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\begin{document}

\section{Rules for extending sudoku diagrams for Equation 1076}

As described in a previous file, we do the following for Equation 1076, which is $x=y\diamond((x\diamond(x\diamond y))\diamond y)$.  Working out the linear-invariant version of Equation 1076, where $x\diamond y=x+f(h)$ where $h=y-x$, we find
\begin{eqnarray*}
x & = & y\diamond((x\diamond [x+f(h)])\diamond y)\\
  & = & y \diamond([x+f^2(h)]\diamond y)\\
  & = & y\diamond [x+f^{2}(h)+f(h-f^{2}(h))]\\
  & = & x+h+f(f(h-f^2(h))-(h-f^2(h))),
\end{eqnarray*}
and so we want to choose $f$ satisfying the functional equation
\[
f[f(h-f^2(h))-(h-f^2(h))]=-h.
\]
We take our carrier set to be a free abelian group $A$ on $\aleph_0$ generators.

Let $\mathscr{E}$ be the collection of sets $E\subseteq A^2$ where the following properties holds.
\begin{itemize}
    \item[(1)] $E$ is finite.
    \item[(2)] $E$ is a function (i.e., no element of $A$ occurs as the first coordinate of two different ordered pairs in $E$).
    \item[(3)] There are enough pairs in $E$ so that treating $E$ as a partial definition of $f$, the functional equation holds when $h=0$.
    \item[(4)] If $(a,b),(b,c)\in E$, then $(a-c,d),(-a+c+d,-a)\in E$ for some $d\in A$. (Thus, once we know that $E$ defines both $f(a)$ and $f^2(a)$, the functional equation holds when $h=a$.)
\end{itemize}

We partially order $\mathscr{E}$ under set inclusion.  The set $\mathscr{E}$ is nonempty, by noting $\{(0,0)\}\in \mathscr{E}$.  Alternatively, taking $p,q,r,s\in A$ to be independent generators, one can quickly check that
\[
E_0:=\{(0,p),(p,q),(-q,r),(q+r,0),(-p+q+r,s),(p-q-r+s,-q-r)\}
\]
also belongs to $\mathscr{E}$.  We will primarily work with extensions of $E_0$ in $\mathscr{E}$, in order to guarantee that $f(0)\neq 0$.

\begin{lemma}\label{Lemma:1076Extension}
For each $E\in \mathscr{E}$ and for each $z\in A$, there is an extension $E\subseteq E'\in \mathscr{E}$ where $E'$ satisfies the functional equation when $h=z$.
\end{lemma}
\begin{proof}
{\bf Case 1}:  Assume that $z$ is the first coordinate of some pair in $E$.

Fix $b$ to be the (unique, by condition (2)) second coordinate.  If $b$ is the first coordinate of some pair in $E$, then by condition (4), the functional equation already holds for $z$ by taking $E'=E$.  So, for the rest of this case, assume that $b$ is not a first coordinate of any pair.

Since $E$ is finite by (1), $b$ appears as the second coordinate of only finitely many pairs.  Fix $a_1,\ldots, a_n$ to be the list of the distinct first coordinates for such pairs.  (So one of the $a_i$ is $z$.)  Note that none of these $a_i$ is zero, by condition (3), since $f^2(a_i)$ is not defined.

For each integer $i\in [1,n]$, the element $-a_i\in A$ can appear as the first coordinate in at most one ordered pair in $E$.  Let $S$ be the set of indices where there is such a pair, and for each $i\in S$ fix $b_i'$ to be the unique element where $(-a_i,b_i')\in E$.  Let $T=\{i\in S\, :\, b_i'=0\}$.

Now, let $c$, $d_i$ ($i\in [1,n]$), and $e_j$ ($j\in S-T$) be a set of independent generators of $A$ that do not appear anywhere in the support of $E$.  We take
\begin{eqnarray*}
E' & := & E\cup\{(b,c)\}\cup\{(a_i-c,d_i),(-a_i+c+d_i,-a_i)\}_{i\in [1,n]}\\
    &   & \quad \cup \{(-a_j-b_j'+c+d_j,e_j),(a_j+b_j'-c-d_j+e_j,a_j-c-d_j)\}_{j\in S-T}\\
    &   & \quad\cup\{(-c-d_k,a_k-c-d_k)\}_{k\in T}.
\end{eqnarray*}

A finite check verifies each of the conditions, showing that $E'\in \mathscr{E}$ and that $E'$ satisfies the functional for each $a_i$ (hence for $z$).

{\bf Case 2}: Assume that $z$ is not the first coordinate of some pair in $E$.

If $z$ appears as the second coordinate of some pair in $E$, say $(z',z)$, by applying Case 1 to $z'$, we can extend $E$ to a new set $F$ so that the functional equation holds for $z'$.  Now, $z$ appears as a first coordinate of a pair in $F$, so using Case 1 again we can extend $F$ so that the functional equation holds for $z$.

