hcju/proofwikiqa huggingface dataset:
Let the unit $A$ measure any (natural) number $BC$. Let another (natural) number $D$ measure any other (natural) number $EF$ the same number of times. We need to show that $A$ measures $D$ the same number of times that $BC$ measures $EF$. :400px We have that $A$ measures $BC$ the same number of times that $D$ measures $EF$. So as many units as there are in $BC$ there are numbers equal to $D$ in $EF$. Let $BC$ be divided into the units in it: $BG, GH, HC$. Let $EF$ be divided into the numbers in it equal to $D$: $EK, KL, LF$. So the multitude of $BG, GH, HC$ will equal the multitude of $EK, KL, LF$. We have that $BG = GH = HC$ and $EK = KL = LF$. So $BG : EK = GH : KL = HC : LF$. So from {{EuclidPropLink|book=VII|prop=12|title=Ratios of Numbers is Distributive over Addition}}, $BG : EK = BG + GH + HC : EK + KL + LF$. That is, $BG : EK = BC : EF$. But $BG = A$ and $EK = D$. So $A : D = BC : EF$. So $A$ measures $D$ the same number of times that $BC$ measures $EF$. {{qed}} {{Euclid Note|15|VII}}
Let the number $A$ by multiplying the two numbers $B, C$ to make $D, E$. We need to show that: : $B : C = D : E$ :450px We have that: : $A \times B = D$ Therefore $B$ measures $D$ according to the units in $A$. But the unit $F$ also measures $A$ according to the units in it. Therefore $F$ measures $A$ the same number of times that $B$ measures $D$. So from {{EuclidDefLink|VII|20|Proportional}}: : $F : A = B : D$ For the same reason: : $F : A = C : E$ Therefore also: : $B : D = C : E$ So from {{EuclidPropLink|book=VII|prop=13|title=Proportional Numbers are Proportional Alternately}}: : $B : C = D : E$ {{qed}} {{Euclid Note|17|VII}}
Let two (natural) numbers $A, B$ by multiplying any number $C$ make $D, E$. Then we need to show that: : $A : B = D : E$ :350px We have that: : $A \times C = D$ So from Natural Number Multiplication is Commutative, also: : $C \times A = D$ For the same reason: : $C \times B = E$ Therefore from {{EuclidPropLink|book=VII|prop=17|title=Multiples of Ratios of Numbers}}: : $A : B = D : E$ {{qed}} {{Euclid Note|18|VII}}
Let $A, B, C, D$ be four (natural) numbers in proportion, so that $A : B = C : D$. Let $A \times D = E$ and $B \times C = F$. We need to show that $E = F$. :300px Let $A \times C = G$. Then $A \times C = G$ and $A \times D = E$. So from {{EuclidPropLink||book=VII|prop=17|title=Multiples of Ratios of Numbers}}: : $C : D = G : E$ But we have: : $C : D = A : B$ : $A : B = G : E$ : $A \times C = G$ : $B \times C = F$ Then from {{EuclidPropLink|book=VII|prop=18|title=Ratios of Multiples of Numbers}}: : $A : B = G : F$ Further, we have that: : $A : B = G : E$ Thus: : $G : F = G : E$ So from {{EuclidPropLink|book=V|prop=9|title=Magnitudes with Same Ratios are Equal}}: : $E = F$ {{qed|lemma}} Now suppose that $E = F$. We need to show that: : $A : B = C : D$ Using the same construction, from {{EuclidPropLink|book=V|prop=7|title=Ratios of Equal Magnitudes}}: : $G : E = G : F$ But from {{EuclidPropLink|book=VII|prop=17|title=Multiples of Ratios of Numbers}}: : $G : E = C : D$ Then from {{EuclidPropLink|book=VII|prop=18|title=Ratios of Multiples of Numbers}}: : $G : F = A : B$ So: : $A : B = C : D$ {{qed}} {{Euclid Note|19|VII}}
