that for any $x, z_1, z_2 \in \mathbb{N}^#$, if $(x, z_1) \in \mathcal{F}$ and $(x, z_2) \in \mathcal{F}$, then $z_1 = z_2$.

We shall prove this by **hyper infinite induction** (1) on $x$. Let

(5) $A = \{x | x \in \mathbb{N}^# \text{ and for all } z_1, z_2 \in \mathbb{N}^#, \text{ if } (x, z_1) \in \mathcal{F} \text{ and } (x, z_2) \in \mathcal{F} \\ \text{then } z_1 = z_2\}$.

We shall show $A = \mathbb{N}^#$ by applying hyper infinite induction (1). First we have (6) $1 \in A$.

To prove (6), it suffices to show that for any $z$, if $(1, z) \in \mathcal{F}$ then $z = c$.

We prove this by contradiction; in other words, suppose to the contrary that there is some $z$ with $(1, z) \in \mathcal{F}$ but $z \neq c$. Consider the relation $W = \mathcal{F}\{(1, z)\}$. Since $(1, c) \in \mathcal{F}$ and $(1, c) \neq (1, z)$, it follows that $(1, c) \in W$. Moreover, whenever $(u, y) \in W$ then $(u, y) \in \mathcal{F}$ and hence $(\mathbf{Sc}(u), G(y)) \in \mathcal{F}$ but $\mathbf{Sc}(u) \neq 1$, so $(\mathbf{Sc}(u), G(y)) \neq (1, z)$, and hence $(\mathbf{Sc}(u), G(y)) \in W$. Thus $W$ satisfies both conditions (ii') and (iii'); in other words, $W \in \mathbf{M}$. But then it follows from (2) that $\mathcal{F} \subseteq W$ however this is clearly false since $(1, z) \in \mathcal{F}$ and $(1, z) \notin W$. Thus our hypothesis has led us to a contradiction, and hence (6) is proved. Next we show that

(7) for any $x \in \mathbb{N}^#$ if $x \in A$ then $\mathbf{Sc}(x) \in A$.

Suppose that $x \in A$, so that whenever $(x, z_1) \in \mathcal{F}$ and $(x, z_2) \in \mathcal{F}$ then $z_1 = z_2$. We must show that whenever $(\mathbf{Sc}(x), w_1) \in \mathcal{F}$ and $(\mathbf{Sc}(x), w_2) \in \mathcal{F}$ then $w_1 = w_2$. To prove this, it suffices to show that

(8) whenever $(\mathbf{Sc}(x), w) \in \mathcal{F}$ then there exists a $z$ with $w = G(z)$ and $(x, z) \in \mathcal{F}$. For if (8) is true, we would have for the given $w_1, w_2$ some $z_1 = z_2$ with $w_1 = G(z_1)$, $w_2 = G(z_2)$, $(x, z_1) \in \mathcal{F}$ and $(x, z_2) \in \mathcal{F}$. Then, since $x \in A$, $z_1 = z_2$ and hence $G(z_1) = G(z_2)$, that is, $w_1 = w_2$.

Now to prove (8) suppose, to the contrary, that it is not true; in other words, suppose that we have some $w$ with $(\mathbf{Sc}(x), w) \in \mathcal{F}$ but such that for all $z$ which $(x, z) \in \mathcal{F}$ we have $w \neq G(z)$. Consider the relation $W = \mathcal{F}\{(\mathbf{Sc}(x), w)\}$.

We shall show that $W \in \mathbf{M}$. First of all $(1, c) \in \mathcal{F}$ and $(1, c) \neq (\mathbf{Sc}(x), w)$; hence $(1, c) \in W$. Suppose tbat $(u, y) \in W$; then $(u, y) \in \mathcal{F}$ and $(\mathbf{Sc}(u), G(y)) \in \mathcal{F}$.

