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\begin{center}
Notes and Comments on the Addition Theorem
\end{center}

\normalsize
\begin{center}
by Leonard Blackburn
\end{center}

\paragraph{} 
The proof of the Addition Theorem given in the Spring '02 3283W class
notes entitled ``The Peano Postulates, version 4'' has a fundamental
error. The Theorem is correct. The proof uses a maneuver that is correct,
but that cannot be used without proof. These Notes and Comments explain
why the argument is incomplete.  A correct proof, using a
set--theory/Peano theorem called The Recursion Theorem, begins on
page 3.
\paragraph{The Addition Theorem.}
There exists exactly one function $p:\mathbb{N} \times \mathbb{N}
\rightarrow
\mathbb{N}$ such that

\begin{tabular}{rl}
(A) & $(\forall \, n \in \mathbb{N})\,[p(n,0) = n]$;\\
(B) & $(\forall \, n \in \mathbb{N})(\forall \, m \in \mathbb{N})
      \,[p(n,s(m)) = s(p(n,m))]$.
\end{tabular}

\paragraph{Incorrect Proof of Addition Theorem (Outline).}
First, suppose that $p:\mathbb{N} \times \mathbb{N} \rightarrow
\mathbb{N}$ and $q:\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$
satisfy (A) and (B).  Let $n_0 \in \mathbb{N}$. Let
$S := \{m \in \mathbb{N} : p(n_0,m) = q(n_0,m)\}$.  Using induction,
one easily shows that $S = \mathbb{N}$.  Hence $p = q$.  Therefore,
there is at most one function $p:\mathbb{N} \times \mathbb{N} \rightarrow
\mathbb{N}$ that satisfies (A) and (B).

Now, define $E$ to be the set of all $n \in \mathbb{N}$ such that there
exists
a function $p_n:\mathbb{N} \rightarrow \mathbb{N}$ such that

\begin{tabular}{rl}
(i) & $p_n(0) = n$, and\\
(ii) & $(\forall \, m \in \mathbb{N})\,[p_n(s(m)) = s(p_n(m))]$.
\end{tabular}

Now, we will use induction to show that $E = \mathbb{N}$.  Define
$$p_0(m) := m$$
for all $m \in \mathbb{N}$.  Then, one easily checks
that $p_0$ satisfies (i) and (ii).  Hence, $0 \in E$.  Now, suppose
$n \in E$.  Then define
$$p_{s(n)}(m) := s(p_n(m))$$
for all $m \in \mathbb{N}$.  Again, one easily checks that $p_{s(n)}$
satisfies (i) and (ii).  Hence, $s(n) \in E$.  Therefore, by induction,
$E = \mathbb{N}$.  Now we simply define
$$p(n,m) := p_n(m)$$
for all natural numbers $n$ and $m$.  Then (A) and (B) clearly hold.
\hfill $\square$

\paragraph{Remarks.}
First of all, the proof that there is at most one function
$p:\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ that satisfies
(A) and (B) is entirely correct.  There are two main errors in the
existence part of the proof.  One of these errors is easily understood
and easily fixed.  Let's discuss that one first.\\
\hspace*{.25in}When we are trying to prove by induction  that $E =
\mathbb{N}$,
we correctly prove that $0 \in E$.  Then we assume $n \in E$.  This means
that there exists \textit{at least one} function $p_n$ satisfying (i) and
(ii).
Therefore our definition of $p_{s(n)}$ in terms of $p_n$ is ambiguous
(perhaps
there is more than one function $p_n$ satisfying (i) and (ii)---which
are we using?).  It turns out, actually, that for all
$n \in \mathbb{N}$ there is \textit{exactly one} function $p_n$ satisfying
(i) and (ii).  Once we prove this, then we have solved this difficulty.
This is not hard to do.  However, this is all irrelevant since the
second error in the proof invalidates the entire proof, including the
proof that $E = \mathbb{N}$ (even after fixed as noted above).\\
\hspace*{.25in}The second error is much more serious, and much harder to
explain.  In fact, the error is fatal, i.e. we won't be able to salvage
the main structure of our proof.  We will have to scrap the proof and
start
again.  The problem is that the author of this proof has failed to
understand
that \textit{definition} by induction is not the same as \textit{proof}
by induction.  Let me explain.  We all know what a proof by induction is.
We begin with some set, say $S$, and we prove that $S = \mathbb{N}$
by showing that $0 \in S$,
and that for all natural numbers $n$, whenever $n \in S$, then $s(n) \in
S$.
What is a \textit{definition} by induction?  A definition by induction is
a means of defining a sequence of objects (sets), $A_0, A_1, A_2,
\ldots$, i.e. of defining a function $f$ whose domain is the set of
natural
numbers (or some other index set) such that for all natural numbers $n$,
$f(n) = A_n$.  We \textit{define} this sequence by first explicitly
defining
$A_0$.  Then, for an arbitrary $n \in \mathbb{N}$, we define $A_{s(n)}$ in
terms of $A_{n}$.  Thus, our definition takes the following form:

