Because \(f\) maps \(E_1 \backslash E_0\) bijectively to \(E_2 \backslash E_1\), there are three cases:

• \(h\) is equal to an element \(h_0\) of \(G \backslash E_2\).

• \(h\) is equal to an element \(h_0\) of \(E_1 \backslash E_0\).

• \(h\) is equal to \(h_1 = f(h_0)\) for some \(h_0 \in E_1 \backslash E_0\), so that \(h_1 \in E_2 \backslash E_1\).

We deal with these three cases in turn.

First suppose that \(h = h_0\in G \backslash E_2\). We perform the following construction.

• Choose an element \(h_1 \in \mathbb {Z}\) that does not lie in \(E_2 \cup \{ h_0\} \); this is possible because \(E_2\) is finite.

• Choose an element \(h_2 \in \mathbb {Z}\) such that \(h_2, h_0+h_1+h_2\), and \(-h_1-h_2\) are all distinct from each other and lie outside of \(E_2 \cup \{ h_0, h_1\} \); this is possible because \(E_2\) is finite.

• Promote \(h_0\) to \(E_0\), promote \(h_1, h_0+h_1+h_2\) to \(E_1\), and promote \(h_2, -h_1-h_2\) to \(E_2\), creating new sets

  \begin{align*} E’_0 & := E_0 \cup \{ h_0\} \\ E’_1 & := E_1 \cup \{ h_0, h_1, h_0+h_1+h_2\} \\ E’_2 & := E_2 \cup \{ h_0, h_1, h_0+h_1+h_2, h_2, -h_1-h_2\} . \end{align*}

  Clearly we still have nested finite sets \(E'_0 \subset E'_1 \subset E'_2\).

• Extend \(f : E_1 \to E_0\) to a function \(f': E'_1 \to E'_0\) by defining

  \begin{align*} f’(h_0) & := h_1 \\ f’(h_1) & := h_2 \\ f’(h_0+h_1+h_2) & := -h_1-h_2 \end{align*}

  while keeping \(f'(h)=f(h)\) for all \(h \in E_1\).

It is then a routine matter to verify that \((E'_0,E'_1,E'_2,f')\) is a partial solution to Equation 1 extending \((E_0,E_1,E_2,f)\) and that \(E'_0\) contains \(h_0\), as required.

Now suppose that \(h = h_0 \in E_1 \backslash E_0\), then the quantity \(h_1 := f(h_0)\) lies in \(E_2 \backslash E_1\). We perform the following variant of the above construction:

• Choose an element \(h_2 \in \mathbb {Z}\) such that \(h_2, h_0+h_1+h_2\), and \(-h_1-h_2\) are all distinct and lie outside of \(E_2\). This is possible because \(E_2\) is finite.

• Promote \(h_0\) to \(E_0\), promote \(h_1\) and \(h_0+h_1+h_2\) to \(E_1\), and promote \(h_2, -h_1-h_2\) to \(E_2\), thus creating new sets

  \begin{align*} E’_0 & := E_0 \cup \{ h_0\} \\ E’_1 & := E_1 \cup \{ h_1, h_0+h_1+h_2\} \\ E’_2 & := E_2 \cup \{ h_0+h_1+h_2, h_2, -h_1-h_2\} . \end{align*}

  Clearly we still have nested finite sets \(E'_0 \subset E'_1 \subset E'_2\).

• Extend \(f : E_1 \to E_0\) to a function \(f': E'_1 \to E'_0\) by defining

  \begin{align*} f’(h_1) & := h_2 \\ f’(h_0+h_1+h_2) & := -h_1-h_2 \end{align*}

  while keeping \(f'(h)=f(h)\) for all \(h \in E_1\).

It is then a routine matter to verify that \((E'_0,E'_1,E'_2,f')\) is a partial solution to Equation 1 extending \((E_0,E_1,E_2,f)\) and that \(E'_0\) contains \(h_0\), as required.

Finally, suppose that \(h = h_1 = f(h_0)\) for some \(h_0 \in E_1 \backslash E_0\), so that \(h_1 \in E_2 \backslash E_1\). Then we perform the following algorithm.

• Choose an element \(h_2 \in \mathbb {Z}\) such that \(h_2, h_0+h_1+h_2\), and \(-h_1-h_2\) are all distinct and lie outside of \(E_2\). This is possible because \(E_2\) is finite.

