Why define addition with successor?

Question

I'm reading Russell's Introduction to Mathematical Philosophy Russell defines the sum of two numbers in terms of successors. I don't understand why:

Suppose we wish to define the sum of two numbers. Taking any number $m$, we define $m+0$ as $m$, and $m+(n+1)$ as the successor of $m+n$. In virtue of (5) this gives a definition of the sum of $m$ and $n$, whatever number $n$ may be.

Is this any different than doing $m+ (n+1) = (m+n) +1$ ?

What is the purpose of $+1$? We might as well write $m+n$.

Edit:

Russell cites Peano's 5. Axiom as proof. How does Peano's 5. Axiom (mathematical induction axiom) prove addition here?

The successor function is defined separately from and before addition. — anomaly, Oct 27 '23 at 14:00
This may become clearer if you write "successor of $n$" not as $n+1$ but as $S(n)$, for example. Then $n+1=S(n)$ (i.e., "adding one to a number is the same as taking the successor of the number") is a theorem, and the recursive definition of addtion is by setting $m+0:=m$ and $m+Sn:=S(m+n)$. -- Then in hindsight, this still becomes equivalent to $m+(n+1)=(m+n)+1$, but that's because that is exactly what we want to achieve — Hagen von Eitzen, Oct 27 '23 at 14:04
If you simply write $m+n$ before defining addition, we can't say what it means. — David K, Oct 27 '23 at 14:34
@HagenvonEitzen But it seems to me that "addition" or "+" is hidden in the successor function S. Because, $n+1$ is not a function, it adds 1 to $n$, it does nothing more. S is a function that repeats $n+1$, that is, $S(n+1)=$ "Add 1 to $x$ repeatedly". What am I missing here? — zeynel, Oct 28 '23 at 10:09
@anomaly The successor function is defined by using addition, $S(n)=n+1$ not before addition. Successor function assumes addition by using the "+" sign. And Russell does not use a "successor function" he uses addition as $m+0=m$ Isn't this addition? — zeynel, Nov 08 '23 at 07:23
@DavidK But Russell uses $m+0=m$ before he defines addition. $m+0=m$ is not a definition of addition, it is a statement about identity property of zero by using addition. So Russell assumes addition. And I'm not sure that addition needs definition. Euclid doesn't define addition because it is included in its definition of unit and number. Russell rejects Euclid's definition of number as multitude of units but he is using numbers with units to define additon. — zeynel, Nov 08 '23 at 08:28
The successor function is part of the five Peano axioms mentioned two or three paragraphs above the passage in the OP; it's just something abstract that's taken as part of the definition of numbers. It isn't defining using addition; addition is defined using it. — anomaly, Nov 08 '23 at 12:44
You are misreading Russell. The fact that $m+0=m$ is the first part of the definition of addition. He even uses the word “define”. — David K, Nov 08 '23 at 13:21
@DavidK It can the first part, doesn't matter. Russell uses the "+" sign to define the "+" sign. Did Russell define the "+" before using here? No. — zeynel, Nov 09 '23 at 07:53
@DavidK If the left hand side of $m+0=m$ is the expression to be defined and the RHS is the definition, here Russell defines addition as "$m$". This doesn't make sense. — zeynel, Nov 09 '23 at 07:54
@DavidK The word "define" refers to the definition of the identity property of zero, not to addition.
Russell never defined the "+" sign but he is using it to define the same "+" sign. Russell did not define "=" sign either, but he is using that too. — zeynel, Nov 09 '23 at 07:54
@DavidK Also, Russell is not constructing $\mathbb{N}$ with a "successor function" as modern mathematicians do, he assumes $\mathbb{N}$. But $\mathbb{N}$ is an ordered list of numbers and addition is built into it. Each number is created by adding 1 unit to its predecessor. So, Russell is using the list he built by addition to define addition. But he is trying to hide the fact that he is using addition by using the non-mathematical word "successor." — zeynel, Nov 09 '23 at 07:55
I still think you are misreading Russell. The piece of text you have quoted is a perfectly acceptable step in a definition of addition according to the Peano axioms (aside from some possibly old-fashioned notation). But now that I have looked at the passage in its original context (the parts you did not quote), I think this passage is merely a reminder of what he did in Principia Mathematica, which you must read if you want the actual definition. All this painstaking analysis of this passage out of that context is flensing a flea. — David K, Nov 12 '23 at 19:09
In short, it is true that what Russell presented here is not a complete definition of addition. He never said it was. — David K, Nov 12 '23 at 19:12