If $z$ doesn't appear as either a first or a second coordinate, then by adding a new pair $(z,z')$, where $z'$ is any element of $A$ not appearing in the support of $E$, we again reduce to Case 1.
\end{proof}

By running over a well-ordering of $A$, we can always extend any initial seed in $\mathscr{E}$ to a full function on $A$ satisfying the functional equation for 1076.  The following special fact will be quite useful to us.

\begin{cor}\label{Cor:0NotInImage}
Each $X\in \mathscr{E}$ extends to a function $f\colon A\to A$ satisfying the functional equation for 1076.
\end{cor}
\begin{proof}
Fix a well-ordering $a_0,a_1,a_2\ldots$ of $A$, of order type $\omega$.  Put $X_{-1}:=X$, and for each integer $i\geq 0$ recursively let $X_i$ be the extension of $X_{i-1}$ guaranteed by Lemma \ref{Lemma:1076Extension} where the functional equation holds at $h=a_i$.  Taking $f=\bigcup_{i\in \N}X_i$ we are done.
\end{proof}

\section{Equation 47}

Equation 47 is $x=x\diamond(x\diamond(x\diamond x))$.  The relation we get (and want to fail) is $f^3(0)=0$.  Applying the algorithm described in Lemma \ref{Lemma:1076Extension} to the pair $(E_0,p)$ we get the set
\[
E_0\cup\{(q,t),(p-t,u),(-p+t+u,-p)\}\in \mathscr{E},
\]
where $t,u$ are independent generators, also independent of $p,q,r,s$.  Extend this set to a full function by Corollary \ref{Cor:0NotInImage}.  It does not satisfy $f^3(0)=0$.

\section{Equation 99}

Equation 99 is $x=x\diamond ((x\diamond x)\diamond x)$.  The relation is $0=f(f(0)+f(-f(0)))$.  Extend $E_0\cup\{(-p,t),(p+t,u)\}$.

\section{Equation 151}

Equation 151 is $x=(x\diamond x)\diamond (x\diamond x)$.  The relation is $2f(0)=0$. There is no 2-torsion in our carrier group, so this reduces to $f(0)=0$, which is handled by simplify extending $E_0$.

\section{Equation 203}

Equation 203 is $x=(x\diamond (x\diamond x))\diamond x$.  The relation is $f(-f^2(0))+f^2(0)=0$.  This already fails in $E_0$, so we finish as before.

\section{Equation 255}

Equation 255 is $x=((x\diamond x)\diamond x)\diamond x$.  The relation is $f(0)+f(-f(0))+f(-f(0)-f(-f(0)))=0$.  Taking $t,u$ to be independent generators of $A$, also independent from $p,q,r,s$, then $E_0\cup\{(-p,t),(-p-t,u)\in \mathscr{E}$ and fails this new relation.  We finish by extending this set to a full function.

\section{Equation 411}

Equation 411 is $x=x\diamond (x\diamond (x\diamond (x\diamond x)))$.  The relation is $0=f^4(0)$.  Applying the algorithm of Lemma \ref{Lemma:1076Extension} to the set in the section on Equation 47, we can define $f(t)$, and we get
\begin{eqnarray*}
E_0  & \cup & \{(q,t),(p-t,u),(-p+t+u,-p),(t,v),\\ && (q-v,w),(-q+v+w,-q),(-q-r+v+w,x),(q+r-v-w+x,q-v-w))\}
\end{eqnarray*}
which fails this new relation.  Extend the given set by Corollary \ref{Cor:0NotInImage}.

\section{Equation 614}

Equation 614 is $x=x\diamond (x\diamond ((x\diamond x)\diamond x))$.  The relation is $0=f^2(f(0)+f(-f(0)))$.  Extend $E_0\cup\{(-p,t),(p+t,u),(u,v),(p+t-v,w),(-p-t+v+w,-p-t)\}$.

\section{Equation 817}

Equation 817 is $x=x\diamond ((x\diamond x)\diamond (x\diamond x))$.  The relation is $0=f(2f(0))$.  Extend $E_0\cup\{(2p,t)\}$.

\section{Equation 1223}

Equation 1223 is $x=x\diamond (((x\diamond x)\diamond x)\diamond x)$.  The relation is $0=f(f(0)+f(-f(0))+f(-f(0)-f(-f(0))))$.  Extend $E_0\cup\{(-p,t),(-p-t,u),(p+t+u,v)\}$.

\section{Equation 1426}

Equation 1426 is $x=(x\diamond x)\diamond (x\diamond (x\diamond x))$.  The relation is $f(0)+f(f^2(0)-f(0))$=0.  Extending $E_0\cup\{(q-p,t)\}$ to a full function, we are done.