• Choose an element \(h_3 \in \mathbb {Z}\) such that \(h_3, h_1+h_2+h_3\), and \(-h_2-h_3\) are all distinct and lie outside of \(E_2 \cup \{ h_2, h_0+h_1+h_2,-h_1-h_2\} \). This is possible because \(E_2\) is finite.

• Promote \(h_0, h_1\) to \(E_0\), promote \(h_2, h_0+h_1+h_2, h_1+h_2+h_3\) to \(E_1\), and promote \(h_3, -h_1-h_2,-h_2-h_3\) to \(E_2\), creating new sets

  \begin{align*} E’_0 & := E_0 \cup \{ h_0, h_1 \} \\ E’_1 & := E_1 \cup \{ h_1, h_2, h_0+h_1+h_2, h_1+h_2+h_3 \} \\ E’_2 & := E_2 \cup \{ h_2, h_3, h_0+h_1+h_2, h_1+h_2+h_3, -h_1-h_2, -h_2-h_3\} . \end{align*}

  Clearly we still have nested finite sets \(E'_0 \subset E'_1 \subset E'_2\).

• Extend \(f : E_1 \to E_0\) to a function \(f': E'_1 \to E'_0\) by defining

  \begin{align*} f’(h_1) & := h_2 \\ f’(h_0+h_1+h_2) & := -h_1-h_2 f’(h_2) & := h_3 \\ f’(h_1+h_2+h_3) & := -h_2-h_3 \end{align*}

  while keeping \(f'(h)=f(h)\) for all \(h \in E_1\).

It is then a routine matter to verify that \((E'_0,E'_1,E'_2,f')\) is a partial solution to Equation 1 extending \((E_0,E_1,E_2,f)\) and that \(E'_0\) contains \(h_0\), as required.

If we arbitrarily well-order the integers, and iterate Lemma 7.2 to add the least element of \(\mathbb {Z}\backslash E_0\) in this well-ordering to \(E_0\), we obtain an increasing sequence \((E^{(n)}_0, E^{(n)}_1, E^{(n)}_2, f^{(n)})\) of partial solutions to Equation 1, where the \(E^{(n)}_0\) exhaust \(\mathbb {Z}\): \(\bigcup _{n=1}^\infty E^{(n)}_0 = \mathbb {Z}\). Taking limits, we obtain a full solution \(\tilde f\).

Select integers \(h_0,h_1,h_2,h'_0,h'_1,h'_2\) such that the quantities

are all distinct, but such that

(there are many assignments of variables that accomplish this). Then set

and define \(f: E_1 \to E_2\) by the formulae

One can then check that \((E_0, E_1, E_2, f)\) is a partial solution to Equation 1, and by construction \(h \mapsto h + f(h)\) is not injective on \(E_1\). Using Lemma 7.2 to extend this partial solution to a full solution, we obtain the claim.

Note that \(L_y (y+h) = y + h + f(h)\), so the injectivity of the left-multiplication maps is equivalent to the injectivity of the map \(h \mapsto h + f(h)\). The non-injectivity then follows from Corollary 7.4. Note that the Obelix law clearly expresses \(x\) as a function of \(y\) and \(L_y x = y \diamond x\), forcing injectivity of left-multiplication, so the Obelix law fails.

From Definition 2.22 we see the map \(z \mapsto y \diamond z\) is surjective, hence injective on a finite magma; thus the magma is left-cancellative. Replacing \(x\) by \(y \diamond x\) in this law, we see that

using injectivity, we conclude

which is Definition 2.31.

Case 1: Assume \((a,b)\in E\) for some \(b\in G\).

If \(b\in {\rm dom}(E)\), then by condition (4) we are already done. So reduce to the case when \(b\notin {\rm dom}(E)\). In particular, by (2) and (3) we know \(a,b\neq 1\), and also by (6) we know that \(ab^{-1}\notin {\rm dom}(E)\cup {\rm im}(E)\) (and is not \(1\)).

Take \(c\) to be a generator of \(G\) not appearing in the reduced form of any entry in \(E\), and set \(E':=E\cup \{ (b,c),(cb^{-1},ab^{-1}\} \). Conditions (1), (3), and (5) are immediate. Condition (2) is also clear, where injectivity needs \(ab^{-1}\notin {\rm im}(E)\). Condition (4) is also easy to check, using the fact that \(E\) is injective, \(cb^{-1}\neq c\), and \(ab^{-1}\notin {\rm dom}(E)\).

