Prove uniqueness and existence of $\theta$ such that $\theta\left(x,0\right)=x$ and $\theta\left(x,y^{\prime}\right)=\theta\left(x,y\right)^{\prime}.$

Question

The domain of consideration is the set of whole numbers, $\mathbb{N}_0$.

The following theorem (see facsimile below) appears in The Number System, by H.A. Thurston:

3. Theorem: There is just one operation $\theta$ such that, for every $x$ and $y,$ \begin{align*} (i)\text{ }\theta\left(x,0\right)=&x\text{ and}\\ (ii)\text{ }\theta\left(x,y^{\prime}\right)=&\theta\left(x,y\right)^{\prime}. \end{align*} Proof: If $\theta$ exists, let $\phi$ be an operation such that \begin{align*} (iii)\text{ }\phi\left(x,0\right)=&x\text{ for every }x\text{ and}\\ (iv)\text{ } \phi\left(x,y^{\prime}\right)=&\phi\left(x,y\right)^{\prime}\text{ for every }x\text{ and }y. \end{align*} Let $M$ be the set of $y$ for which $\phi\left(x,y\right)=\theta\left(x,y\right)$ for every $x.$ Then $0\in M,$ because $\phi\left(x,0\right)=x=\theta\left(x,0\right)$ for every $x,$ by $(iii)$ and $(i).$ \begin{align*} \text{If }y\in M\text{, then }\phi\left(x,y^{\prime}\right)= & \phi\left(x,y\right)^{\prime}\text{ by }(iv)\\ = & \theta\left(x,y\right)^{\prime}\text{ because }y\in M\\ = & \theta\left(x,y^{\prime}\right)\text{ by }(ii). \end{align*} Therefore $y^{\prime}\in M.$ Therefore, by [the principle of induction], $M$ contains all whole numbers, and so $\phi\left(x,y\right)=\theta\left(x,y\right)$ for every $x$ and $y;$ that is, there is at most one operation satisfying $(i)$ and $(ii).$ It remains to prove that there is at least one. Let $M$ be the set of $x$ for which there is, for each $y,$ a whole number $\theta\left(x,y\right)$ such that $(i)$ and $(ii)$ are true. If we let \begin{align*} (v)\text{ }\theta\left(0,y\right)= & y\text{ for every }y,\text{then}\\ \theta\left(0,0\right)= & 0\text{ by }(v). \end{align*} Therefore $(i)$ is true when $x=0.$ And \begin{align*} \theta\left(0,y^{\prime}\right)= & y^{\prime}\text{ by }(v)\\ = & \theta\left(0,y\right)^{\prime}\text{ by }(v). \end{align*} Therefore $(ii)$ is true when $x=0.$ Therefore $0\in M.$ If $z\in M,$let \begin{align*} (vi)\text{ }\theta\left(z^{\prime},y\right)= & \theta\left(z,y\right)^{\prime}\text{ for every }y. \end{align*} ($\theta\left(x,y\right)$ is defined, because $z\in M.$) Then \begin{align*} \theta\left(z^{\prime},0\right)= & \theta\left(z,0\right)^{\prime}\text{ by }(vi)\\ = & z^{\prime}\text{ by }(i) \end{align*} because $z\in M.$ Therefore $(i)$ is true when $x=z^{\prime}.$ And \begin{align*} \theta\left(z^{\prime},y^{\prime}\right)= & \theta\left(z,y^{\prime}\right)^{\prime}\text{ by }(vi)\\ = & \theta\left(z,y\right)^{\prime\prime}\text{ by }(ii)\text{ because }z\in M\\ = & \theta\left(z^{\prime},y\right)^{\prime}\text{ by }(vi). \end{align*} Therefore $(ii)$ is true when $x=z^{\prime},$ and so $z^{\prime}\in M.$ Therefore, by [the principal of induction], $M$ contains every whole number, and so $(i)$ and $(ii)$ are true for every $x$ and $y.$

I contend that to prove the existence of $\theta$ it suffices to show that for arbitrary $x\in\mathbb{N}_0$ the function $\alpha_x\left(y\right)=\theta\left(x,y\right)$ satisfying $(i)$ and $(ii)$ is well defined. So, for all $x,y\in\mathbb{N}_0$ we define \begin{align*} \alpha_x\left(y\right)\in&\mathbb{N}_0,\\ \alpha_x\left(0\right)=&x,\\ \alpha_x\left(y^{\prime}\right)=&\alpha_x\left(y\right)^{\prime}.\\ \end{align*}

The only condition on $x$ is that it is a whole number. So $\alpha_x$ is defined for $x=0$ and since $x\in\mathbb{N}_0$ then $x^\prime\in\mathbb{N}_0.$ So $\alpha_{x^\prime}$ is equally well defined. Similarly,since $y,\alpha_x\left(y\right)\in\mathbb{N}_0$ then $y^\prime,\alpha_x\left(y\right)^\prime\in\mathbb{N}_0.$

Is this sufficient to prove the existence of $\theta$ as defined in Thurston's theorem quoted above?

