We may assume that \(E\) does not already contain any pair of the form \((a_0,c)\), since we are done otherwise. By Axiom 1 implies that \(a_0 \neq 0\). Let \(b_1,\dots ,b_n\) denote all the integers \(b_i\) such that \((b_i,a) \in E\), then \(n \geq 0\) is finite, and by Axiom 2 all the \(b_i\) are non-zero and not equal to \(-a\). Let \(c_0\) be a sufficiently large integer to be chosen later. We then add the pairs \((a_0,c_0)\) and \((c-b_i,-b_i)\) to \(E\) for all \(i\). Furthermore, if \(i\) is such that \((-b_i,d_i) \in E\) for some (necessarily unique and non-zero) \(d_i\), we also add \((d_i+b+i-c_0, b_i-c_0)\) to \(E\). Let \(E'\) be the resulting set of pairs. Axioms 4, 5 for \(E\) ensure that (for \(c_0\) large enough) the addition of these pairs do not cause a violation of Axioms 2 or 3 for \(E'\), and of course Axiom 1 for \(E'\) will also be retained. As for Axiom 4, one can check that the only new pairs of pairs \((',b), (b,c)\) that would trigger these axioms either take the form \((b_i,a_0), (a_0,c_0)\) or \((c_0-b_i,-b_i), (-b_i,d_i)\), and in either case we see from construction that Axiom 4 remains in effect for \(E'\). Finally, the new quadruples of pairs \((b,a), (b',a), (-b, d), (-b',d') \in E'\) only arise when either \((-b,d)\), \((-b',d')\) is equal to \((a_0,c_0)\) and the other three pairs in the quadruple were already in \(E\), and for \(c_0\) large enough we see that Axiom 5 remains valid.

If we choose \(a\) sufficiently large, and set \(a_i = i a\) (say) for \(i=1, 2, 3, 4\), the claim simply follows by adding \((a_i,a_i+h)\) to \(E\) and verifying that none of the axioms are violated.

By Lemma 17.2, Lemma 17.3 and the greedy algorithm starting with the seed consisting of \((0,0)\), \((-1,2)\), \((3,1)\), \((-10,20)\), \((30,10)\), we obtain a graph \(\{ (a,f(a)): a \in \mathbb {Z}\} \) of a function \(f:\mathbb {Z}\to \mathbb {Z}\) with \(f(0)=0\), \(f(-1)=2\), \(f(3)=1\), \(f(-10) = 20\), \(f(30)=10\) and the property that if \(f(a)=b\) and \(f(b)=c\), then \(f(c-a)=-a\), and also with the property that for every non-zero \(h\) there are distinct \(a_{h,1}, a_{h,2}, a_{h,3}, a_{h,4}\) with \(f(a_{h,i})=a_{h,i}+h\) for \(i=1, 2, 3, 4\). We then have that Equation 3 holds. If we then define the magma operation \(\diamond \) by Equation 2, we obtain Equation 1, and since \(f(0)=0\) we have (i). Also, from construction we see that \(R_d (d - a_{h,i}) = d+h\) for any \(d,h,i\), giving (ii). Finally, from construction we have \(R_a (a+1) = a+1+f(-1) = a+3\), so \(L_a R_a (a+1) = a + f(3) = a+1\), and similarly \(L_a R_a (a+10) = a+10\), giving (iii).

Let \(y = (a,c,n)\). By Corollary 17.4 (iii), we can find \(c_{y,a} \neq a,c\) such that \(L_a R_a c_{y,a} = c_{y,a}\), or equivalently \(L_a L_{c_{y,a}} a = c_{y,a}\). Next, for any \(b \neq a\), we can apply Corollary 17.4 (ii) to find \(c_{y,b} \neq b, c, c_{y,a}\) such that \(R_a c_{y,b} = b\); note that as \(R_a a = Sa = a \neq b\), this also implies that \(c_{y,b} \neq a\), and that the \(c_{y,b}\) are distinct as \(b\) variables. We now have \(L_a L_{c_{y,b}} b = L_{Sa} L_{c_{y,b}} L_{c_{y,b}} a = c_{y,b}\) for such \(b\), giving the claim.

We define \(L_{c'} y\) for \(y = (a,c,n) \in G'\) as follows:

• (i) If \(c' = a\) and \(n=0\), then \(L_{c'} y := a\).

• (ii) If \(c' = a\) and \(n \neq 0\), then \(L_{c'} y := (a,c,0)\).

• (iii) If \(c' = c_{y,b}\) for some \(b\), then \(L_{c'} y := b\).

• (iv) In all other cases, \(L_{c'} y\) is undefined.

We of course also define \(L_{c'} y\) for \(y \in \mathbb {Z}\) using Corollary 17.4.

From Lemma 17.5 we see that this is a well-defined operation, with \(L_c y\) undefined for any \(y = (a,c,n) \in G'\). So properties (a)-(e) and (g) are clear from construction. Property (e) is also clear from construction in each of the three cases (i), (ii), (iii) (for (i), (ii) we also use the fact that \(L_a a = a\)).