Finally, for condition (6), a finite check works. The main case is checking that \(ca^{-1}\notin {\rm dom}(E')\cup {\rm im}(E')\). This is clear since \(a\neq 1\) and \(a\neq b\) (by condition (5) for \(E\)). One should also note that by condition (5) (and the fact that \(E\) is a function), there is no pair \((x,y)\in E\) with \((x,y)\neq (a,b)\) and \(xy^{-1}=ab^{-1}\), so we don’t ruin condition (6) for pairs already in \(E\).

Case 2: Assume \(a\notin {\rm dom}(E)\). If \((x,a)\in E\) for some (unique by (2)) \(x\in G\), then applying Case 1 to \(x\), we reduce to the case when \(a\in {\rm dom}(E)\).

Thus, we may consider the case when \(a\notin {\rm dom}(E)\cup {\rm im}(E)\). If \((x,y)\in E\) with \(xy^{-1}=a\), then by applying Case 1 to \(x\), we get \(a\in {\rm im}(E)\), and reduce to a previously considered case. So, we may assume there is no such pair \((x,y)\). Fixing \(b\) to be a generator of \(G\) not appearing in the reduced forms for the entries in \(E\), nor in \(a\), then after passing to \(E\cup \{ (a,b)\} \in \mathscr {E}\) we again reduce to Case 1.

This requires picking an initial set that still obeys the correct initial closure properties. Letting \(x_1\), \(x_2\), \(x_3\) and \(x_4\) be independent elements, then taking the function

already has elements that contradict the Asterix equation, and it’s a valid partial solution. By extending it, we can get an Obelix that does not obey the Asterix equation.

Suppose first that \(b=a\), so \(a \diamond a\) is undefined, then by Law 4 we have \(d \diamond a \neq a\) for all \(d\) with \(d \diamond a\) defined. Set \(a_0=a\) and find nine distinct elements \(a_1,a_2,a_3,a_4,b_0,b_1,b_2,b_3,b_4\) not previously appearing in the domain or range of \(\diamond \). Extend the indices periodically by setting \(a_{i+4} = a_i,b_{i+4} = b_i\) for \(i \ge 1\), and define \(a_i \diamond ' a_i = a_{i+1}\) (so \(a_i = S^i a\)), \(a_{i+1} \diamond ' a_i = b_i\), \(b_i \diamond ' a_i = a_{i+2}\), \(a_{i+3} \diamond ' a_{i+1} = a_{i+1}\), \(a_{i+1} \diamond ' a_{i+2} = a_{i+1}\), \(a_{i+1} \diamond ' b_i = a_i\). We also define \(x \diamond ' y = x \diamond y\) whenever the right-hand side is defined. It is easy to see that laws 2,3,4 are preserved (though for law 2 note that we have not defined \(a_2 \diamond ' a_0\) due to the way indices are extended). Now we case check Law 1.

• Case 1. \(y\) is fresh, that is \(y \in \{ a_1,a_2,a_3,a_4,b_0,b_1,b_2,b_3,b_4\} \). If \(x \diamond ' y\) is not defined, we are done, so assume this is so. In particular, \(x \in \{ a_1,a_2,a_3,a_4,b_0,b_1,b_2,b_3,b_4\} \). If \(x,y \in \{ a_1,a_2,a_3,a_4\} \) we are done by construction. If \(x = b_i\) for some \(i\), then \(y = a_i\) and we are again done by construction. Finally, if \(y = b_i\) for some \(i\), then \(x = a_{i+1}\) and \((x \diamond ' y) \diamond ' y\) is undefined.

• Case 2. \(x\) is fresh, that is \(x \in \{ a_1,a_2,a_3,a_4,b_0,b_1,b_2,b_3,b_4\} \). If we are not in Case 1 and \(x \diamond ' y\) is defined, we must have \(y = a_0\) and \(x \in \{ a_1, b_0\} \). The case \(x = a_1\) holds by construction, and the case \(x = b_0\) holds as \(a_2 \diamond ' a_0\) is not defined.

• Case 3. \(x,y \not\in \{ a_1,a_2,a_3,a_4,b_0,b_1,b_2,b_3,b_4\} \) and \(x \diamond y\) was undefined. Then \(x \diamond ' y\) is undefined unless \(x=y=a_0\), in which case Law 1 holds by construction.

• Case 4. \(x,y \not\in \{ a_1,a_2,a_3,a_4,b_0,b_1,b_2,b_3,b_4\} \), \(x \diamond y\) was defined, and \((x \diamond y) \diamond y\) was not defined. Then \((x \diamond ' y) \diamond ' y\) is undefined unless \((x \diamond y) = a = y\), but this is impossible by Law \(4\) as already observed.