Answer 1

I reject Thurston's entire proof, including the uniqueness part. Changing the name of the operator proves nothing. His proposed proof of existence shows that $(v),(vi)$ provide equivalent definitions of $\theta.$ That begs the question.

Instead I proceed as follows: $\theta$ can be viewed as a three-place relation of whole numbers. Thus it can be treated as the set

$$\vartheta=\left\{\langle{x,y,z}\rangle\backepsilon z=\theta\left(x,y\right)\right\}\subset\mathbb{N}_0^3$$

generated recursively using $(i),(ii).$ That $\vartheta\neq\mathbb{N}_0^3$ follows from $\langle{0^\prime,0,0}\rangle\notin\vartheta.$

To show that this generation deterministically produces exactly $\vartheta,$ we could use induction on $x$ with each step being induction on $y.$ However, since $(i),(ii)$ are defined identically for all $x,$ there is no need for induction on $x$. Thus we consider the case of an arbitrarily chosen $x.$

Assume $x\in\mathbb{N}_0.$ Rule $(i)$ provides the base case. Assume $y=0.$

$$(i)\land y=0\implies\langle x,y,x\rangle\in\vartheta$$

So $\theta$ is uniquely determined for each $x$ and $y=0.$

Rule $(ii)$ provides the $y\mapsto y^\prime$ induction step. For arbitrary $y\in\mathbb{N}_0$ assume $z=\theta\left(x,y\right)\in\mathbb{N}_0$ is uniquely determined.

$$(ii)\land\langle x,y,z\rangle\in\vartheta\implies\langle x,y^\prime,z^\prime\rangle\in\vartheta$$

Uniqueness and closure follow from the defined uniqueness and closure of the successor of a whole number.

Thus we have shown that for all $x$ we have that $\theta\left(x,y\right)\in\mathbb{N}_0$ is uniquely determined for $y=0,$ and if $\theta\left(x,y\right)\in\mathbb{N}_0$ is uniquely determined for $y\in\mathbb{N}_0$ then $\theta\left(x,y^\prime\right)\in\mathbb{N}_0$ is uniquely determined.

We conclude that $\theta:\mathbb{N}_0\times\mathbb{N}_0\mapsto\mathbb{N}_0$ is well defined. That is, $\theta$ gives exactly one whole number for each pair of whole number arguments.

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To regain Thurston's implicit results that $\theta\left(0,y\right)=y$ and $\theta\left(x^\prime,y\right)=\theta\left(x,y^\prime\right)$ we proceed as follows:

By $(i)$ we have $y=0\implies\theta\left(0,y\right)=y=0.$

By $(ii)$ and the induction hypothesis that for arbitrary $y$ we assume $\theta\left(0,y\right)=y$ we obtain $\theta\left(0,y^\prime\right)=y^\prime.$

So we have produced Thurston's $(v)$

$$y\in\mathbb{N}_0\implies\theta\left(0,y\right)=y.$$

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We now show $$\theta\left(x^\prime,y\right)=\theta\left(x,y^\prime\right).$$

Assume $x$ is arbitrary and $y=0.$ By $(i)$ we have the first two equalities in $$y=0\implies\theta\left(x^\prime,y\right) =x^\prime=\theta\left(x,y\right)^\prime =\theta\left(x,y^\prime\right).$$ The third equality follows from $(ii).$

Assume that for arbitrary $x,y$ we have $$\theta\left(x^\prime,y\right)=\theta\left(x,y^\prime\right).$$

The induction step $y\mapsto y^\prime$ is as follows: Using $(ii)$ to obtain the first and third equalities, and our assumption to obtain the second, we have $$\theta\left(x^\prime,y^\prime\right) =\theta\left(x^\prime,y\right)^\prime =\theta\left(x,y^\prime\right)^\prime =\theta\left(x,y^{\prime\prime}\right).$$

Using $(ii)$ we recover Thurston's $(vi)$

$$\theta\left(x^\prime,y\right)=\theta\left(x,y\right)^\prime.$$