Let $CD, EF$ be the least (natural) numbers of those which have the same ratio with $A, B$. We need to show that $CD$ measures $A$ the same number of times that $EF$ measures $B$. :250px Suppose $CD$ is an aliquant part of $A$. Then from {{EuclidPropLink|book = VII|prop = 13|title = Proportional Numbers are Proportional Alternately}} and {{EuclidDefLink|VII|20|Proportional}}, $EF$ is also the same aliquant part of $B$ that $CD$ is of $A$. Therefore as many aliquant parts of $A$ as there are in $CD$, so many aliquant parts of $B$ are there also in $EF$. Let $CD$ be divided into the aliquant parts of $A$, namely $CG, GD$ and $EF$ into the aliquant parts of $B$, namely $EH, HF$. Thus the multitude of $CG, GD$ will be equal to the multitude of $EH, HF$. We have that the numbers $CG, GD$ are equal to one another, and the numbers $EH, HF$ are also equal to one another,  Therefore $CG : EH = GD : HF$. So from {{EuclidPropLink|book = VII|prop = 12|title = Ratios of Numbers is Distributive over Addition}}, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents. Therefore $CG : EH = CD : EF$. Therefore $CG, EH$ are in the same ratio with $CD, EF$ being less than they. This is impossible, for by hypothesis $CD, EF$ are the least numbers of those which have the same ratio with them. So $CD$ is not an aliquant part of $A$. Therefore from {{EuclidPropLink|book = VII|prop = 4|title = Natural Number Divisor or Multiple of Divisor of Another}}, $CD$ is an aliquot part of $A$. Also, from {{EuclidPropLink|book = VII|prop = 13|title = Proportional Numbers are Proportional Alternately}} and {{EuclidDefLink|VII|20|Proportional}}, $EF$ is the same aliquot part of $B$ that $CD$ is of $A$. Therefore $CD$ measures $A$ the same number of times that $EF$ measures $B$. {{Qed}} {{Euclid Note|20|VII}}
Let $A, B$ be (natural) numbers which are prime to one another. We need to show that $A$ and $B$ are the least of those which have the same ratio with them. :250px Suppose this were not the case. Then there would be some numbers $C, D$ which are less than $A, B$ such that $A : B =  C : D$. From {{EuclidPropLink|book=VII|prop=20|title=Ratios of Fractions in Lowest Terms}} we have that $C$ measures $A$ the same number of times that $D$ measures $B$. Now, as many times as $C$ measures $A$, let there be so many units in $E$. Then from {{EuclidPropLink|book=VII|prop=20|title=Natural Number Multiplication is Commutative}} $E$ also measures $A$ according to the units in $C$. For the same reason $E$ also measures $B$ according to the units in $D$. Therefore $E$ measures $A$ and $B$, which are prime to one another, which is contrary to our initial hypothesis. Hence the result. {{qed}} {{Euclid Note|21|VII|{{EuclidNoteConverse|prop=22|title=Numbers forming Fraction in Lowest Terms are Coprime}}}}
By definition: :$\forall a, b \in A_F: \norm {a \oplus b} = \norm a \norm b$ So: :$\norm {1_A} = \norm {1_A \oplus 1_A} = \norm {1_A} \norm {1_A}$ So $\norm {1_A} \in \R$ is idempotent under real multiplication. From Idempotent Elements of Ring with No Proper Zero Divisors, only $0 \in \R$ and $1 \in \R$ fit that bill. But $\norm {1_A}$ can not be $0$ as that would make: :$\forall a \in A_F: \norm a = \norm {1_A \oplus a} = \norm {1_A} \norm a = 0 \norm a = 0$ which contradicts the definition of the norm. Hence the result. {{Qed}}
For the result to hold, we need to show that $a \oplus \dfrac {a^*} {\left \Vert {a}\right \Vert^2} = 1 = \dfrac {a^*} {\left \Vert {a}\right \Vert^2} \oplus a$. {{begin-eqn}} {{eqn | o =        | r = a \oplus \dfrac {a^*} {\left \Vert {a}\right \Vert^2}       | c =  }} {{eqn | r = a \oplus a^* \cdot \dfrac 1 {\left \Vert {a}\right \Vert^2}       | c =  }} {{eqn | r = \left \Vert {a}\right \Vert^2 \cdot \dfrac 1 {\left \Vert {a}\right \Vert^2}       | c = Definition of nicely normed algebra }} {{eqn | r = 1       | c =  }} {{eqn | r = \dfrac 1 {\left \Vert {a}\right \Vert^2} \cdot \left \Vert {a}\right \Vert^2       | c =  }} {{eqn | r = \dfrac 1 {\left \Vert {a}\right \Vert^2} \cdot a^* \oplus a       | c = Definition of nicely normed algebra }} {{eqn | r = \dfrac {a^*} {\left \Vert {a}\right \Vert^2}\oplus a       | c =  }} {{end-eqn}} {{qed}} Note that this construction works whether $\oplus$ is associative or not.