• Case 5. \(x,y \not\in \{ a_1,a_2,a_3,a_4,b_0,b_1,b_2,b_3,b_4\} \) and \(x \diamond y\) and \((x \diamond y) \diamond y\) were defined. Then the claim follows from Law 1 applied to \(\diamond \).

Now suppose that \(a \diamond b\) is undefined for some \(a \neq b\). By applying the previous construction, we may assume without loss of generality that \(Sb\) is defined. We may assume that \(a \diamond b\) remains undefined after doing so, since we are done otherwise. By Law 3, \(a \neq Sb \diamond b\). By Law 2, we have \(d \diamond b = a\) for at most one \(d\), which is necessarily different from \(Sb\). Choose a \(c\) not previously appearing in the domain or range of \(\diamond \), and set \(a \diamond ' b = c\). If a \(d\) exists as before, we additionally set \(Sb \diamond ' c = d\). Finally we define \(x \diamond ' y = x \diamond y\) whenever the right-hand side is defined. It is again routine to check that \(\diamond '\) obeys Laws 2,3,4. Now we case check Law 1.

• Case 1. \(x=c\) or \(x \diamond ' y = c\). Not possible since no left multiplication by \(c\) is defined.

• Case 2. \(y=c\). This would force \(x = Sb\) and \(d = Sb\), but this is not possible since \(d\) is different from \(Sb\).

• Case 3: \((x \diamond ' y) \diamond ' y = c\). This forces \(x=d\) and \(y=b\), and the claim follows from construction.

• Case 4: None of the terms in \((x \diamond ' y) \diamond ' y\) are equal to \(c\). Then the claim follows from Law 1 applied to \(\diamond \).

First suppose that \(a \diamond a\) is not defined. Set \(a_0 := a\) and select three new elements \(a_1,a_2,a_3\). Define \(a_0 \diamond a_0 = a_1\), \(a_1 \diamond a_0 = a_2\), \(a_0 \diamond a_2 = a_3\), \(a_0 \diamond a_1 = a_3\), \(a_0 \diamond a_3 = a_0\). It is a routine but tedious matter to check that Laws 1,2,3 are preserved by these operations. (The trickiest case is the \(x=a_0\) case of Law 1; here we need Law 2 to prevent \(y \diamond a_0\) from equalling \(a_0\).)

Now suppose that \(a \diamond b\) is undefined for some \(a \neq b\). Let \(d_1,d_2,\dots ,d_n\) be all the elements with \(d_i \diamond b = a\). Note from Law 3 that none of the \(d_i\) can be equal to \(b\). Choose a new element \(c\), as well as \(c_i\) for each \(d_i\), and set \(a \diamond b = c\), \(d_i \diamond c = c_i\), and \(d_i \diamond c_i = b\). It is easy to see that this preserves Laws 2,3. For Law 1 ,we do the usual case analysis:

• Case 1: \(y = c\), \(y = c_i\), \(y \diamond x = c\), or \(y \diamond x = c_i\). This case cannot occur because we have not defined left multiplication with \(c\) or \(c_i\).

• Case 2: \(x = c\) or \(x = c_i\). Because \(d_i \neq b\), it is easy to see that \((y \diamond x) \diamond x\) is not defined for any \(x\).

• Case 3: \((y \diamond x) \diamond x = c\) or \((y \diamond x) \diamond x = c_i\). This is only possible if \(x=b\) and \(y = d_j\) for some \(j\), and the claim then follows from construction.

• Case 4: None of the terms in \((y \diamond x) \diamond x\) are equal to \(c\) or \(c_i\). This follows from the previous seed.

If \(a \diamond b\) is currently undefined, introduce a new element \(c\) and define \(a \diamond b = c\) and \( c \diamond c = c.\) If furthermore \(a = (d \diamond b) \diamond b\) for some d, then also define \(b \diamond c = d\) (This last step is well defined by injectivity of \(R_b\)).

We now check that the three axioms still hold. Law 3 is obvious, law 2 follows since \(d \not= c\). For law 1, we first make the observation that \(a \not= b\), for otherwise \(a \diamond b\) would have been defined. This has the consequence that if \(d\) exists, then \(d \not= b\), for otherwise \(a = (b \diamond b) \diamond b = b.\) In particular, the product \(d \diamond c\) remains undefined after running the algorithm once.