Write \(y = (a,c,n)\). Because \(L_a: \mathbb {Z}\to \mathbb {Z}\) and \(L_{c'}: \mathbb {Z}\to \mathbb {Z}\) are surjective, we can find \(b \in \mathbb {Z}\) such that \(L_a L_{c'} b = c'\). If we had \(b=c'\), then \(L_a c' = c'\), which from 1 (and \(Sc' = c'\)) implies that \(a = c'\), which contradicts the undefined nature of \(L_{c'} y\) thanks to properties (b), (c) of a partial solution. Hence, \(b \neq c'\). We then set \(L_{c'} y := b\). It is then clear that all the properties (a)-(g) of a partial solution are maintained.

Write \(y = (a,c,n)\). There are several cases.

• Case 1: \(L_{c'} y = w\) for some \(w \in G'\). By Lemma 17.5, \(c_{w,c'}\) is distinct from \(c'\), and from property (d) of a partial solution, \(L_{c_{w,c'}} w = c'\). By property (e) of a partial solution, we can find \(z = (c_{w,c'}, c', n')\) such that \(L_{c'} z\) is currently undefined. Then set \(L_{c'} z := y\), so that we have \(L_{c_{w,c'}} L_{c'} L_{c'} z = c'\). One then verifies that all the axioms (a)-(g) of a partial solution are valid in this case.

• Case 2: \(c' \neq a\) and \(L_{c'} y = b\) for some \(b \in \mathbb {Z}\). By property (g) of a partial solution, \(c' \neq b\). By Corollary 17.4, we can find \(a'\) such that \(L_{a'} b = R_b a' = c'\) and \(a' \neq c'\). By property (e) of a partial solution, we can find \(z = (a', c', n')\) such that \(L_{c'} z\) is currently undefined. Then set \(L_{c'} z = y\), so that we have \(L_{a'} L_{c'} L_{c'} z = c'\). One then verifies that all the axioms (a)-(g) of a partial solution are valid in this case.

• Case 3: \(c' \neq a\) and \(L_{c'} y\) is currently undefined. Use Lemma 17.7 set \(L_{c'} y\) equal to some \(b \in \mathbb {Z}\), then apply the Case 2 analysis.

• Case 4: \(c' = a\). Because we are excluding the case \(n=0\), we are now in Case 1 thanks to property (c) of a partial solution.

By iterating Lemma 17.7 and Lemma 17.8 in alternation, we can find an increasing chain of partial solutions with the property that for any \(y = (a,c,n) \in G'\) and \(c' \in \mathbb {Z}\) and any natural number \(k\), excluding the case where \(c'=a\) and \(n=0\), that one can find a partial solution on this chain for which \(L_{c'} y\) is defined, and such that \(L_{c'} z = y\) for at least \(k\) distinct choices of \(z \in G'\). Taking the limit of this chain, we can then find a fully defined operation \(L_{c'} y\) for \(c' \in \mathbb {Z}\) and \(y \in G\) satisfying the properties (a)-(d), (f), (g) (but not (e)) of a partial solution, with the property that \(y = (a,c,n) \in G'\) and \(c' \in \mathbb {Z}\), excluding the case \(c'=a\) and \(n=0\), that there are infinitely many \(z \in G'\) such that \(L_{c'} z = y\). The excluded case \(c'=a, n=0\) is also covered by property (b) of a partial solution. The claim follows.

We can work with a single \(x \in G'\). Our task is to find a function \(L_x\) for which

and

for all \(y \in G\) (note that \(L_{Sy}\) is already fully constructed).

Define a seed to be an injective partial function \(L_x\) defined on finitely many values and satisfying Equation 4, as well as Equation 5 whenever \(L_x L_x y\) is defined. If there is a \(y \in G\) for which \(L_x y\) is currently undefined, then by hypothesis \(y\) is equal to \(L_x z\) for at most one \(z\). If such a \(z\) exists, we set \(L_x y\) to be an element of \(\{ w: L_{Sz} w = x \} \) different from \(y\) that is not already in the domain or range of \(L_x\); we are able to do so thanks to the infinite surjectivity (Corollary 17.4 (ii)) and the fact that the domain and the range of \(L_x\) are finite. If no such \(z\) exists, we set \(L_x y\) to be arbitrary element of \(G\) not already in the domain or range. In either case, we see that the seed property is preserved. Starting from the seed in which \(L_x\) is only defined on \(x\) and maps it to \(Sx\), we obtain the claim.

Let us define the element \(x_0 := (0, 1, 0) \in G'\). Since \(S x_0 = 0\), we know that \(0 \diamond x_0 = 0\) (see the first part of the proof of Proposition 17.9). Therefore, \(x_0\) does not satisfy the 255 equation, as we have