Let $a, b \in A$. Then all of $a, b, a^*, b^*$ can be generated by $\Im \left({a}\right)$ and $\Im \left({b}\right)$. {{explain|Prove the above statement.}} So as $A$ is an alternative algebra, it follows that $\oplus$ is associative for $a, b, a^*, b^*$. So: {{begin-eqn}} {{eqn | l = \left \Vert{a b}\right \Vert^2       | r = \left({a \oplus b}\right) \oplus \left({a \oplus b}\right)^*       | c = Definition of norm in nicely normed $*$-algebra }} {{eqn | r = a \oplus b \oplus \left({b^* \oplus a^*}\right)       | c = Definition of conjugate }} {{eqn | r = a \oplus \left({b \oplus b^*}\right) \oplus a^*       | c = Associativity of $\oplus$ (from above) }} {{eqn | r = \left \Vert{a}\right \Vert^2 \left \Vert{b}\right \Vert^2       | c =  }} {{end-eqn}} {{qed}}
When $A$ is  So suppose that: :$\forall a, b \in A: \paren {a \oplus a} \oplus b = a \oplus \paren {a \oplus b}$ :$\forall a, b \in A: \paren {b \oplus a} \oplus a = b \oplus \paren {a \oplus a}$ Then: :$\sqbrk {a, a, b} = 0$ :$\sqbrk {b, a, a} = 0$ where $\sqbrk {a, a, b}$ denotes the associator of $a, b \in A_R$. Now let us compute, using the linearity of $\oplus$ and the two suppositions: {{begin-eqn}} {{eqn | l = \paren {a - b} \oplus \paren {\paren {a - b} \oplus a}       | r = a \oplus \paren {\paren {a - b} \oplus a} - b \oplus \paren {\paren {a - b} \oplus a} }} {{eqn | r = a \oplus \paren {a \oplus a} - a \oplus \paren {b \oplus a} - b \oplus \paren {a \oplus a} + b \oplus \paren {b \oplus a} }} {{eqn | l = \paren {\paren {a - b} \oplus \paren {a - b} } \oplus a       | r = \paren {a \oplus a} \oplus a - \paren {a \oplus b} \oplus a - \paren {b \oplus a} \oplus a + \paren {b \oplus b} \oplus a }} {{eqn | ll= \leadsto       | l = 0       | r = \sqbrk {a - b, a - b, a} }} {{eqn | r = \sqbrk {a, a, a} - \sqbrk {a, b, a} - \sqbrk {b, a, a} + \sqbrk {b, b, a}       | c = Subtract the two expressions }} {{eqn | r = \sqbrk {a, b, a} }} {{end-eqn}} {{explain|This suffices, but why it does could do with explanation (although trivial). Also, where do I use that $A$ isn't Boolean? Maybe better use talk page}} {{qed}} {{Namedfor|Emil Artin|cat = Artin}}
Let the field of real numbers be denoted $\struct {\R, +, \times}$. From Real Vector Space is Vector Space, we have that $\struct {\R^n, +, \cdot}$ is a vector space, where: :$\mathbf a + \mathbf b = \tuple {a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n}$ :$\lambda \cdot \mathbf a = \tuple {\lambda \times a_1, \lambda \times a_2, \ldots, \lambda \times a_n}$ where: :$\mathbf a, \mathbf b \in \R^n$ :$\lambda \in \R$ :$\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$ :$\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$ When $n = 1$, the vector space degenerates to: :$\mathbf a + \mathbf b = \tuple {a + b}$ :$\lambda \cdot \mathbf a = \tuple {\lambda \times a}$ where: :$\mathbf a, \mathbf b \in \R$ :$\lambda \in \R$ :$\mathbf a = \tuple a$ :$\mathbf b = \tuple b$ Thus it can be seen that the vector space $\struct {\R^1, +, \cdot}$ is identical with the field of real numbers denoted by $\struct {\R, +, \times}$. {{qed}}