Clearly if $u \neq x$ then $(\mathbf{Sc}(u), G(y)) \neq (\mathbf{Sc}(x), w)$, so that in this case $(\mathbf{Sc}(u), G(y)) \in W$. On the other hand, if $u = x$ and $(\mathbf{Sc}(u), G(y)) = (\mathbf{Sc}(x), w)$, then $w = G(y)$, where $(x, y) \in \mathcal{F}$, contrary to the choice of $w$ henee $(\mathbf{Sc}(u), G(y)) \neq (\mathbf{Sc}(x), w)$, so again $(\mathbf{Sc}(u), G(y)) \in W$. Thus whenever $(u, y) \in W$, also $(\mathbf{Sc}(u), G(y)) \in W$. Now that we have shown $W \in \mathbf{M}$ we see by (2) that $\mathcal{F} \subseteq W$ but this is false since $(\mathbf{Sc}(x), w) \in \mathcal{F}$ and $(\mathbf{Sc}(x), w) \notin W$. Thus our hypothesis that (8) is incorrect has led to a contradiction, and now (8) is proved. Since (7) follows from (8), we have by **hyper infinite induction** from (6) that $A = \mathbb{N}^#$. Hence

(9) $\mathcal{F}$ is a function.

We have still to prove that $\mathcal{F}$ satisfies, condition (i); we must show that for each $x \in \mathbb{N}^#$ there is a $y$ with $(x, y) \in \mathcal{F}$. Since $\mathcal{F} \subseteq \mathbb{N}^# \times S$, it will then follow that $\text{dom}(\mathcal{F}) = \mathbb{N}^#$ and $\text{range}(\mathcal{F}) \subseteq S$. Let $B = \text{dom}(\mathcal{F})$, that is, (10) $B = \{x | x \in \mathbb{N}^# \text{ and for some } y, (x, y) \in \mathcal{F}\}$.

We prove now by **hyper infinite induction** that $B = \mathbb{N}^#$. First, $1 \in B$, sínce $(1, c) \in \mathcal{F}$ by (3). Next, if $x \in B$, pick some $y$ with $(x, y) \in \mathcal{F}$; then by (4), $(\mathbf{Sc}(x), G(y)) \in \mathcal{F}$, and henee $\mathbf{Sc}(x) \in B$.

Thus concludes the first part of the proof, that there is at least one function $\mathcal{F}$ satisfying conditions (i)-(iii).
**Part 2.** We prove that there cannot be more than one such function.
Suppose that $\mathcal{F}_1$ and $\mathcal{F}_2$ both satisfy the conditions (i)-(iii) we wish to show
$\mathcal{F}_1 = \mathcal{F}_2$, i.e., that for all $x \in \mathbb{N}^{\#}, \mathcal{F}_1(x) = \mathcal{F}_2(x)$. Thus
is proved by **hyper infinite induction** on $X$. By (ii), $\mathcal{F}_1(1) = c$ and $\mathcal{F}_2(1) = c$, so
$\mathcal{F}_1(1) = \mathcal{F}_2(1)$. Suppose that $\mathcal{F}_1(x) = \mathcal{F}_2(x)$; then $\mathcal{F}_1(\mathbf{Sc}(x)) = G(\mathcal{F}_1(x))$
and $\mathcal{F}_2(\mathbf{Sc}(x)) = G(\mathcal{F}_2(x))$, so $\mathcal{F}_1(\mathbf{Sc}(x)) = \mathcal{F}_2(\mathbf{Sc}(x))$.

**Theorem 2.** Let $S$ be a set, $c \in S$ and $G: S \times \mathbb{N}^{\#} \to S$ is a binary function with $\text{dom}(G) = S \times \mathbb{N}^{\#}$ and $\text{range}(G) \subseteq S$.

Then there exists a function $\mathcal{F}: \mathbb{N}^{\#} \to S$ such that:

(i) $\text{dom}(\mathcal{F}) = \mathbb{N}^{\#}$ and $\text{range}(\mathcal{F}) \subseteq S$;

(ii) $\mathcal{F}(1) = c$;

(iii) for all $x \in \mathbb{N}^{\#}, \mathcal{F}(\mathbf{Sc}(x)) = G(\mathcal{F}(x), x)$.

We omit the proof of the Theorem 2 since it can be given by simple modification of the proof to Theorem 1.