We now work by cases on the values of the variables \(x,y\) appearing in law 1:

• Case 1: \(y=c\). In this case, for the product \(x \diamond y\) to be defined, we require \(x =b\) or \(x = c\). In the former case either \(b \diamond c\) or \((b \diamond c) \diamond c = d \diamond c\) is not defined. In the latter case, law 1 holds as \(c\) is idempotent.

• Case 2: \(y=b\). For \(((x \diamond b) \diamond b) \diamond b\) to be defined now but not have been defined before, we require one of \(x= a, (x \diamond b) = a, ((x \diamond b) \diamond b) = a.\) The first two cases are eliminated as \(c \diamond b\) is not yet defined. The last case holds as then \(d\) exists and equals \(x\), so \(b \diamond (((x \diamond b) \diamond b) \diamond b) = b \diamond (a \diamond b) = b \diamond c = x.\)

• Case 3: \(y \not= c, y \not= b\). We have introduced no new products with a righthand side other than \(c\) or \(b\), so if \(((x \diamond y) \diamond y) \diamond y\) was undefined before it remains undefined now.

If \(a \diamond b\) is undefined, set it equal to a new element \(c\). If we also have \(b = d \diamond a\) for some \(d\) (unique by Law 2, and only possible for \(a \neq b\) by Law 3), set \(a \diamond c = d\).

It is obvious that Laws 2, 3 are preserved. Now we case check Law 1:

• Case 1: \(x=c\) or \(y=c\). Not possible since no left multiplication with \(c\) is defined.

• Case 2: \(x \diamond y = c\). Only possible when \(x = a\), \(y = b\), but then \(y \diamond (x \diamond y)\) is undefined since \(y = b \neq a\) if \(d\) is defined.

• Case 3. \(y \diamond (x \diamond y) = c\). Only possible when \(y=a\) and \(x=d\), and holds in this case.

• Case 4: \(x, y, x \diamond y, y \diamond (x \diamond y) \neq c\): this case is already covered by the previous seed.

If \(a = b\), set \(a \diamond ' a = a\). This clearly does not destroy Laws 3 or 4, and because of Law 4, Law 2 will also be retained. Finally, using Law 4, we can check that this addition does not break Law 1.

Now suppose \(a \neq b\). Let \(d_1,d_2,..., d_n\) be all the elements with \(b = a \diamond d_i\), and note from Law 3 and \(a \neq b \) that we have the crucial inequality \(d_i \neq a\) for all \(i\). If \(d_i \diamond a\) is defined, let \(e_i = d_i \diamond a.\) From Law 4 and \(d_i \neq a\), we have \(e_i \neq d_i\). From Law 2 and \(a \diamond d_i = a \diamond d_j\), we further have \(e_i \neq e_j\) and \(e_i \neq d_j\) for \(i \neq j\).

For the algorithm, we set \(a \diamond ' b = c\) for some new \(c\), \(d_i \diamond ' c = a\), and \(c \diamond ' e_i = d_i\) (This is well defined as we have just argued \(e_i \neq e_j\) for \(i \neq j\)).

This creates a new partial function. Law 2 is verified as follows: The case \(x=c\) cannot occur because \(c \diamond ' e_i \neq c \diamond ' e_j\) for \(i \neq j\). If \(y=c\), then \(x = d_i\) for some \(i\), but then \(y \diamond ' x = c \diamond ' d_i\) is undefined as \(d_i \neq e_j\), so this case is vacuous. If \(z=c\) and \(x,y \neq c\), then similarly \(z \diamond ' x\) is undefined, and \(y \diamond ' x\) is not equal to \(c = z\), so we are ok in this case also.

Law 3 is clearly unaffected, and it is also straightforward to check that Law 4 is also not invalidated (here it is essential that \(d_i \neq a\)).

Now we verify Law 1.

• Case 1: \(x=c\). Then for \(x \diamond ' y\) to be defined we must have \(y=e_i\) for some \(i\), but then \(x \diamond ' (x \diamond ' y) = c \diamond ' d_i\) is undefined as \(d_i \neq e_j\) for all \(j\).

• Case 2: \(y=c\). Then for \(x \diamond ' y\) to be defined we must have \(x = d_i\) for some \(i\). Then \(d_i \diamond ' (d_i \diamond ' c)\) is either undefined or is equal to \(e_i\), in which case Law 1 holds as \(c \diamond ' e_i = d_i\).

• Case 3: \(x \diamond ' y = c\). Then \(x=a\) and \(y=b\). Since \(d_i \neq a\) for all \(i\), \(x \diamond (x \diamond y)\) is undefined.