Suppose that the collection of all ordinals is a set. Let this set be denoted as $\On$. From the corollary of ordinals are well-ordered, it is seen that $\Epsilon {\restriction_{\On} }$ is a strict well-ordering on $\On$. By Element of Ordinal is Ordinal, it is seen that $\On$ is transitive. And so $\On$ is itself an ordinal. Since $\On$ is an ordinal, it follows from hypothesis that $\On \in \On$. Since $\On$ is an ordinal, it follows from Ordinal is not Element of Itself that $\On \notin \On$. But this is a contradiction. Therefore a paradox has been formed. {{qed}}
The complex numbers $\C$ are formed by the Cayley-Dickson Construction from the real numbers $\R$. From Real Numbers form Algebra, we have that $\R$ forms: :$(1): \quad$ An associative algebra. :$(2): \quad$ A commutative algebra. :$(3): \quad$ A normed division algebra. :$(4): \quad$ A nicely normed $*$-algebra whose $*$ operator is the identity mapping. :$(5): \quad$ A real $*$-algebra. From Cayley-Dickson Construction forms Star-Algebra, $\C$ is a $*$-algebra. From Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, $\C$ is a nicely normed $*$-algebra. From Cayley-Dickson Construction from Real Algebra is Commutative, $\C$ is a commutative algebra. From Cayley-Dickson Construction from Commutative Associative Algebra is Associative, $\C$ is an associative algebra. However, from Algebra from Cayley-Dickson Construction Never Real, $\C$ is not a real algebra. === Proof of Normed Division Algebra === Consider the element $\left({1, 0}\right)$ of $\R^2$. We have: {{begin-eqn}} {{eqn | o =        | r = \left({x_1, x_2}\right) \times \left({1, 0}\right)       | c =  }} {{eqn | r = \left({x_1 \times 1 - 0 \times x_2, x_1 \times 0 + x_2 \times 1}\right)       | c =  }} {{eqn | r = \left({x_1, x_2}\right)       | c =  }} {{end-eqn}} As $\times$ has already been shown to be commutative, it follows that $\left({1, 0}\right) \times \left({x_1, x_2}\right) = \left({x_1, x_2}\right)$. So $\left({1, 0}\right) \in \R^2$ functions as a unit. That is, $\left({\R^2, \times}\right)$ is a unitary algebra. <!-- From Inverses for Real Multiplication, every element of $\left({\R, \times}\right)$ except $0$ has a multiplicative inverse. So $\left({\R, \times}\right)$ is a division algebra and hence a unitary division algebra. --> We define a norm on $\left({\R^2, \times}\right)$ by: :$\forall \mathbf a = \left({a_1, a_2}\right) \in \R^2: \left \Vert {\mathbf a} \right \Vert = \sqrt {a_1^2 + a_2^2}$ This is a norm because: : $(1): \quad \left \Vert \mathbf x \right \Vert = 0 \iff \mathbf x = \mathbf 0$ : $(2): \quad \left \Vert \lambda \mathbf x \right \Vert = \left \vert \lambda \right \vert \left \Vert x \right \Vert$ : $(3): \quad \left \Vert x - y \right \Vert \le \left \Vert x - z \right \Vert + \left \Vert z - y \right \Vert$ It also follows that: : $\left \Vert x \times y \right \Vert = \left \vert x \times y \right \vert = \left \vert x \right \vert \times \left \vert y \right \vert = \left \Vert x \right \Vert \times \left \Vert y \right \Vert$ and so $\left({\R^2, \times}\right)$ is a normed division algebra. {{proofread|Make sure the above are all proved properly.}} {{qed}}