\begin{tabular}{rcl}
$A_0$ & $:=$ & $B$\\
$A_{s(n)}$ & $:=$ & $F(A_n)$.
\end{tabular}

In what sense is this a \textit{definition} of $A_0, A_1, A_2, \ldots$?
The fact that such a sequence exists is not an axiom of set theory, nor
is it a Peano Postulate.  Thus we are required to prove that this sort
of definition by induction is allowed.  Let me now explain where the
above incorrect proof of the Addition Theorem assumes that definition
by induction is OK.  The set $E$ is defined to be the set of all
$n \in \mathbb{N}$ such that there exists (``exists exactly one,''
perhaps)
a function $p_n: \mathbb{N} \rightarrow \mathbb{N}$ that has certain
given properties (these properties are irrelevant for the present
discussion).  Now, the definition of this set $E$ is perfectly OK.
It is a well-defined set.  In fact, it equals $\mathbb{N}$ (this can be
proven, but not in the manner of the above incorrect proof).  Since
$E$ is a subset of $\mathbb{N}$, it makes sense to try to use induction
to show that $E = \mathbb{N}$.  Here is the structure of the incorrect
induction proof given above:  We first showed that $0 \in E$ (this is
fine) by constructing a particular function (set) $p_0$.  Then we assumed
$n \in E$, i.e., we assumed that a certain $p_n$ existed, and we proceeded
to define a $p_{s(n)}$ in terms of $p_n$.  In other words, we defined,
by \textit{induction}, a sequence of functions (sets)
$$p_0, p_1, p_2, p_3, \ldots,$$
in order to simultaneously prove that $E = \mathbb{N}$.  The definition
of our sequence took the following form:

\begin{tabular}{rcl}
$p_0$ & $:=$ & $B$\\
$p_{s(n)}$ & $:=$ & $F(p_n)$
\end{tabular}

In fact, notice how the ``definition''
of the sequence $p_0, p_1, p_2, \ldots$ looks very much like the
``definition'' of one of the individual functions $p_n$.
If we allow ourselves to define the sequence $p_0, p_1, p_2, ...$ in this
way, then why did we bother proving that the functions $p_n$ exist in the
first place, since they are defined inductively in this very same way
($p_n(0) := n, p_n(s(m)) := s(p_n(m))$)!

If this doesn't convince the reader that the above proof of the
Addition Theorem is flawed, let us point out the exact point where
an ``incorrect'' statement was made.  It occurred exactly when we
said:  ``Then define $p_{s(n)}(m) := s(p_n(m))$''.  This definition is
simply not allowed (and this would be painfully obvious if we were working
entirely within \textit{axiomatic} set theory).
We cannot define the set
$p_{s(n)}$ in terms of a set $p_n$ which we don't know exists in the
first place.  I know some readers would object to this by saying that
we are, in the course of our proof by induction, \textit{assuming} that
the set $p_n$ exists.  But this is just an attempt to cleverly disguise
a definition by induction.  Readers who are still not convinced are
invited to read  Leon Henkin's paper
``On Mathematical Induction,''  in which Henkin emphasizes how definition
by induction works.  The paper can be found in the American Mathematical
Monthly, vol. 67 (1960), pp. 323-338.  It may also be helpful to consult
two books by Herbert Enderton, one of which is \textit{A Mathematical
Introduction to Logic}, in which definition by induction is discussed
early in the book.  The other book's title escapes me.  It is a book
on set theory and it is on reserve in the mathematics library.


