Is this proof of the principle of recursion theorem proving anything?

Question

The following is presented as an example to motivate my question. It's my paraphrase of the principle of recursion theorem and proof. From Fundamentals of Mathematics Foundations of Mathematics: The Real Number System and Algebra

Let $c$ be a number and let $F$ be a function of two arguments defined in $\mathbb{N}$ and with values in $\mathbb{N}.$ Then there exists exactly one function $f$ defined in $\mathbb{N}$ such that \begin{align*} f\left(1\right)= & c;\\ \forall_{x}f\left(x^{\prime}\right)= & F\left(x,f\left(x\right)\right) \end{align*} Proof: Express $f$ as the set $\mathcal{P}$ of ordered pairs $\left\langle x,y\right\rangle $ where $y=f\left(x\right)$ and having only the recursively defined elements given by \begin{align*} \left\langle 1,c\right\rangle \in & \mathcal{P};\\ \left\langle x,y\right\rangle \in & \mathcal{P}\implies\left\langle x^{\prime},F\left(x,y\right)\right\rangle \in\mathcal{P}. \end{align*} We need to prove that for every number $x\in\mathbb{N}$ there is exactly one number $y$ such that $\left\langle x,y\right\rangle \in\mathcal{P}.$ Since $\left\langle 1,c\right\rangle \in\mathcal{P}$ and all other pairs have a successor as the first component, $\left\langle 1,y\right\rangle \in\mathcal{P}\implies\left\langle 1,y\right\rangle =\left\langle 1,c\right\rangle .$ For a specific $x$ assume that $$\left\langle x,y_{1}\right\rangle \in\mathcal{P}\land\left(\left\langle x,y_{2}\right\rangle \in\mathcal{P}\implies y_{1}=y_{2}\right).$$ It follows from the recursion rule that $\left\langle x^{\prime},F\left(x,y_{1}\right)\right\rangle \in\mathcal{P}.$ This means that $$ \left\langle x^{\prime},z_{1}\right\rangle \in\mathcal{P}\land\left\langle x^{\prime},z_{2}\right\rangle \in\mathcal{P}\implies z_{1}=z_{2} $$ because $F$ is a function, and the recursive construction requires $z_{1}=F\left(x,y_{1}\right)$ and $z_{2}=F\left(x,y_{2}\right)$ with $\left\langle x,y_{1}\right\rangle \in\mathcal{P}$ and $\left\langle x,y_{2}\right\rangle \in\mathcal{P}$ but $y_{1}=y_{2}$ by our assumption.

I'm left wondering what the point of that part of the proof is. Suppose we define a function $H$ of two variable arguments as $H\left(x,y\right)=0.$ Then let $h$ be the function corresponding to $f$ above. \begin{align*} h\left(1\right)= & 0,\\ \forall_{x}h\left(x^{\prime}\right)= & H\left(x,h\left(x\right)\right) \end{align*} Set up the recursion: \begin{align*} \left\langle 1,0\right\rangle \in & \mathcal{P}\\ \left\langle x,y\right\rangle \in & \mathcal{P}\implies\left\langle x^{\prime},H\left(x,y\right)\right\rangle \in\mathcal{P} \end{align*}

But that amounts to

\begin{align*} \left\langle 1,0\right\rangle \in & \mathcal{P}\\ \left\langle x,0\right\rangle \in & \mathcal{P}\implies\left\langle x^{\prime},0\right\rangle \in\mathcal{P} \end{align*}

So it seems that what really needs to be shown is that the recursive definition produces exactly one pair for each number with that number as the first component. It doesn't matter what the second component is as long as it's well defined.

The second part of the proof is this:

To show that $f$ is unique, assume with have a function $g$ defined by \begin{align*} g\left(1\right)= & c,\\ \forall_{x}g\left(x^{\prime}\right)= & F\left(x,g\left(x\right)\right) \end{align*} Then \begin{align*} g\left(1\right)= & f\left(1\right)=c,\text{ and }\\ g\left(x\right)= & f\left(x\right),\text{ implies }\\ g\left(x^{\prime}\right)= & F\left(x,g\left(x\right)\right)=F\left(x,f\left(x\right)\right)=f\left(x^{\prime}\right) \end{align*}

This seems completely pointless. All it says is that the letter we used to define the function doesn't matter.