{{questionable|Inelegant and unclear to the point of soliciting purging}} By Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous: :$\left({\exists y: y = x \land \forall y: \left({y = x \implies P \left({x}\right)}\right)}\right) \implies \exists y: \left({y = x \land P \left({x}\right)}\right)$ Then: {{begin-eqn}} {{eqn | o =        | r = x = x       | c = Equality is Reflexive }} {{eqn | o = \implies       | r = \exists y: y = x       | c = Existential Generalisation }} {{eqn | o = \implies       | r = \left({ \forall y: \left({ y = x \implies P \left({ x }\right) }\right) \implies \exists y: \left({ y = x \land P\left({ x }\right) }\right) }\right)       | c = Modus Ponendo Ponens }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \left({ y = x \land P \left({ y }\right) }\right)       | o = \implies       | r = P \left({x}\right)       | c = Substitutivity of Equality }} {{eqn | n = 1       | ll= \implies       | l = \exists y: \left({ y = x \land P \left({y}\right) }\right)       | o = \implies       | r = P \left({x}\right)       | c = Universal Generalisation }} {{eqn | n = 2       | ll= \implies       | l = \forall y: \left({ y = x \implies P \left({y}\right) }\right)       | o = \implies       | r = P \left({x}\right)       | c = Hypothetical Syllogism with first lemma }} {{end-eqn}} {{wtd|Theorem needs to connect the first implications.  Universal Generalisation is not sufficient.}} {{qed|lemma}} Similarly: {{begin-eqn}} {{eqn | l = P \left({x}\right)       | o = \implies       | r = \left({ y = x \implies P \left({y}\right) }\right)       | c = Substitutivity of Equality }} {{eqn | n = 3       | o = \implies       | r = \forall y: \left({ y = x \implies P \left({y}\right) }\right)       | c = Universal Generalisation }} {{eqn | n = 4       | o = \implies       | r = \exists y: \left({ y = x \land P \left({y}\right) }\right)       | c = First lemma }} {{end-eqn}} The above two statements comprise the other direction of the biconditional assertions. Together, $(1)$, $(2)$, $(3)$, and $(4)$ prove the two assertions. {{qed}}
From the definition of the universal class: :$\Bbb U = \set {x: x = x}$ From this, it follows immediately that: :$\forall x: \paren {x \in \Bbb U \iff x = x}$ From Equality is Reflexive, $x = x$ is a tautology. Thus the asserted statement is also tautologous. {{qed}}
{{begin-eqn}} {{eqn | l = \exists x       | o = :       | r = \forall y: \paren {\neg \paren {y \in x} }       | c = Axiom of the Empty Set }} {{eqn | ll= \leadsto       | l = \exists x       | o = :       | r = \forall y: \paren {y \in x \iff y \ne y}       | c = Equality is Reflexive }} {{eqn | ll= \leadsto       | l = \exists x       | o = :       | r = x = \set {y: y \ne y}       | c =  }} {{end-eqn}} {{qed|lemma}} Then: {{begin-eqn}} {{eqn | l = A \in U       | o = \iff       | r = \exists x: \paren {x = A \land x \in U}       | c = {{Defof|Class Membership}} }} {{eqn | l = x       | o = \in       | r = U       | c = Fundamental Law of Universal Class }} {{eqn | ll= \leadsto       | l = A \in U       | o = \iff       | r = \exists x: x = A       | c =  }} {{end-eqn}} {{qed|lemma}} Hence: {{begin-eqn}} {{eqn | l = \set {y: y \ne y}       | o = \in       | r = U       | c =  }} {{eqn | ll= \leadsto       | l = \O       | o = \in       | r = U       | c = {{Defof|Empty Set}} }} {{end-eqn}} {{qed}}
Let the conjugation operator on $A$ be $*$. Suppose $A'$ is a real algebra whose conjugation operator  is $*'$. Then by definition: : $\forall a \in A': {a^*}' = a$ Let $a = \left({x, y}\right) \in A'$. Then by definition of the Cayley-Dickson construction: : $x, y \in A$ By definition of the conjugation operator: : ${\left({x, y}\right)^*}' = \left({x^*, -y}\right)$ So for ${a^*}' = a$ it must be that $x = x^*$ and $y = -y$ from definition of Equality of Ordered Pairs. That is, $y = 0$. But $A'$ is a $*$-algebra, which is by definition a unitary division algebra. So $\exists y \in A: y = 1 \ne 0$. From that contradiction, it follows that $A'$ can not be a real algebra. {{qed}}