\paragraph{Further Remarks.}
As further evidence that the first proof of the Addition Theorem is
incorrect, take note of the fact that the proof does not use the
postulates that $0 \notin S(\mathbb{N})$ and that $s$ is one-to-one (the
proof of the Recursion Theorem does use these postulates), and as Henkin
points out in his paper, the recursion theorem is in general false
for systems $(N,0,s)$ that satisfy the induction postulate but do not
satisfy the postulates that $0 \notin s(\mathbb{N})$ and $s$ is
one-to-one.
\pagebreak
\paragraph{A correct Proof of the Addition Theorem.}  Now
we do not assume the validity of definition by induction in order
to prove that a certain definition by induction is valid.  The proof
begins with the Recursion Theorem as a Lemma.  The proof below closely
follows that given by Enderton in his set theory book [2], which closely
follows the proof given by Henkin [1] in the paper mentioned above.

\paragraph{The Recursion Theorem.}  Let $A$ be any nonempty set, $a \in
A$, and suppose that
$F:A \rightarrow A$ is a function.  Then there exists exactly one function
$h:
\mathbb{N}
\rightarrow A$ such that

\begin{tabular}{rl}
(1) & $h(0) = a$;\\
(2) & $h(s(n)) = F(h(n))$ for all $n \in \mathbb{N}$.
\end{tabular}

\paragraph{Proof of The Recursion Theorem.}
First, suppose that $g:\mathbb{N} \rightarrow A$ and $h:\mathbb{N}
\rightarrow
A$ both satisfy (1) and (2).  Let $S := \{n \in \mathbb{N} : g(n) =
h(n)\}$.
Note that $g(0) = a = h(0)$, and so $0 \in S$.  Suppose that $n \in S$.
Then $g(s(n)) = F(g(n)) = F(h(n)) = h(s(n))$, and so $s(n) \in S$.
Therefore
by mathematical induction, $S = \mathbb{N}$.  Therefore, $g = h$, and so,
there is at most one function $h:\mathbb{N} \rightarrow A$ that satisfies
(1) and (2).

Now, we prove that there is a function $h:\mathbb{N} \rightarrow A$
that satisfies (1) and (2).  Call a function $f$ \textit{acceptable}
iff $dom \, f \subseteq \mathbb{N}$, $ran \, f \subseteq A$, and
if the following conditions hold:

\begin{tabular}{rl}
(i) & If $0 \in dom \, f$, then $f(0) = a$;\\
(ii) & If $s(n) \in dom \, f$ (where $n \in \mathbb{N}$), then
       $n \in dom \, f$ and $f(s(n)) = F(f(n))$.
\end{tabular}

Let $\mathcal{H}$ be the collection of all acceptable functions, and let
$h = \bigcup \mathcal{H}$ (i.e., $h$, as a set of ordered pairs, is the
union of all acceptable functions, each of which is a set of ordered
pairs).
Thus,

\begin{tabular}{lrcl}
(iii)$\,\,\,\,\,\,\,\,$ & $(n,y) \in h$ & iff & $(n,y)$ is a member of
  some acceptable $f$\\
& & iff & $f(n) = y$ for some acceptable $f$.
\end{tabular}

We claim that $h$ meets the demands of the theorem.  We break the proof
into
three parts, which involve showing that (I) $h$ is a function, (II) $h$ is
acceptable, and (III) $dom \, h$ is all of $\mathbb{N}$.