Is there an example showing that either part of the proof does more than I have claimed?

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Added: comment moved to answer.

Answer 1

The original title of my post was

Why is it so hard for mathematicians to say exactly one?

The proof under critique seems to be an elaborate way of saying: given a function $s$ let $a=s\left(x\right)$ and $b=s\left(y\right)$ then $x=y\implies a=b.$ Which is the definition of a function.

The way I see it, what needs to be shown is that for each number $x^\prime$ there is exactly one pair $\langle{x^\prime,f\left(x^\prime\right)}\rangle.$ That is, $f\left(x^\prime\right)=F\left(x,f\left(x\right)\right)$ has exactly one value. It is axiomatic that there is exactly one value of $x$ for each $x^\prime$ value. So we are obligated to show that for each $x$ value there is exactly one value given by $F\left(x,f\left(x\right)\right).$

The notation $\underline{\exists}$ means there exists exactly one. Some authors write $\exists!,$ but as G. H. Hardy said: "Beauty is the first test: there is no permanent place in the world for ugly mathematics."

Base Case of $x=1$

For the exceptional case of $x=1$ we have exactly one number value of $F\left(1,f\left(1\right)\right)=F\left(1,c\right),$ because $F$ is defined to be a function.

$x=1\implies\underline{\exists} F\left(1,f\left(1\right)\right)\in\mathbb{N}$

Induction hypothesis of $x=p^\prime$

Let $x=p^\prime$ and assume there is exactly one value of $f\left(p\right).$ Then $f\left(x\right)=F\left(p,f\left(p\right)\right)$ has exactly one value.

$\left(x=p^\prime\land\underline{\exists}f\left(p\right)\in\mathbb{N} \right)\implies\underline{\exists}f\left(x\right)= F\left(p,f\left(p\right)\right)\in\mathbb{N}$

Induction step from $x$ to $x^\prime$

It follows that $f\left(x^\prime\right)=F\left(x,f\left(x\right)\right)$ has exactly one value.

$\underline{\exists}f\left(x\right)\in\mathbb{N} \implies\underline{\exists}f\left(x^\prime\right)= F\left(x,f\left(x\right)\right)\in\mathbb{N}$

QED

Assume we are using the set $\mathcal{P}$ discussed above.

Let $\left[x\right]$ return the value of the second component of the ordered pair $\langle x,y\rangle.$ We now have $F\left(x,\left[x\right]\right).$ And the second part of the proof amounts to saying we could have used a different kind of bracketing. What is at issue is whether the recursive definition deterministically specifies a set of ordered pairs in which every natural number appears exactly once. The uniqueness of the function referred to as $f$ depends on the values in $\mathcal{P}$ not on the name given to the function. The values in $\mathcal{P}$ depend on the definitions of $F$ and $f$ not on the names they are given. It is not inconsistent for $f$ to appear in both roles. That is the nature of recursion.

Comment: By defining $F$ as a function we are saying: $$\forall_{n}\left\{\underline{\exists}F\left(n,f\left(n\right)\right)\iff \left(n=m\implies F\left(n,f\left(n\right)\right)=F\left(m,f\left(m\right)\right)\right)\right\}.$$

The proof in the original post restates this fact in the form on the right hand side but with an additional level of indirection. That completely obscures the essential point and adds nothing to the argument that isn't given in the recursive definition. It may not be logically incorrect, but it doesn't make sense to do it that way.