**Appendix C. General associatíve and commutative laws.**

**Definition 1.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^{\#} \mathbb{N}$ be a hyperfinite sequence of elements of $\mathbb{R}_c^{\#}$.
Then $\text{Ext}-\sum_{k=m}^{n} x_k$ and $\text{Ext}-\prod_{k=m}^{n} x_k$ are defined for any $n, m \in \mathbb{N}^{\#}$ by the recursions

(i) $\text{Ext}-\sum_{k=m}^{n} x_k = 0$ and $\text{Ext}-\prod_{k=m}^{n} x_k = 1$ if $n < m$;

(ii) $\text{Ext}-\sum_{k=m}^{n} x_k = \left( \text{Ext}-\sum_{k=m}^{n-1} x_k \right) + x_n$ and

(iii) $\text{Ext}-\prod_{k=m}^{n} x_k = x_n \times \left( \text{Ext}-\prod_{k=m}^{n} x_k \right)$ if $m < n$.

The condition (ii) of the above definition is justified by recursive definition, see
Appendix B.

**Definition 2.** Let $\langle x_1, \dots, x_j, \dots \rangle, j \in \mathbb{N}$ be a countable sequence of elements of $\mathbb{R}_c^\#$.
Then $\omega$-sum $\text{Ext}-\sum_{j=m}^\omega x_k$ and $\omega$-product $\text{Ext}-\prod_{j=m}^\omega x_k$ are defined for any $m \in \mathbb{N}$ by

(iv) $\text{Ext}-\sum_{j=m}^{\omega} x_j \triangleq \text{Ext}-\sum_{j=m}^{n} y_j$, where $\langle y_1, \dots, y_j, \dots, y_n \rangle, n \in \mathbb{N}^{\#} \mathbb{N}$ is a hyperfinite sequence
such that $x_j = y_j$ for all $j \in \mathbb{N}$ and $y_j = 0$ for all $j \in \mathbb{N}^{\#} \mathbb{N}$;

(v) $\text{Ext}-\prod_{j=m}^{\omega} x_j \triangleq \text{Ext}-\prod_{j=m}^{\omega} y_j$, where $\langle y_1, \dots, y_j, \dots, y_n \rangle, n \in \mathbb{N}^{\#} \mathbb{N}$ is a hyperfinite sequence
such that $x_j = y_j$ for all $j \in \mathbb{N}$ and $y_j = 1$ for all $j \in \mathbb{N}^{\#} \mathbb{N}$.

**Theorem 1.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^{\#} \mathbb{N}$ be a hyperfinite sequence of elements of $\mathbb{R}_c^\#$.
Then we have

$$ \text{Ext}-\sum_{k=m}^{n} x_k = \text{Ext}-\sum_{k=m}^{n-m+q} x_{k+m-q} \quad (1) $$

and
$$z \times \left( \mathrm{Ext} - \sum_{k=m}^{n} x_k \right) = \mathrm{Ext} - \sum_{k=m}^{n} z \times x_k, \qquad (2)$$

$z \in \mathbb{R}_c^#.$

Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^\# \mathbb{N}$ be an hyperfinite sequence of elements of $\mathbb{R}_c^#$. Consider now any infinite nonnegative integers $n_1, n_2, \dots, n_i, \dots, n_t, n_i \in \mathbb{N}^\# \mathbb{N}, 1 \le i \le t$, and set

$$n = n_1 + n_2 + \dots + n_t. \qquad (3)$$

Given $x_1, \dots, x_n$, we can group these as:

$$x_1, \dots, x_{n_1}; \ x_{n_1+1}, \dots, x_{n_1+n_2}; \ x_{n_1+n_2+1}, \dots, x_{n_1+n_2+n_3}; \dots x_{n_1+n_2+\dots+n_{i+1}}, \dots, x_{n_1+n_2+\dots+n_{i+1}}; \dots \quad (4)$$

Here, if $n_i = 0$, the corresponding subsequence is regarded as being empty.

**Theorem 1.** Let $\langle x_1, \dots, x_k, \dots \rangle$ be an hyperinfinite sequence of elements of $\mathbb{R}_c^#$.

Let $\langle n_1, \dots, n_t \rangle$ be a sequence of nonnegative integers. For each $i = 1, \dots, t \in \mathbb{N}^#,$

let $m_i = \sum_{j=1}^{i-1} n_j$ and let $n = m_t + n_t$. Then

$$\mathrm{Ext} - \sum_{k=1}^{n} x_k = \sum_{i=1}^{t} \left( \mathrm{Ext} - \sum_{k=1}^{n_i} x_{m_i + k} \right) \qquad (5)$$

and

$$\mathrm{Ext} - \prod_{k=1}^{n} x_k = \prod_{i=1}^{t} \left( \mathrm{Ext} - \prod_{k=1}^{n_i} x_{m_i + k} \right). \qquad (6)$$

**Definition 3.** A function $F$ is said to be a permutation of a set $S$ if it is one-to-one and $\mathrm{dom}(F) = \mathrm{range}(F) = S$.

