathbf b \mathbf c}\right)$ is the cardinal associated with $A \times \left({B \times C}\right)$. Consider the mapping $f: A \times \left({B \times C}\right) \to \left({A \times B}\right) \times C$ defined as: :$\forall \left({a, \left({b, c}\right)}\right) \in A \times \left({B \times C}\right): f \left({a, \left({b, c}\right)}\right) = \left({\left({a, b}\right), c}\right)$ Let $\left({a_1, \left({b_1, c_1}\right)}\right), \left({a_2, \left({b_2, c_2}\right)}\right) \in A \times \left({B \times C}\right)$ such that: :$f \left({a_1, \left({b_1, c_1}\right)}\right) = f \left({a_2, \left({b_2, c_2}\right)}\right)$ That is: :$\left({\left({a_1, b_1}\right), c_1}\right) = \left({\left({a_2, b_2}\right), c_2}\right)$ It follows from Equality of Ordered Tuples that $a_1 = a_2, b_1 = b_2$ and $c_1 = c_2$. That is: :$f \left({a_1, \left({b_1, c_1}\right)}\right) = f \left({a_2, \left({b_2, c_2}\right)}\right) \implies \left({a_1, \left({b_1, c_1}\right)}\right) = \left({a_2, \left({b_2, c_2}\right)}\right)$ and so $f$ is injective. Now let $\left({\left({a, b}\right), c}\right) \in \left({A \times B}\right) \times C$. By definition: :$\left({\left({a, b}\right), c}\right) = f \left({a, \left({b, c}\right)}\right)$ and so $f$ 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 $A \times \left({B \times C}\right)$ and $\left({A \times B}\right) \times C$. It follows by definition that $A \times \left({B \times C}\right)$ and $\left({A \times B}\right) \times C$ are equivalent. The result follows by definition of cardinal. {{qed}}
Let $x, y \in X$. Then: {{begin-eqn}} {{eqn|l = x \circ y      |r = \left({x \circ y}\right) \circ 0      |c = Axiom $(A2)$ for $B$-algebras }} {{eqn|r = x \circ\left({0 \circ\left({ 0 \circ y}\right)}\right)      |c = Axiom $(A3)$ for $B$-algebras }} {{end-eqn}} Hence the result. {{qed}} Category:B-Algebras bb8lemdoykob1m09fx6ldvqmx6w27dm
Let $x, y, z \in X$. Then: {{begin-eqn}} {{eqn | l = \left({x \circ \left({0 \circ z}\right)}\right) \circ y       | r = x \circ \left({y \circ \left({0 \circ \left({0 \circ z}\right)}\right)}\right)       | c = Axiom $(A2)$ for $B$-algebras }} {{eqn | r = x \circ \left({y \circ z}\right)       | c = Identity: $y \circ \left({0 \circ \left({0 \circ z}\right)}\right) = y \circ z$ }} {{end-eqn}} Hence the result. {{qed}}
=== Sufficient Condition === Suppose that $x = y$. Then by Axiom $(A1)$ for $B$-algebras: :$x \circ y = x \circ x = 0$ {{qed|lemma}} === Necessary Condition === Let $x, y \in X$ such that $x \circ y = 0$. Then: {{begin-eqn}} {{eqn | l = x \circ y       | r = y \circ y       | c = Axiom $(A1)$ for $B$-algebras }} {{eqn | ll= \leadsto       | l = x       | r = y       | c = $B$-algebra operation is right cancellable }} {{end-eqn}} Hence the result. {{qed}} Category:B-Algebras 3sij1geqwyvxjnuwn9w1x1czf8u0gr5
{{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
