Doubt: Proof of existence of the summation of natural numbers in Landau

Question

With reference to the Definition of Addition in Landau's Foundations of Analysis,

the author, in proving the existence of a natural number $(x+y)$, takes for granted that $x' + y = (x+y)'$ where $x'$is defined below. This seems to me a use of the Commutative law which has not been proven yet. And in the proof of Commutativity there is a reference to this particular statement which I can't help but think is a circular argument. Please correct me if I am wrong. This is what I arrived at as a way out of it. But it does pose a small issue.

Requirement: To prove that $(x+y) \in \bf{\mathbb{N}}; \quad \forall{x,y\in \bf{\mathbb{N}}} $

If $ y = 1,$ then $(x+1) = x' \in \mathbb{N} $ since $x\in \mathbb{N}$ by Axiom 2(see below). If not by Theorem 3, $\exists {y_1} \in \mathbb{N}$ such that $y = y_1'$ and $(x+y)=(x+ y_1') = (x+y_1)'$

If $y_1 = 1$ then $(x+y) = (x+1)' = (x')' \in\mathbb{N} $ since the successor of the successor of $x$ must be a natural number since $x$ is one.

Continuing in this fashion for some $n$ we have that $y_n = 1$ and $(x +y) = (((x +y_n)')'...)'= (((x')')'...)' \in \mathbb{N} $

But is $n$ finite or countable? And so how long is the process? Will that be a problem? Is it possible to think that as soon as $y$ is prescribed the ensuing process is finite. Please comment. Any help is appreciated.

Answer 1

This is a subtle point, and your question is definitely a fair one, but the argument is not circular. The key passage is the one which states (it's important to quote verbatim):

Let $x$ belong to $\mathfrak{M}$, so that there exists an $x+y$ for all $y$. Then the number $$x' + y = (x+y)'$$ is the required number for $x'$, since [...]

The equation above is not an assertion about two existing quantities, so it isn't assuming anything about commutativity. At this point in the proof, $x'+y$ has not even been defined yet, so this line should be taken as a definition of $x'+y$ for any $y$. The remainder of the paragraph justifies this definition by showing that if we define $x'+y$ in this way, then it indeed satisfies the two addition axioms for that particular $x'$ and arbitrary $y$.

As for your alternative argument, the main problem is that it appeals to poorly-defined concepts like "eventually" and "$\cdots$". I'm afraid these go against the very aim of the book which tries to lay precise foundations using first-order logic and a minimal number of arithmetic axioms. To fully achieve this goal, it's important to express such concepts rigorously in terms of the axiom of induction. The fact that it's unclear to you whether $n$ is finite is a good indication that you haven't extracted a rigorous definition of $n$.

The proof in the book escapes this trap by not attempting to define $x+y$ by repeated application of the successor operation. Instead, it shows that $1+y$ can be defined, and from that $2+y$ can be defined, and by the process of induction $x+y$ can be defined for all $x \in \mathbb N$. This isn't strictly necessary — it probably would have been more obvious to show for each $x$ that the set of $y$ for which $x+y$ is defined is all of $\mathbb N$ (this seems the natural result of formalizing the argument you gave in the question). But I'd guess that doing it in the way the book did streamlines the proof of commutativity in Theorem 6.

Answer 2

Commutativity still needs to be proven; it is not the case that $x'+y=(x+y)'=x+y'$ as a matter of definition, and certainly not the case that $x+y=y+x$ simply as a matter of definition.

But you can show $x+y=y+x$ to be true for all $(x,y)$ by induction on $x$ and $y$. Clearly it is true for $x=y=1$. Now assume it's true for $(x,1)$ for some $x$. Then $$ 1+x'=(1+x)'=(x+1)'=x''=x'+1, $$ so it's true for $(x',1)$. By induction on $x$, we now know it's true for $(x,1)$ for all $x$. Now assume it's true for $(x,y)$ for all $x$ and some particular $y$. Then $$ 1+y'=(1+y)'=(y+1)'=y''=y'+1, $$ so it's true for $(1,y')$. Now, in addition, assume it's true for $(x,y')$ for some $x$ (and the same fixed $y$). Then $$ x'+y'=(x'+y)'=(y+x')'=(y+x)''=(x+y)''=(x+y')'=(y'+x)'=y'+x'; $$ so it's true for $(x',y')$. We conclude that it's true for $(x,y')$ for all $x$ (by induction on $x$), and therefore it's true for $(x,y)$ for all $x$ and $y$ (by induction on $y$). $\square$

Once commutativity has been proven, of course, $$ x'+y=y+x'=(y+x)'=(x+y)'=x+y' $$ is trivial.