**Definition 4.** Let $[1, n]$ a set $\{k | k \in \mathbb{N}^# \wedge (1 \le k \le n)\}$

**Theorem 2.** Let $\langle x_1, \dots, x_n \rangle, n \in \mathbb{N}^\# \mathbb{N}$ be an hyperfinite external sequence of elements of $\mathbb{R}_c^#$. Then for any $n \in \mathbb{N}^#$ and any permutation $F$ of $[1, n]$ following holds

$$\mathrm{Ext} - \sum_{k=1}^{n} x_k = \mathrm{Ext} - \sum_{k=1}^{n} x_{F(k)}. \qquad (\quad)$$

The same holds if we replace $\mathrm{Ext} - \sum$ by $\mathrm{Ext} - \prod$.

**Proof.** The proof is by hyper infinite induction on $n \in \mathbb{N}^#$. For $n = 1$ it is trivial.

Suppose that it is true for $n$. Let $G$ be a permutation of $[1, n+1]$. Then $G(m) = n+1$ for a unique $m$, such that $1 \le m \le n+1$. Then by Eq.()

$$\mathrm{Ext} - \sum_{k=1}^{n+1} x_{G(k)} = \mathrm{Ext} - \sum_{k=1}^{m-1} x_{G(k)} + x_{n+1} + \mathrm{Ext} - \sum_{k=m+1}^{n+1} x_{G(k)} \quad (\quad)$$

and by Eq.()

$$\mathrm{Ext} - \sum_{k=1}^{m-1} x_{G(k)} + x_{n+1} + \mathrm{Ext} - \sum_{k=m+1}^{n+1} x_{G(k)} = \mathrm{Ext} - \sum_{k=1}^{m-1} x_{G(k)} + \mathrm{Ext} - \sum_{k=m}^{n} x_{G(k+1)} + x_{n+1}. \quad (\quad)$$

Thus by Eq.()-Eq.() we obtain
$$ Ext-\sum_{k=1}^{n+1} x_{G(k)} = Ext-\sum_{k=1}^{m-1} x_{G(k)} + Ext-\sum_{k=m}^{n} x_{G(k+1)} + x_{n+1}. \quad (\quad) $$

To reduce this to the inductive hypothesis, we wish to rewrite the external sum of the first

two terms as $Ext-\sum_{k=1}^{n} x_{F(k)}$ for suitable $F$. Define $F$ by

$$ F(k) = \begin{cases} G(k) & \text{if } 1 \le k < m \\ G(k+1) & \text{if } m \le k \le n \end{cases} \quad (\quad) $$

Since all values of $G(k)$ for $k \ne m$, we have for all $k \le n$

$$ 1 \le F(k) \le n \quad (\quad) $$

Now we claim that

$F$ is a permutation of $[1, n]$. \quad (\quad)

By (2), (3) we need only check that $F$ is one-to-one. Suppose that $F(k_1) = F(k_2)$. If both $k_1, k_2$ are $< m$ or both are $\ge m$, it follows from (2) and the fact that $G$ is a permutation that $k_1 = k_2$. If, say, $k_1 < m \le k_2$, we have $G(k_1) = G(k_2 + 1)$, hence $k_1 = k_2 + 1$, which contradicts our assumption. Thus neither this case nor, by symmetry, the case $k_2 < m \le k_1$ can occur. We have from (1) and (2) that

$$ Ext-\sum_{k=1}^{m+1} x_{G(k)} = Ext-\sum_{k=1}^{m-1} x_{F(k)} + Ext-\sum_{k=m}^{n} x_{F(k)} + x_{n+1} = Ext-\sum_{k=1}^{n} x_{F(k)} + x_{n+1} \quad (\quad) $$

by (4) and inductive hypothesis

$$ Ext-\sum_{k=1}^{n} x_{F(k)} + x_{n+1} = Ext-\sum_{k=1}^{n} x_{k} + x_{n+1} = Ext-\sum_{k=1}^{n+1} x_{k} \quad (\quad) $$

This equality completes the inductive step and hence the proof of the theorem.