• Case 4: \(x \diamond ' (x \diamond ' y) = c\). Then \(x = a\) and \(y = d_i\) for some \(i\), and Law 1 holds as \(d_i \diamond ' c = a\).

• Case 5: None of \(x, y, x \diamond ' y, x \diamond ' (x \diamond ' y)\) are equal to \(c\). Then this case is covered by the previous seed.

Set \(a \diamond ' b = c\) for some new element \(c\). If \(d_1,d_2,...,d_n\) are the elements for which \(a \diamond d_i = b\), set \(c \diamond ' d_i = c'_i\) for some new element \(c'_i\), and also set \(d_i \diamond ' c'_i = a\). If \(d_i \diamond a\) is equal to some \(e_i\), set \(e_i \diamond ' c'_i = c''_i\) and \(c'_i \diamond ' c''_i = d_i\), where \(c''_i\) is equal to \(a\) if \(e_i = d_i\), or is some new element otherwise.

One can verify that no collisions are caused by this construction (this is why we have to take \(c''_i=a\) when \(e_i=d_i\)), and that Law 2 is preserved.

We need to verify that these constructions preserve Law 1. There are several cases:

• \(y=c\). Not possible since there are no right multiplications by \(c\) defined.

• \(x=c\). Not possible since it forces \(y = d_i\), \(x \diamond y = c'_i\) for some \(i\).

• \(y=c'_i\), \(x = d_i\). True by construction.

• \(y=c'_i\), \(x \neq d_i\). This forces \(x = e_i \neq d_i\) and \(x \diamond ' y = c''_i\), but then \(x \diamond ' (x \diamond ' y)\) is undefined.

• \(x=c'_i\). Not possible because this forces \(x \diamond ' y = d_i\), and \(x \diamond ' (x \diamond ' y)\) will be undefined. (Here we use the fact that \(e_i=d_i\) can only occur when \(d_i \neq a\), by Law 2.)

• \(x = c''_i\) is new. Not possible as no left multiplications by \(c''_i\) are defined.

• \(y=c''_i\) is new. This forces \(x = c'_i\) and \(x \diamond ' y = d_i\), but then \(x \diamond ' (x \diamond ' y)\) is undefined.

• \(x\) and \(y\) are not new and \(x \diamond y\) was undefined. This requires \(x=a, y=b\), but then \(x \diamond ' (x \diamond ' y)\) is undefined since right multiplication by \(c\) is undefined.

• \(x\), \(y\) are not new and \(x \diamond y\) was defined, but \(x \diamond (x \diamond y)\) was undefined. This requires \(x=a\), \(x \diamond y = b\), and hence \(y = d_i\) for some \(i\). The claim now follows from construction.

• \(x\), \(y\), are not new and \(x \diamond y\), \(x \diamond (x \diamond y)\) were defined. This then follows from the previous seed.

This will be a greedy construction, but we have to introduce a rather large number of additional elements to ensure that certain non-degeneracy conditions are maintained (in particular, the new elements \(h'\) introduced have to be such that \(f^2(h') - h'\) avoids \(E_2\) and hence \(\mathbb {Z}^3\), which requires a certain amount of complexity in the construction).

Let \(\tilde E_2\) denote the set \(E_2\) together with all the elements of the form \(-h'\), \(h' - f^2(h')\), and \(f^2(h') - h'\) for \(h' \in E_2 \backslash E_1\); this is \(\mathbb {Z}^3\) with a finite number of additional elements.

Let \(a\) be an element of \(\mathbb {Z}^3\) such that the quantities

are distinct from each other and from

this is possible since \(T, T+1\) are invertible.

Next, we pick an \(x \in \mathbb {Z}^3 \times \mathbb {Z}\) such that the quantities

are all distinct from each other and from \(\tilde E_2\); indeed, from the choice of \(a\) (and the non-zero nature of \(h\), \(f(h)\), \(h-f(h)\), and \(f^2(h)-h\)) there are only finitely many \(x\) for which a collision can occur between any pair of these expressions.

We now promote \(\pm h, \pm x, \pm x + Ta, \pm x - Ta\) to \(E_1\), and also promote \(h+x\), \(h - f^2(h)\), \(\pm 2x + Ta\), \(\pm x + Ta + a\) to \(E_2\). That is to say, we define

and

We then define an extension \(f'\) of \(f\) to \(E'_2\) by setting

It is easy to see that \((E'_1,E'_2,f')\) obeys properties (a), (b), (d). By construction, we have

for

giving property (c). By construction, the quantities

are distinct from each other and lie outside of \(\tilde E_2 \cup E'_2\) for

while these quantities are distinct from each other and lie in \(\tilde E_2 \backslash E'_2\) for \(h' \in E_2 \backslash E'_1\) (note that this forces \(h' \neq \pm h\)). This gives property (e).

