Left Quasigroup if (2-3) Parastrophe of Magma is Magma

== Theorem ==
Let $\struct {S, \circ}$ be a magma, where
:$S$ is a set
:$\circ : S \times S \to S$ is a binary operation on $S$
Let the $\paren {2 - 3}$ parastrophe of $\struct {S, \circ}$ be a magma.
Then $\struct {S, \circ}$ is a left quasigroup.
== Proof 1 ==
By the definition of a left quasigroup it must be shown that:
:$\forall a, b \in S: \exists ! x \in S: a \circ x = b$
 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.
== Proof 2 ==
Let $\struct {S, \circ}$ be a magma such that $S$ is an empty set.
Then it is vacuously true that if $\paren {2 - 3}$ parastrophe of $\struct {S, \circ}$ is a magma, then $\struct {S, \circ}$ is a left quasigroup.
Suppose $S$ is a non-empty set.
Let $\struct {S, \circ}$ be a magma on $S$.
Let the $\paren {2 - 3}$ parastrophe of $\struct {S, \circ}$ be a magma on $S$, denoted by $\struct {S, *}$.
That is:
:$\forall x_1, x_2, x_3 \in S: x_1 \circ x_2 = x_3 \iff x_1 * x_3 = x_2$
Let $a \in S$ be arbitrary:
Let the mapping $\lambda_a: S \to S$ denote the left regular representation:
:$\forall x \in S: \map {\lambda_a} x = a \circ x$
Since $\struct {S, \circ}$ is a magma, for any $a, x \in S$, let:
:$b = a \circ x$
By the definition of $\paren {2 - 3}$ parastrophe:
:$b = a \circ x \iff x = a * b$
Since $\lambda_a$ is a left-total relation:
:$\forall x = a * b \in S: \exists b = a \circ x \in \Dom {\lambda_a}: \tuple {x, b } \in \lambda_a$
Therefore:
:$\forall b = a \circ x \in S: \exists x = a * b \in \Dom {\lambda_a} : \tuple {x, b } \in \lambda_a$
So $\lambda_a$ is a Surjection from a set $S$ to itself.
Also, we note that for any $a \in S$:
:$\forall x_1, \map {\lambda_a} {x_1} \in S: a \circ x_1 = \map {\lambda_a} {x_1} \iff a * \map {\lambda_a} {x_1}= x_1$
and:
:$\forall x_2, \map {\lambda_a} {x_2} \in S: a \circ x_2 = \map {\lambda_a} {x_2} \iff a * \map {\lambda_a} {x_2}= x_2$
So:
:$\forall x_1 \in S: (x_1, a \circ x_1) \in \lambda_a \iff a * \map {\lambda_a} {x_1} = x_1$
:$\forall x_2 \in S: (x_2, a \circ x_2) \in \lambda_a \iff a * \map {\lambda_a} {x_2} = x_2$
By definition of a mapping, the relation $\lambda_a \subseteq S \times S$ is many-to-one.
So:
:$\forall x \in \Dom {\lambda_a}: \forall y_1, y_2 \in \Cdm {\lambda_a}: \tuple {x, y_1} \in \lambda_a \land \tuple {x, y_2} \in \lambda_a \implies y_1 = y_2$
Thus:
:$\forall x_1, x_2 \in S: (x_1, a \circ x_1) \in \lambda_a \wedge (x_2, a \circ x_2) \in \lambda_a \iff x_1 = a * \map {\lambda_a}{x_1}= a * \map {\lambda_a} {x_2} = x_2$
Therefore:
:$\forall x_1, x_2 \in S: (x_1, a \circ x_1) \in \lambda_a \wedge (x_2, a \circ x_2) \in \lambda_a \implies x_1 = x_2$
So $\lambda_a$ is an Injection from a set $S$ to itself.
It follows that $\lambda_a$ is a bijection from a set $S$ to itself.
Then:
:for all $a \in S$, the left regular representation $\lambda_a$ is a permutation on $S$.
Hence, the result follows from the definition of a left quasigroup for a non-empty set $S$.
Hence, the result.
Category:Magmas
Category:Quasigroups
Category:Parastrophes