Proof by induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\forall m \in \N_{> 0}, m \le n: \forall x \in X: x^n \circ x^m = x^{n-m}$ === Basis for the Induction === $P \left({1}\right)$ is true, as this just says: : $x \circ x = 0$ which follows from the definition of the zeroth power in $B$-algebra. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\forall m \in \N_{> 0}, m \le k: \forall x \in X: x^k \circ x^m = x^{k-m}$ Then we need to show: :$\forall m \in \N_{> 0}, m \le k+1: \forall x \in X: x^{k+1} \circ x^m = x^{k+1-m}$ === Induction Step === This is our induction step: First we show that: :$\forall x \in X: x^{k+1} \circ x = x^k$ Thus: {{begin-eqn}} {{eqn|l = x^{k+1} \circ x      |r = \left({x^k \circ \left({0 \circ x}\right)}\right) \circ x      |c = Definition of power of an element in $B$-algebra }} {{eqn|r = x^k \circ \left({x \circ \left({0 \circ \left({0 \circ x}\right)}\right)}\right)      |c = Axiom $(A3)$ for B-algebras }} {{eqn|r = x^k \circ \left({\left({x \circ x}\right) \circ 0}\right)      |c = Axiom $(A3)$ for B-algebras }} {{eqn|r = x^k \circ \left({x \circ x}\right)      |c = Axiom $(A2)$ for B-algebras }} {{eqn|r = x^k \circ 0      |c = Axiom $(A1)$ for B-algebras }} {{eqn|r = x^k      |c = Axiom $(A2)$ for B-algebras }} {{end-eqn}} By induction, it follows that: :$\forall n \in \N_{>0}: \forall x \in X: x^n \circ x = x^{n-1}$ Now let $1 \le m \le k$. We have: {{begin-eqn}} {{eqn|l = x^{k+1} \circ x^{m+1}      |r = x^{k+1} \circ \left({x^m \circ \left({0 \circ x}\right)}\right)      |c = Definition of power of an element in $B$-algebra }} {{eqn|r = \left({x^{k+1} \circ x}\right) \circ x^m      |c = Axiom $(A3)$ for B-algebras }} {{eqn|r = x^k \circ x^m      |c = by the previous lemma }} {{eqn|r = x^{k-m} \circ x^0      |c = By induction }} {{eqn|r = x^{k-m} \circ 0      |c = Definition:B-Algebra zeroth power of an element }} {{eqn|r = x^{k-m}      |c = Axiom $(A2)$ for B-algebras }} {{end-eqn}} By induction, it follows that: :$\forall x \in X: x^n \circ x^m = x^{n-m}$ for all $n, m \in N$ such that $n \ge m$. {{qed}} {{wtd|Add a proof which uses the correspondence between B-Algebra and Group, and exploit the index laws for groups.}} Category:B-Algebras goukl8l1qpmbioh24zxqx4hqe3eihse
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 naib8zyo64h8dlh1o9uldrrddvl5e71
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 $\left({X, \circ}\right)$ 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}}
As all groups are quasigroups we will use a small group as a counterexample. 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 $(A2)$ does not hold as $1 \circ 1 \ne 0$. {{qed}}
By the definition of a quasigroup it must be shown that: :$\forall g \in G$ the left and right regular representations of $\lambda_g$ and $\rho_g$ are permutations on $G$. This follows immediately from Regular Representations in Group are Permutations. {{qed}} Category:Group Theory Category:Quasigroups prup3816df6j4i87n58rd7zuepd7j6t
First we show that: $\forall x \in X: x \circ x^2 = 0 \circ x$: {{begin-eqn}} {{eqn | l = x \circ x^2       | r = x \circ \left({x^1 \circ \left({0 \circ x}\right)}\right)       | c = {{Defof|B-Algebra Power of Element}} }} {{eqn | r = x \circ \left({x \circ \left({0 \circ x}\right)}\right)       | c = First Power of Element in B-Algebra }} {{eqn | r = \left({x \circ x}\right) \circ x       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = 0 \circ x       | c = Axiom $(A1)$ for $B$-algebras }} {{end-eqn}} :{{qed|lemma}} Now we show that: $\forall x \in X: m \in \N_{>0}: x \circ x^m = 0 \circ x^{m-1}$: {{begin-eqn}} {{eqn | l = x \circ x^m       | r = x \circ \left({x^{m-1} \circ \left({0 \circ x}\right)}\right)       | c = {{Defof|B-Algebra Power of Element}} }} {{eqn | r = \left({x \circ x}\right) \circ x^{m-1}       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = 0 \circ x^{m-1}       | c = Axiom $(A1)$ for $B$-algebras }} {{end-eqn}} :{{qed|lemma}} Now proving the original proposition using a proof by induction. The base case for $n=1, m=2$ was established in the first lemma.  Assuming $x^m \circ x^n = 0 \circ x^{n-m}$ for some $n, m \in N_{>0}$:  {{begin-eqn}} {{eqn | l = x^{m+1} \circ x^n       | r = \left({x^m \circ \left({0 \circ x}\right)}\right) \circ x^n       | c = {{Defof|B-Algebra Power of Element}} }} {{eqn | r = x^m \circ \left({x^n \circ x}\right)       | c = Identity: $x \circ \left({y \circ z}\right) = \left({x \circ \left({0 \circ z}\right)}\right) \circ y$ }} {{eqn | r = x^m \circ x^{n-1}       | c = $B$-Algebra Power Law }} {{eqn | r = x^0 \circ x^{n-m-1}       | c = Induction using the previous lemma }} {{eqn | r = 0 \circ x^{n-m-1}       | c = Definition of Zeroth Power of Element }} {{end-eqn}} Hence the result. {{qed}} Category:B-Algebras gxovz1bnfqs7i4g2h7ce69kewyr0url
For $m \ge n$ the result follows from $B$-Algebra Power Law. For $n > m$ the result follows from $B$-Algebra Power Law with Zero. {{qed}} Category:B-Algebras gu7po00ciykwftwg0aegzzw2bgxvj5n
Let $S$ be a set. Let $A$ be co-countable in $S$, and let $B$ be such that $A \subseteq B \subseteq S$. From Relative Complement inverts Subsets, it follows that: :$\complement_S \left({B}\right) \subseteq \complement_S \left({A}\right)$ As $A$ is co-countable, $\complement_S \left({A}\right)$ is countable. By Subset of Countably Infinite Set is Countable, it follows that $\complement_S \left({B}\right)$ is also countable. Therefore, $B$ is also co-countable, and the result follows. {{qed}}
We have that: :$\forall x \in G: x * x = x \circ x^{-1} = e$ by definition of inverse element. Let $0 := e$. Then it can be seen that: :$(A0): \quad \exists 0 \in G$ :$(A1): \quad \forall x \in G: x * x = 0$ Next note that: : $0^{-1} = e^{-1} = e = 0$ and so: :$\forall x \in G: x * 0 = x \circ 0^{-1} = x \circ e = x$ demonstrating that: :$(A2): \quad \forall x \in G: x * 0 = x$ Finally, let $x, y, z \in G$: {{begin-eqn}} {{eqn | l=\left({x * y}\right) * z       | o= =       | r=\left({x \circ y^{-1} }\right) \circ z^{-1}       | c=by definition of $*$ }} {{eqn | r=x \circ \left({y^{-1} \circ z^{-1} }\right)       | c=$G$ is a group, so $\circ$ is associative }} {{eqn | r=x \circ \left({z \circ y}\right)^{-1}       | c=Inverse of Group Product }} {{eqn | r=x \circ \left({z \circ \left({y^{-1} }\right)^{-1} }\right)^{-1}       | c=Inverse of Group Inverse }} {{eqn | r=x * \left({z * y^{-1} }\right)       | c=by definition of $*$ }} {{eqn | r=x * \left({z * \left({0 \circ y^{-1} }\right) }\right)       | c=as $0 = e$ }} {{eqn | r=x * \left({z * \left({0 * y}\right) }\right)       | c=by definition of $*$ }} {{end-eqn}} This demonstrates: :$(A3): \quad \forall x, y, z \in G: \left({x * y}\right) * z = x * \left({z * \left({0 * y}\right)}\right)$ All axioms of a $B$-algebra have been demonstrated to hold. Hence the result. {{qed}} Category:B-Algebras Category:Group Theory 63bgtbs4t6yzloqrl14rh5zbdp0wkwn