\section{Equation 1629}

Equation 1629 is $x=(x\diamond x)\diamond((x\diamond x)\diamond x)$.  The relation is $f(0)+f^2(-f(0))=0$.  Using the algorithm in Lemma \ref{Lemma:1076Extension} on the pair $(E_0,-p)$ we get
\[
E_0\cup\{(-p,t),(t,u),(-p-u,v),(p+u+v,p),(p-q+u+v,w),(-p-q-u-v+w,-p-u-v)\}\in \mathscr{E}
\]
where $t,u,v,w$ are independent generators of $A$, independent of $p,q,r,s$ also.  The relation now fails on this set, so extending to a full function we finish as before.

\section{Equation 1832}

Equation 1832 is $x=(x\diamond (x\diamond x))\diamond(x\diamond x)$.  The relation is $f^2(0)+f(f(0)-f^2(0))=0$.  Extending $E_0\cup\{(p-q,t)\}$, we finish as before.

\section{Equation 2035}

Equation 2035 is $x=((x\diamond x)\diamond x)\diamond (x\diamond x)$.  The relation is $f(0)+f(-f(-f(0)))=0$.  Extend $E_0\cup\{(-p,t),(-t,u)\}\in \mathscr{E}$.

\section{Equation 2238}

Equation 2238 is $x=(x\diamond (x\diamond (x\diamond x)))\diamond x$.  The relation is $f^3(0)+f(-f^3(0))=0$.  Extend $E_0\cup\{(q,t),(p-t,u),(-p+t+u,-p)\}\in \mathscr{E}$.

\section{Equation 2441}

Equation 2441 is $x=(x\diamond((x\diamond x)\diamond x))\diamond x$.  The relation is
\[
f(f(0)+f(-f(0)))+f(-f(f(0)+f(-f(0))))=0.
\]
Extend $E_0\cup\{(-p,t),(p+t,u),(-u,v)\}$.

\section{Equation 2644}

Equation 2644 is $x=((x\diamond x)\diamond (x\diamond x))\diamond x$. The relation is $2f(0)+f(-2f(0))=0$.  Extend $E_0\cup \{(-2p,t)\}\in \mathscr{E}$.

\section{Equation 2847}

Equation 2847 is $x=((x\diamond (x\diamond x))\diamond x)\diamond x$.  The relation is
\[
f^2(0)+f(-f^2(0))+f(-f^2(0)-f(-f^2(0)))=0.
\]
Extend $E_0\cup\{(-q-r,t)\}$.

\section{Equation 3050}

Equation 3050 is $x=(((x\diamond x)\diamond x)\diamond x)\diamond x$.  The relation is
\[
f(0)+f(-f(0))+f(-f(0)-f(-f(0)))+f(-f(0)-f(-f(0))-f(-f(0)-f(-f(0)))).
\]
Extend $E_0\cup\{(-p,t),(-p-t,u),(-p-t-u,v)\}$.

\section{Equation 3253}

Equation 3253 is $x\diamond x=x\diamond (x\diamond (x\diamond x))$.  The relation is $f(0)=f^3(0)$.  Extending the same set as for Equation 2238 works.

\section{Equation 3456}

Equation 3456 is $x\diamond x=x\diamond ((x\diamond x)\diamond x)$.  The relation is $f(0)=f(f(0)+f(-f(0)))$.  Extending the same as for Equation 2441 (even without the last pair) works.

\section{Equation 3659}

Equation 3659 is $x\diamond x=(x\diamond x)\diamond (x\diamond x)$.  The relation is $f(0)=2f(0)$, or in other words $f(0)=0$.  This is handled by Corollary \ref{Cor:0NotInImage}.

\section{Equation 3862}

Equation 3862 is $x\diamond x = (x\diamond (x\diamond x))\diamond x$.  The relation is $f(0)=f^2(0)+f(-f^2(0))$.  Extend $E_0\cup\{((-p,t))\}$.

\section{Equation 4065}

Equation 4065 is $x\diamond x= ((x\diamond x)\diamond x)\diamond x$.  The relation is $f(0)=f(0)+f(-f(0))+f(-f(0)-f(-f(0)))$, which simplifies to $f(-f(0))+f(-f(0)-f(-f(0)))=0$.  Extend $E_0\cup\{(-p,t),(-p-t,u)\}$.

\section{Equation 4380}

Equation 4380 is $x\diamond (x\diamond x)=(x\diamond x)\diamond x$.  The relation is $f^2(0)=f(0)+f(-f(0))$.  Extend $E_0\cup\{(-p,t)\}$.

\section{Other stuff}

The other non-implications now follow by known implications.



\end{document} 