(I)  Define $S := \{n \in \mathbb{N} : \textup{for at most
one } y, (n,y) \in h\}$.
We use induction to show that $S = \mathbb{N}$.  Suppose
$(0,y_1) \in h$ and $(0,y_2) \in h$.  By (iii), there exist acceptable
$f_1$
and $f_2$ such that $f_1(0) = y_1$ and $f_2(0) = y_2$.  But by (i), it
follows
that $y_1 = a = y_2$.  Thus $0 \in S$.  Next suppose that $n \in S$.
Suppose
that $(s(n),y_1) \in h$ and $(s(n),y_2) \in h$.  Again there must exist
acceptable $f_1$ and $f_2$ such that $f_1(s(n)) = y_1$ and $f_2(s(n)) =
y_2$.
By (ii) it follows that $y_1 = f_1(s(n)) = F(f_1(n))$ and
$y_2 = f_2(s(n)) = F(f_2(n))$.  But since $n \in S$, we have $f_1(n) =
f_2(n)$
(since $(n,f_1(n))$ and $(n,f_2(n))$ are in $h$).  Therefore,
$y_1 = F(f_1(n)) = F(f_2(n)) = y_2$.  So, $s(n) \in S$, and hence,
$S = \mathbb{N}$.  Consequently, $h$ is a function.

(II)  First, it is clear from (iii) that $dom \, h \subseteq \mathbb{N}$
and
$ran \, h \subseteq A$.  If $0 \in h$, then there must be some acceptable
$f$ with $f(0) = h(0)$.  Since $f(0) = a$, we have $h(0) = a$.  So, (i)
holds
for $h$.  Next, assume $s(n) \in dom \, h$.  There must be some acceptable
$f$ with $f(s(n)) = h(s(n))$.  Since $f$ is acceptable we have $n \in
dom \, f$ (and $f(n) = h(n)$) and $h(s(n)) = f(s(n)) = F(f(n)) = F(h(n))$.
Thus $h$ satisfies (ii) and so it is acceptable.

(III)  We use induction to show that $dom \, h = \mathbb{N}$.  The
function
$\{(0,a)\}$ is acceptable and hence $0 \in dom \, h$.  Suppose that
$k \in dom \, h$.  We will show that $s(k) \in dom \, h$.  Define
$f := h \cup \{(s(k),F(h(k)))\}$.  Then $f$ is a function,
$dom \, f \subseteq \mathbb{N}$, and $ran \, f \subseteq A$.  We will show
that $f$ is acceptable.  Condition (i) holds since $f(0) = h(0) = a$.
For condition (ii) there are two cases.  If $s(n) \in dom \, f$ where
$s(n) \neq s(k)$, then $s(n) \in dom \, h$ and $f(s(n)) = h(s(n)) =
F(h(n)) = F(f(n))$.  The other case occurs if $s(n) = s(k)$.  Then $n =
k$,
since $s$ is one-to-one.  By assumption, $k \in dom \, h$.  Thus
$f(s(k)) = F(h(k)) = F(f(k))$ and (ii) holds.  Hence $f$ is acceptable.
Therefore $f \subseteq h$, so that $s(k) \in dom \, h$.  Therefore
$dom \, h = \mathbb{N}$.
\hfill $\square$

\paragraph{First Correct Proof of the Addition Theorem.}
The proof that there is at most one function $p:\mathbb{N} \times
\mathbb{N} \rightarrow \mathbb{N}$ that satisfies (A) and (B) is the
same as above. We just prove the existence part here. The argument is
Enderton's

Now, by the Recursion Theorem, for all $n \in \mathbb{N}$, there exists
exactly one function $p_n:\mathbb{N} \rightarrow \mathbb{N}$ such that
$p_n(0) = n$ and $p_n(s(m)) = s(p_n(m))$ for all $m \in \mathbb{N}$.
(We are using $A = \mathbb{N}$ and $F = s$.)  So, define
$p(n,m) = p_n(m)$ for all natural numbers $n$ and $m$.  It is clear
that this $p$ satisfies (A) and (B).  \hfill $\square$