From Group Induces $B$-Algebra, $\left({G, *}\right)$ is a $B$-algebra. As in the proof Group Induces $B$-Algebra, we let: :$0 := e$ Now we demonstrate $0$-commutativity. Let $x, y \in G$: {{begin-eqn}} {{eqn | l=x * \left({0 * y }\right)       | r=x \circ \left({e \circ y^{-1} }\right)^{-1}        | c=by definition of $*$ and $0$ }} {{eqn | r=x \circ (y^{-1})^{-1}       | c=by definition of identity element }} {{eqn | r=x \circ y       | c=Inverse of Group Inverse }} {{eqn | r=y \circ x       | c=by definition of abelian group }} {{eqn | r=y \circ \left({x^{-1} }\right)^{-1}       | c=Inverse of Group Inverse }} {{eqn | r=y \circ \left({e \circ x^{-1} } \right)^{-1}       | c=by definition of identity element }} {{eqn | r=y * \left({0 * x}\right)       | c=by definition of $*$ and $0$ }} {{end-eqn}} Hence the result. {{qed}} Category:B-Algebras Category:Abelian Groups 9cys3fduhaj05rcakpzha68h2xkv8xm
Let $\mathbf a = \operatorname{Card} \left({A}\right)$ for some set $A$. From the definition of (cardinal) one, $\mathbf 1$ is the cardinal associated with a singleton set, say, $\left\{{\varnothing}\right\}$. We have by definition of product of cardinals that $\mathbf 1 \mathbf a$ is the cardinal associated with $\left\{{\varnothing}\right\} \times A$. Consider the mapping $f: \left\{{\varnothing}\right\} \times A \to A$ defined as: :$\forall \left({\varnothing, a}\right) \in \left\{{\varnothing}\right\} \times A: f \left({\varnothing, a}\right) = a$ Let $\left({\varnothing, a_1}\right), \left({\varnothing, a_2}\right) \in \left\{{\varnothing}\right\} \times A$ such that: :$f \left({\varnothing, a_1}\right) = f \left({\varnothing, a_2}\right)$ That is: :$a_1 = a_2$ It follows from Equality of Ordered Pairs that $\left({\varnothing, a_1}\right) = \left({\varnothing, a_2}\right)$. That is: :$f \left({\varnothing, a_1}\right) = f \left({\varnothing, a_2}\right) \implies \left({\varnothing, a_1}\right) = \left({\varnothing, a_2}\right)$ and so $f$ is injective. Now let $a \in A$. By definition: :$a = f \left({\varnothing, a}\right)$ and so $f$ 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 $\left\{{\varnothing}\right\} \times A$ and $A$. It follows by definition that $\left\{{\varnothing}\right\} \times A$ and $A$ are equivalent. The result follows by definition of cardinal. {{qed}}
Let $\mathbf a = \operatorname{Card} \left({A}\right)$ for some set $A$. From the definition of (cardinal) zero, $\mathbf 0$ is the cardinal associated with the empty set $\varnothing$. We have by definition of product of cardinals that $\mathbf 0 \mathbf a$ is the cardinal associated with $\varnothing \times A$. But from Cartesian Product is Empty iff Factor is Empty: :$\varnothing \times A = \varnothing$ Hence the result. {{qed}}
Let $\mathbf a = \map \Card A$ and $\mathbf b = \map \Card B$ for some sets $A$ and $B$ such that $A \cap B = \O$. Then: {{begin-eqn}} {{eqn | l = \mathbf a + \mathbf b       | r = \map \Card {A \cup B}       | c = {{Defof|Sum of Cardinals}} }} {{eqn | r = \map \Card {B \cup A}       | c = Union is Commutative }} {{eqn | r = \mathbf b + \mathbf a       | c = {{Defof|Sum of Cardinals}} }} {{end-eqn}} {{qed}}