Let the conjugation operator on $A$ be $*$. Let $\left({a, b}\right), \left({c, d}\right) \in A'$. Let $A$ be a real algebra. {{begin-eqn}} {{eqn | l = \left({a, b}\right) \oplus' \left({c, d}\right)       | r = \left({a \oplus c - d \oplus b^*, a^* \oplus d + c \oplus b}\right)       | c =  }} {{eqn | r = \left({a \oplus c - d \oplus b^*, a^* \oplus d + c \oplus b}\right)       | c =  }} {{eqn | r = \left({a \oplus c - d \oplus b, a \oplus d + c \oplus b}\right)       | c = As $A$ is real, $b^* = b$ and $a^* = a$ }} {{eqn | r = \left({c \oplus a - b \oplus d, c \oplus b + a \oplus d}\right)       | c = Real Star-Algebra is Commutative, Real Addition is Commutative }} {{eqn | r = \left({c \oplus a - b \oplus d^*, c^* \oplus b + a \oplus d}\right)       | c = As $A$ is real, $d = d^*$ and $c = c^*$ }} {{eqn | r = \left({c, d}\right) \oplus' \left({a, b}\right)       | c =  }} {{end-eqn}} So $A'$ is a commutative algebra. {{qed|lemma}} Let $A'$ be a commutative algebra. By picking apart the above equations, it is clear that for $A'$ to be a commutative algebra it is necessary for $A$ to be both real and commutative. Hence the result. {{qed}}
Let the conjugation operator on $A$ be $*$. Let $\left({a, b}\right), \left({c, d}\right), \left({e, f}\right) \in A'$. In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation: :$a \oplus b =: a b$ :$x \oplus' y =: x y$ The context will make it clear which is meant. Suppose $A$ is commutative and associative. Then: {{begin-eqn}} {{eqn | r = \left({\left({a, b}\right) \left({c, d}\right)}\right) \left({e, f}\right)       | o =        | c =  }} {{eqn | r = \left({a c - d b^*, a^* d + c b}\right) \left({e, f}\right)       | c = }} {{eqn | r = \left({\left({a c - d b^*}\right) e - f \left({a^* d + c b}\right)^*, \left({a c - d b^*}\right)^* f + e \left({a^* d + c b}\right)}\right)       | c =  }} {{eqn | r = \left({\left({a c}\right) e - \left({d b^*}\right) e - f \left({a^* d}\right)^* - f \left({c b}\right)^*, \left({a c}\right)^* f - \left({d b^*}\right)^* f + e \left({a^* d}\right) + e \left({c b}\right)}\right)       | c =  }} {{eqn | r = \left({\left({a c}\right) e - \left({d b^*}\right) e - f \left({d^* a}\right) - f \left({b^* c^*}\right), \left({c^* a^*}\right) f - \left({b d^*}\right) f + e \left({a^* d}\right) + e \left({c b}\right)}\right)       | c = Definition of Conjugation }} {{eqn | r = \left({a \left({c e}\right) - a \left({f d^*}\right) - \left({c^* f}\right) b^* - \left({e d}\right) b^*, a^* \left({c^* f}\right) + a^* \left({e d}\right) + \left({c e}\right) b - \left({f d^*}\right) b}\right)       | c = $A$ is commutative and associative }} {{eqn | r = \left({a \left({c e - f d^*}\right) - \left({c^* f + e d}\right) b^*, a^* \left({c^* f + e d}\right) + \left({c e - f d^*}\right) b}\right)       | c =  }} {{eqn | r = \left({a, b}\right) \left({ce - fd^*, c^* f + e d}\right)       | c =  }} {{eqn | r = \left({a, b}\right) \left({\left({c, d}\right) \left({e, f}\right)}\right)       | c =  }} {{end-eqn}} {{qed|lemma}} Now suppose $A'$ is an associative algebra. By picking apart the above series of equations, it is clear that $A$ has to be both commutative and associative in order to assure the associativity of $A'$. {{qed}}
