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This question is regarding the same proposition as is discussed in How should this proof of the associativity of natural number addition be understood?

Here I wish to ask if a much simpler proof than that provided by the book would suffice. We begin with the same starting point as in the other question:

A previously demonstrated result is:

\begin{equation} \left(a+b\right)+c=a+\left(b+c\right). \tag{1} \end{equation}

The following is quoted from The Fundamentals of Mathematics, Volume 1:

By (1) we may therefore omit the parentheses in a sum of three terms. In order to be able to omit them in sums with more than three terms, we first define the expression $\sum_{i=1}^{n}a_{i}$ for a given sequence $\left\{ a_{i}\right\} _{i=1,2,\dots}$ of numbers $a_{i}$ recursively by setting

\begin{equation} \sum_{i=1}^{1}a_{i}=a_{1}, \sum_{i=1}^{n+1}a_{i}=\sum_{i=1}^{n}a_{i}+a_{n+1};\tag{2} \end{equation}

[$\dots$] In particular, we have $\sum_{i=1}^{3}a_{i}=\left(a_{1}+a_{2}\right)+a_{3}=a_{1}+a_{2}+a_{3}$ and $\sum_{i=1}^{4}a_{i}=\left(a_{1}+a_{2}+a_{3}\right)+a_{4},$ for which we again naturally write $=a_{1}+a_{2}+a_{3}+a_{4}.$

[$\dots$]

Can we not simply proceed as follows?

Assume

$S_n=\sum_{i=1}^{n}a_{i}=a_{1}+a_{2}+\dots+a_{n} \text{, and }S_n\in{\mathbb{N}}.$

Then argue

$S_{n+1}=\sum_{i=1}^{n+1}a_{i}=S_n+a_{n+1}=a_{1}+a_{2}+\dots+a_{n+1}.$

Or even more simply by omitting the $\sum$ notation entirely, and argue that the result of the $n^{th}$ iteration is a natural number $S_n$ to which we apply the "we again naturally write" transformation in the $n+1^{th}$ step.

Will this not work? If not, why not?

It may be the case that the author's of the book chose to do things the hard way to illustrate new concepts.


To address the specific question in the comments: what about $\left(a+b\right)+\left(c+d\right)$? I came up with this using the notation I invented to work on this problem:

Define the set of function:

$$ \Phi=\left\{ \varphi_{a\in\mathbb{N}}:\mathbb{N}\to\mathbb{N}\backepsilon\varphi_{a}\left[x^{\prime}\right]\equiv\varphi_{a}\left[x\right]^{\prime}\land\varphi_{a}\left[1\right]\equiv a^{\prime}\right\} . $$

To relate this to conventional notation we also define

$$ \left[\_+\right]:\mathbb{N}\to\Phi $$

so that

$$ \left[a+\_\right]\equiv\varphi_{a}, $$

to be interpreted as $$ a+b:\mathbb{N}\times\mathbb{N}\to\mathbb{N}\text{ where }a+b\equiv\varphi_{a}\left[b\right]. $$

$$ \left(a+b\right)+c=\varphi_{a}\left[b\right]+c=\varphi_{a+b}\left[c\right]=a+b+c. $$

$$ \left(a+b\right)+\left(c+d\right)=\varphi_{a}\left[b\right]+\varphi_{c}\left[d\right] $$

$$ =\varphi_{a+b}\left[\varphi_{c}\left[d\right]\right]=\varphi_{a+b+c}\left[d\right]=a+b+c+d. $$

I'm claiming to have justified treating every thing subscripted to $\varphi$ as a single number.

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    $\begingroup$ The authors of the book are at pains to carefully define what $\sum_{i=1}^na_i$ denotes; you are not. $\endgroup$ – Lord Shark the Unknown Jan 9 at 9:16
  • $\begingroup$ Does you proof cover something like $$(a+b)+(c+d)=a+(b+c)+d=a+b+c+d$$ ? $\endgroup$ – Peter Jan 9 at 9:17
  • $\begingroup$ @Peter That's something that has me puzzled. For natural numbers, their sums will always be natural numbers. So I'm wondering why I can't just replace $(a+b)$ by a natural number called $a+b$. $\endgroup$ – Steven Thomas Hatton Jan 9 at 9:48
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Writing the addition without parentheses is a statement that we have the full power of associativity - that the sum will be the same no matter what we add first. If we don't have that theorem yet, then it's simply a more confusing shorthand for a particular standard order (in your referenced material, from left to right).

Your argument doesn't prove anything. All you have written is a definition. You've addressed how to go from $S_n$ to $S_{n+1}$ by adding one term, but not what happens if we add things in a different order so that it's some previous sum $S_k$ plus a sum of the last $n+1-k$ terms at the outermost layer of parentheses. Of course it will be the same - but we have to actually prove it, and it's not something that will work in one step from the associativity. There's a reason the proof in the book uses strong (modified, in the book's terminology) induction.

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  • $\begingroup$ Superb answer ! $\endgroup$ – Peter Jan 9 at 9:30
  • $\begingroup$ I've pretty much figured out what the authors expect. Now I need to figure out what it really means. $\endgroup$ – Steven Thomas Hatton Jan 9 at 14:45

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