Let \(\tilde E_2\) denote the set \(E_2\) together with all elements of the form

for some \(h' \in E_2 \backslash E_1\); this is \(\mathbb {Z}^3\) with a finite number of elements attached, and is symmetric around the origin.

First suppose that \(h\) lies outside of \(\tilde E_2\). Then we can find \(a \in \mathbb {Z}^3\) such that the quantities

are distinct from each other, and lie outside of \(\tilde E_2\). We now promote \(h\) to \(E_2\), thus setting

and

and define an extension \(f'\) of \(f\) to \(E'_2\) by setting

It is then a routine matter to check that \((E'_1,E'_2,f')\) is an extension of \((E_1,E_2,f)\).

Now suppose that \(h\) lies in \(\tilde E_2\). Then \(h\) is of one of the forms Equation 17 for some \(h' \in E_2 \backslash E_1\). Suppose it is of the form \(-h'\) or \(f^2(h') - h'\). We invoke Lemma 7.18 to promote \(h'\) to \(E_1\) by using a suitable extension \((E'_1,E'_2,f')\); and now \(h\) will lie in \(E'_2\), giving the claim. If instead \(h\) is of the form \(h'-f^2(h')\), then this procedure might not place \(h\) in \(E'_2\); but if it does not, it will be of the form \(-h''\) for \(h'' = f^2(h')-h' \in E'_2 \backslash E'_1\). Applying Lemma 7.18 one more time to now promote \(h''\) to \(E_1\), we will obtain a larger extension \((E''_1,E''_2,f'')\) with \(h \in E''_2\), giving the claim.

Finite check.

We divide into cases.

Case 1: \(h_0 \in E\) and \(f(h_0) \in E\). In this case we are already done thanks to axiom (ii).

Case 2: \(h_0 \in E\) but \(f(h_0) \not\in E\). This is the main case. Let \(H\) be the set of all \(h \in E\) such that \(f(h) = f(h_0)\); this is a finite set containing \(h_0\). Let \(H' \subset H\) be the set of all \(h' \in H\) such that \(-h' \in E\). We make the following observations:

• All elements \(h\) of \(H\) are non-zero. For if \(h=0\) then \(f(h_0)=f(h)=1 \in E\), contradicting the hypothesis \(f(h_0) \not\in E\).

• If \(h' \in H'\) then \(h' \neq h'+f(-h')\). Otherwise we would have \(f(-h')=0\), then by axiom (iii) we have \(h' = -3\), hence \(f(h_0) = f(h') = 4 \in E\) by axiom (i). But this again contradicts the hypothesis \(f(h_0) \not\in E\). (This is the main reason we take \(E_0\) to be so large.)

• If \(h_1 \in H\) and \(h_2 \in H'\) are distinct then \(h_1 \neq h_2+f(-h_2)\). This follows from axiom (iv).

• If \(h_1, h_2\in H'\) are distinct then \(h_1+f(-h_1) \neq h_2+f(-h_2)\). This follows from axiom (v).

From these observations, we see that if we take \(c\) to be a sufficiently large integer, the following claims hold:

• The expressions \(\pm c\), \(\pm (c - h)\) for \(h \in H\), and \(\pm (c-h'-f(-h'))\) for \(h \in H'\) are all distinct from each other.

• \(c\) is not expressible as the sum of four or fewer elements of \(\pm (E \cup f(E))\).

We select such a \(c\). We then promote \(f(h_0)\), \(c - h\) for \(h \in H\), and \(-c+h'+f(-h')\) for \(h' \in H'\), thus setting

We then extend \(f\) to \(f': E' \to \mathbb {Z}\) by setting

for all \(h \in H\) and \(h' \in H'\). Axiom (i) is then obvious. Axiom (ii) needs to be verified for the new elements \(h_0\) and \(c-h'\), \(h' \in H'\) of \(E'\) (the other elements will not obey the hypotheses of this axiom), but this is routine. In particular \(h_0\) obeys the required conclusion of this lemma.

We need to verify (iii) for the new elements of \(E'\), but this is clear from property (a) and claim 2.