Answer 2

To help understnd this proof note that:

(1) The proof is using induction for both the existence and uniqueness steps
(2) The proof is making use of the following argument:

If $<x,y> \in \mathcal{P}$ then $<x,y>$ must have been created from one of the two rules:
a) $<1,c> \in \mathcal{P} $
b) For $x \in \mathbb{N}^+$, $<x,y> \in \mathcal{P} \implies <x',F(x,y)> \in \mathcal{P}$
And so $<x,y> \in \mathcal{P}$ means either $x=1$ and $y=c$ or $ x = x_1'$ and $y=F(x_1,y_1)$ for some $x_1 \in \mathbb{N}^+$ and $<x_1,y_1> \in \mathcal{P}$.

So, to show thet the relation $\mathcal{P}$ is single valued, we use induction.
For the base step, assume $<1,y> \in \mathcal{P}$. Then from (2) either $y=c$ or $1=x_1'$ and $<x_1,y_1> \in \mathcal{P}$. But for the latter case, $<0,y_1> \in \mathcal{P}$. Applying (2) again neither rule applies, a contradiction.
For the inductive step, assume that $\mathcal{P}$ is single valued for $x$. Now suppose $<x',z_1> \in \mathcal{P}$ and $<x',z_2> \in \mathcal{P}$ and $x \gt 1$. Then by (2), $ x' = x_1'$ , $z_1=F(x_1,y_1)$ and $x'=x_2'$ , $z_2=F(x_2,y_2)$ for some $x_1,x_2 \in \mathbb{N}^+$ and $<x_1,y_1> \in \mathcal{P}$ and $<x_2,y_1> \in \mathcal{P}$. But $x'=x_1'=x_2'$ implies $x_1=x_2$ and so, by our IH, $y_1=y_2$. And so as $F$ is a function, $z_1=z_2$. Hence we have shown the case for $x'$ and so by induction, it folllow for all $x \in \mathbb{N}$.

The uniqueness step is a similar induction. We also have to shown the domain of $f$ is $\mathbb{N}$, which I think we can do by induction as well.

Now, what I would like to know, is why this proof (which I claim is the same as the proof in "Fundamentals of Mathematics Foundations of Mathematics: The Real Number System and Algebra") is apparently incorrect..! Most authors use a more sophisticated set theoretic argument (e.g. Enderton, Elements of Set Theory, Recursion theorm on $\omega$). I believe the issue is that step (2) is not justified, however, I would like to know why exactly....

With regards to:

So it seems that what really needs to be shown is that the recursive definition produces exactly one pair for each number with that number as the first component. It doesn't matter what the second component is as long as it's well defined.

Exactly, it is showing that the relation $\mathcal{P}$ is single valued. That is, it is a function.

This seems completely pointless. All it says is that the letter we used to define the function doesn't matter.

It is showing that if there are two functions satisfying the recurrence relation, then they are the same (uniqueness). (I think this needs an explicit inductive argument to be 100% clear).

EDIT: The reason that this proof is incomplete/incorrect is that the author is assuming that there is a relation which satisfies the recurrence equations. He then shows that this relation is a function by showing it is single valued. He hasn't shown that such a relation exists in the first place. That is why the set theoretic version of this proof (e.g. Enderton) is required.

Answer 3

This is really just a too-long comment:

I think it may help to consider cases where the (obvious analogue of the) recursion theorem doesn't hold. Consider something like the following: $$0,1,2,3,...,\infty-2,\infty-1,\infty,\infty+1,\infty+2,...$$ Basically, this is a "$\mathbb{N}+\mathbb{Z}$-shaped" number system. We don't have well-behaved addition or multiplication here, but we do have a successor operation, and that's enough to check whether the recursion theorem holds. And it doesn't: consider the pair of functions $$f_1:x\mapsto\begin{cases} 0 & \mbox{ if $x$ is finite}\\ 1 & \mbox{ otherwise} \\ \end{cases} \quad\mbox{and}\quad f_2: x\mapsto 0.$$ Both of these functions satisfy the rule "$f(0)=0,f(x')=f(x)$," but they are clearly not the same function. Similarly, the rule "$f(0)=0, f(x')=f(x)^2$" has no solutions over all of our new number system.