Let the conjugation operator on $A$ be $*$. Let $\left({a, b}\right), \left({c, d}\right) \in A'$. In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation: :$a \oplus b =: a b$ :$x \oplus' y =: x y$ The context will make it clear which is meant. Let $A$ be a nicely normed associative algebra. Then: {{begin-eqn}} {{eqn | l = \left({\left({a, b}\right) \left({a, b}\right)}\right) \left({c, d}\right)       | r = \left({a a - b b^*, a^* b + a b}\right) \left({c, d}\right)       | c =  }} {{eqn | r = \left({\left({a a - b b^*}\right) c - d \left({a^* b + a b}\right)^*, \left({a a - b b^*}\right)^* d + c \left({a^* b + a b}\right)}\right)       | c =  }} {{eqn | r = \left({\left({a a - b b^*}\right) c - d \left({b^* a + b^* a^*}\right), \left({a^* a^* - b b^*}\right) d + c \left({a^* b + a b}\right)}\right)       | c = from definition of conjugation }} {{eqn | r = \left({a a c - b b^* c - d b^* a - d b^* a^*, a^* a^* d - b b^* d + c a^* b + c a b}\right)       | c = $A$ is associative }} {{eqn | r = \left({a a c - b b^* c - d b^* \left({a + a^*}\right), a^* a^* d - b b^* d + c \left({a^* + a}\right) b}\right)       | c =  }} {{eqn | r = \left({a a c - \left \Vert {b}\right \Vert^2 c - d b^* \left({2 \Re \left({a}\right)}\right), a^* a^* d - \left \Vert {b}\right \Vert^2 d + c \left({2 \Re \left({a}\right)}\right) b}\right)       | c = $A$ is nicely normed }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \left({a, b}\right) \left({\left({a, b}\right) \left({c, d}\right)}\right)       | r = \left({a, b}\right) \left({a c - d b^*, a^* d + c b}\right)       | c =  }} {{eqn | r = \left({a \left({a c - d b^*}\right) - \left({a^* d + c b}\right) b^*, a^* \left({a^* d - c b}\right) + \left({a c + d b^*}\right) b}\right)       | c =  }} {{eqn | r = \left({a a c - a d b^* - a^* d b^* - c b b^*, a^* a^* d + a^* c b + a c b - d b^* b}\right)       | c = $A$ is associative }} {{eqn | r = \left({a a c - \left({a + a^*}\right) d b^* - c \left \Vert {b}\right \Vert^2, a^* a^* d + \left({a^* + a}\right) c b - d b^* b}\right)       | c =  }} {{eqn | r = \left({a a c - \left({2 \Re \left({a}\right)}\right) d b^* - c \left \Vert {b}\right \Vert^2, a^* a^* d + \left({2 \Re \left({a}\right)}\right) c b - d \left \Vert {b}\right \Vert^2}\right)       | c = $A$ is nicely normed }} {{end-eqn}} Thus it can be seen that: : $\left({\left({a, b}\right) \left({a, b}\right)}\right) \left({c, d}\right) = \left({a, b}\right) \left({\left({a, b}\right) \left({c, d}\right)}\right)$ Similarly it can be shown that: : $\left({\left({c, d}\right) \left({a, b}\right)}\right) \left({a, b}\right) = \left({c, d}\right) \left({\left({a, b}\right) \left({a, b}\right)}\right)$ and so $A'$ is seen to be an alternative algebra. It follows from reversing the chain of equalities that if $A'$ is a nicely normed and alternative algebra then $A$ has to be a nicely normed associative algebra. {{qed|lemma}} Then from Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, we have that $A'$ is a nicely normed algebra {{iff}} $A$ is also a nicely normed algebra. Hence the result. {{qed}}