\paragraph{Second Correct Proof of the Addition Theorem.}
Again, we prove the existence part. This argumentm is mine. Define

\begin{tabular}{rl}
$p :=$ & $\{((n,m),k) \in (\mathbb{N} \times \mathbb{N}) \times \mathbb{N}
:
(\exists f \in \mathbb{N}^{\mathbb{N}})[f(0) = n \,\,\wedge$\\
& $(\forall l \in \mathbb{N})(f(s(l)) = s(f(l))) \wedge
f(m) = k]\}$
\end{tabular}

\textit{Proof that p is a function}:  Let $n \in \mathbb{N}$ and
$m \in \mathbb{N}$.  By the Recursion Theorem we can let $f$ be the
unique function such that $f(0) = n$ and $f(s(l)) = s(f(l))$ for all $l
\in
\mathbb{N}$.  Therefore, $((n,m),f(m)) \in p$.  This shows that $p(n,m)$
is defined for all natural numbers $n$ and $m$.  Now suppose that
$((n,m),k_1) \in p$ and $((n,m),k_2) \in p$.  Then, by the
definition of $p$ there are functions
$f \in \mathbb{N}^{\mathbb{N}}$ and $g \in \mathbb{N}^{\mathbb{N}}$ such
that $f(0) = n$, $(\forall l \in \mathbb{N})(f(s(l)) = s(f(l)))$, $f(m) =
k_1$,
$g(0) = n$, $(\forall l \in \mathbb{N})(g(s(l)) = s(g(l)))$,
and $g(m) = k_2$.  But by the uniqueness clause in the Recursion Theorem,
$f = g$.  Therefore $k_1 = k_2$, and so $p$ is a function.

\textit{Proof that p satisfies (A)}:  Let $n \in \mathbb{N}$.  By the
Recursion Theorem, there is a unique function $f:\mathbb{N} \rightarrow
\mathbb{N}$ such that $f(0) = n$ and $f(s(l)) = s(f(l))$ for all $l \in
\mathbb{N}$.  For this $f$, $f(0) = n$.  Thus, $((n,0),n) \in p$, i.e.,
$p(n,0) = n$.

\textit{Proof that p satisfies (B)}:  Let $n \in \mathbb{N}$ and $m \in
\mathbb{N}$.  Again, by the Recursion Theorem, we can let $f$ be the
unique function from $\mathbb{N}$ to $\mathbb{N}$ such that
$f(0) = n$ and
$f(s(l)) = s(f(l))$ for all $l \in \mathbb{N}$.  Therefore,
$((n,m),f(m)) \in p$, i.e. $p(n,m) = f(m)$.  Thus, we need to show that
$p(n,s(m)) = s(f(m))$, i.e. that $((n,s(m)),s(f(m))) \in p$.  But this
is immediate by the definition of $p$ since $f(s(m)) = s(f(m))$.
$\square$

{\bf References}

[1] Enderton, H., {\bf An Introduction
To Set Theory}

[2] Henkin, Leon, 
{\bf On Mathematical Induction,} 
American Mathematical Monthly, vol. 67 (1960), pp. 323-338.

\paragraph{Further Comments on the Recursion Theorem:}
The recursion theorem has many generalizations
that are useful.  For example,  we can
define functions recursively or ``by induction,'' that satisfy

	$f(0) = a,\ $
	$f(s(n)) = g(f(n),n)$

instead of just

	$f(0) = a,\ $
	$f(s(n)) = g(f(n)).$

Also, there are recursion theorems when the domain isn't
the natural numbers.  The domain may be a set generated
from some ``initial elements'' by several functions
(In the case of the natural numbers, 0 is the only initial
element and the successor function s is the only function
that generates all of the natural numbers).  For
generalizations like this, see Enderton's
A Mathematical Introduction to Logic.

There is a more general form of the recursion theorem
for ordinals (transfinite recursion) that can be found
in any graduate-level book on Zermelo-Fraenkel set
theory.

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