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\section{1323 $\rightarrow$ 2744 ?}
We have an arbitrary magma $M$ equipped with a binary operator $\diamond$, such that the property \textbf{1323} holds:
\begin{equation}
x = y \diamond (((y \diamond y)\diamond x)\diamond y).
\end{equation}
Does it then imply the property \textbf{2744}?:
\begin{equation}
x = ((y\diamond y)\diamond (y\diamond x))\diamond y
\end{equation}
We study the diagonal/squaring by setting it first constant $x\diamond x = c$ and include secondly an exception $w\diamond w\neq c$.


\subsection{The diagonal/squaring $y\diamond y$ is not constant}
Assume $y\diamond y = c$ for all $y \in M$ and an arbitrary $c \in M$. Now
\begin{equation}
x = y\diamond((c \diamond x)\diamond y).
\end{equation}
Set $y = c \diamond x$. Now
\begin{equation}
x = (c\diamond x)\diamond c
\end{equation}
and therefore by setting $y = c$ in (1.3) and using (1.4):
\begin{equation}
x = c\diamond((c\diamond x)\diamond c) =c \diamond x 
\end{equation}
which also immediately yields $x = x \diamond c$. Now property 1323 becomes
\begin{equation}
x = y \diamond(x\diamond y).
\end{equation}
and property 2744
\begin{equation}
x = (y \diamond x)\diamond y.
\end{equation}
which are equivalent ("Putnam Law"). Hence if $y\diamond y$ is constant, no counterexample can be derived.

\subsubsection{Relaxation $c\diamond c \neq c$}
We set $c\diamond c = c^2$. Now:
\begin{equation}
c^2 = c\diamond(((c\diamond c) \diamond c^2)\diamond c) = c \diamond c^2.
\end{equation}
Only $x=c^2$ fullfills $c^2 \diamond x = c$:
\begin{equation}
x = c\diamond(((c\diamond c) \diamond x)\diamond c) = c \diamond c^2 = c^2.
\end{equation}
Additionally:
\begin{equation}
c = c^2\diamond (((c^2\diamond c^2)\diamond c)\diamond c^2)=c^2 \diamond c.
\end{equation}
Hence by (1.9) $c = c^2$, which goes against the definition of $c$ and $c^2$.
\subsubsection{Relaxation $\exists\text{!} w: w \diamond w=w^2 \neq c$, $w\neq c$}
Assume $c\diamond w \neq w$. By inserting $x=w$, $y = c\diamond w$ in (1.1) follows:
\begin{equation}
w = (c\diamond w)\diamond c
\end{equation}
and by using this result und setting $x=w$, $y = c$ in (1.1) one derives
\begin{equation}
w = c\diamond (((c\diamond c)\diamond w)\diamond c)= c \diamond w.
\end{equation}
But this is a contradiction. Hence $c\diamond w = w$. Also $x=w$ is the only element of the magma such that $c \diamond x = w$:
\begin{equation}
x = c \diamond(((c \diamond c)\diamond x)\diamond c)=c \diamond(((c \diamond c)\diamond w)\diamond c) = w.
\end{equation}
Therefore for all others $x\neq w$ holds $c\diamond x \neq w$ and with the same analysis as in section 1.1 and our knowledge about $w$ immediately $c\diamond x=x$ follows for \textbf{all} $x\in M$. By inserting $x=w$ in (1.13) follows by this uniqueness
\begin{equation}
w = (c\diamond w)\diamond c = w \diamond c.
\end{equation}
The same holds in general: $x = x \diamond c$ for all $x\in M$.\\
Additionally if $w^2 = w$, then $w^2 \diamond w= w^2 \diamond c$ and with (1.1) one derives $w=c$, a contradiction. Hence $w^2 \neq w$. Additionally
\begin{equation}
c = w\diamond ((w^2\diamond c)\diamond w)=w\diamond (w^2 \diamond w)
\end{equation}
hence $w^2\diamond w \neq w$. And also $w \diamond w^2 \neq w$, since
\begin{equation}
w = w^2\diamond (w\diamond w^2)
\end{equation}
and our latest result. Also trivially $w\diamond w^2, w^2\diamond w \neq c,w^2$. But now with (1.15)
\begin{equation}
w = (w^2\diamond w)\diamond(w\diamond(w^2\diamond w))=(w^2\diamond w)\diamond c=w^2\diamond w
\end{equation}
which is a contradiction.\\
Conclusion: The diagonal/squaring cannot be constant almost everywhere with a single exception, since such a magma doesn't even exist for the law 1323.
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