Now we need to verify (iv). It suffices to do so when at least one of \(a,a',-a\) are new elements of \(E'\). If neither of \(a\), \(-a\) are new, then the only new elements \(a'\) for which \(f(a')\) could equal \(f(a)\) take the form \(c-h\) (thanks to claim 2), but then \(a'\) cannot equal \(a+f(-a)\), again thanks to claim 2. If instead one of \(a, -a\) is new, then from claim 1 the new element must be \(f(h_0)\). There is no old element \(a'\) with \(f(a') = c = f(f(h_0))\), so we must have \(a = -f(h_0)\), but then \(a + f(-a) = -f(h_0)+c\) will not equal \(a'\), again thanks to claim 2.

Finally, we need verify to (v). We may assume that at least one of \(a,a',-a,-a'\) are new elements of \(E'\). As before, this forces the new element to be \(f(h_0)\), and this cannot be \(a\) or \(a'\), so without loss of generality we have \(a = -f(h_0)\) and \(a' \in E\). But then \(a + f(-a) = -f(h_0)+c\) will not equal \(a' + f(-a')\), again thanks to claim 2.

Case 3: \(h_0 \not\in E\) but \(h_0 \in f(E)\). Here we can write \(h_0 = f(h_1)\) where \(h_1\) is of the form in Case 2. Applying the Case 2 construction, we can pass to an extension in which \(f(h_1) = h_0\) lies in \(E\), and so we are now in either Case 1 or Case 2, and so we can again conclude.

Case 4: \(h_0 \not\in E\) and \(h_0 \not\in f(E)\). We let \(c\) be an integer so large that it is not expressible as the sum of four or fewer elements of \(\pm (E \cup f(E))\). We then promote \(h_0\) to \(E\) by setting

and define an extension \(f': E' \to \mathbb {Z}\) by setting \(f'(h_0) := c\). Axioms (i)-(iii) are obvious. For axiom (iv), since \(f'(h_0)\) is not equal to any other value of \(f'\), the only new case introduced is if \(-a = h_0\), but then \(a+f(-a) = -h_0+c\) is distinct from \(a'\) by choice of \(c\). A similar argument yields axiom (v). With this extension, we are now in Case 2, and so we repeat the previous analysis to conclude.

It suffices to find a partial solution that violates injectivity. This can be done for instance by adjoining \(\{ -13, 10\} \) to \(E_0\) and defining \(f(10)= 2 = f(6)\), \(f(-13)=-10\), and performing a finite check to verify that this is still a partial solution.

We begin with some informal motivation. Writing

we conclude that

and hence on iteration

In particular, \(x \diamond x = f(x) \diamond f(x)\), and

If \(f\) is surjective, this would imply

and hence the four laws stated above would hold in this case.

This motivates the use of a non-surjective \(f\). We will take \(G\) to be the space of polynomials \(\mathbb {Z}[t]\) of one variable with integer coefficients, and let \(f: G \to G\) be the multiplication by \(t\) map:

This is of course non-surjective. The magma operation will be constructed as follows:

• If \(P\) is a polynomial with \(P(0) \neq 0\), then \(t^n P \diamond t^n P = t^n P \diamond t^{n+1} P = 2P\) for all \(n \geq 0\), and \(t^{n+m} P \diamond 2t^n P = 2t^n P \diamond t^{n+m+1} P = t^{m+1} P\) for all \(n, m \geq 0\). (This is well-defined since the \(t^n P\), \(2t^n P\) are all distinct polynomials.)

• We define \(P \diamond Q = 0\) if not covered by the above laws.

It is a routine matter to verify Equation 18 and Equation 19, so that equation 3342 holds. However, one can check that the four laws in the conclusion fail with \(x=y=1\).

A first attempt would be the operation \(i \diamond j := j+1\) on \(\mathbb {Z}/3\mathbb {Z}\); this obeys 1437, but unfortunately also obeys 4269. However, an "extension" of this operation will work. Namely, we take the carrier \(G = \mathbb {N}\times \mathbb {Z}/3\mathbb {Z}\) and define \((a,i) \diamond (b,j)\) to equal

• \((a-1,j+1)\) if \(i=j+2\) and \(a{\gt}0\);

• \((b,j+1)\) if \(i=j+2\) and \(a=0\);

• \((a+1,j+1)\) otherwise.

Note that \((a,i) \diamond (a,i) = (a+1,i+1)\), and that \((b,j) \diamond ((c,k) \diamond (a,i))\) will be of the form \((d,i+2)\) for some \(d\). Hence

giving 1437. On the other hand,

and so equation 4269 fails.