Paul Tanenbaum · Answer 1 · 2023-10-31T16:13:49.230

6

Note that he is defining addition. So to avoid circularity he uses recursion, defining $m+(n+1)$ in terms of two things already defined: $m+n$ and the successor operation. So his recursive definition of $m+k$ for any nonnegative integer $k$ is

$m+k=m$ when $k=0$, and
$m+k=\mbox{succ}(m+n)$ when $k>0$, in particular, when $k= n+1$ for any integer $n \geq 0$.

edited Oct 31 '23 at 16:13

answered Oct 27 '23 at 14:25

Paul Tanenbaum

1,450

Thanks. You say, he is defining addition but then you say that he already defined $m+n$ somewhere else. So he must have defined addition when he defined $m+n$. And where was “successor function” defined? To me successor function is the function form of $n+1$ so it also assumes addition. Can you clarify? – zeynel Oct 28 '23 at 10:27
He is in the midst of defining addition, which he does for progressively larger numbers. He starts (in what I numbered as part 1) by defining $m+0$ (to be $m$), then (in part 2) he defines each $m+i$ (whatever positive integer $i$ is) in terms of something smaller, namely $m+(i-1)$. Thus, as one example, $m+3$ gets defined in terms of $m+2$, which in turn is defined in terms of $m+1$, which itself gets defined in terms of $m+0$, which is defined to be $m$. As for the successor function, he’s using that as an axiom. – Paul Tanenbaum Oct 28 '23 at 11:47
It’s like defining the set $P$ of US presidents this way: (1) George Washington is in $P$, and (2) for any $p\in P$, the person who succeeded $p$ is in $P$. In the case of $P$, there are other approaches for defining it, but this recursive version is a valid one. – Paul Tanenbaum Oct 28 '23 at 11:55
Ok. But I still don’t understand. It seems to me that he does not define addition, he assumes addition to define a property of zero. $m+0=m$ is using addition to define a property of zero. Addition is assumed not defined. Where did he define the $+$ sign? What am I missing? – zeynel Oct 30 '23 at 06:12
The example about presidents clarifies the process but how is it related to defining addition of two numbers? And it seems that the definition of “recursion” in math is different than its definition in computer science. I have to look it up. I don’t see a function applied to itself in the presidents example. – zeynel Oct 30 '23 at 06:18
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He’s defining addition out of nothing but the successor function and a thing he calls 0. To understand it, you have to forget that addition exists and forget your intuitions about numbers. This 0 thing of his doesn’t bring with it any properties; its properties emerge from the definitions. So to begin, 0 has a successor, $\mbox{succ}(0)$, which also has a successor of its own, which is $\mbox{succ}(\mbox{succ}(0))$, and so on. And none has any properties that don’t emerge from the definitions. But he calls 0 and everything derived from it by (repeated) use of succ by the name numbers. – Paul Tanenbaum Oct 30 '23 at 09:31
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And here’s a definition of something you can do with pairs of his numbers. You can do what he calls adding them. If you add any such beast, call it $m$, and 0, you get that original beast $m$. But what if you add your beast $m$ and some beast other than 0? Well, anything other than 0 must be if the form $\mbox{succ}(b)$ for some other beast $b$. And he defines adding of $m$ and this $\mbox{succ}(b)$ to result in the successor of what you get from adding $m$ and $b$. Now do you see the recursion? It’s the same as you know from comp. sci. – Paul Tanenbaum Oct 30 '23 at 09:41
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And note that this proposition that adding any $m$ and 0 gives you $m$ is the first part of his definition of what he’s going to mean by this thing he’s calling addition. In particular, he is not, as you put it, “assuming” anything at all. And he uses the symbol + to represent his addition activity that he’s defining. I guess he should have made that notational fact explicit. – Paul Tanenbaum Oct 30 '23 at 09:51
Thanks for this detailed explanation. Now I understand better what Russell does but I still have questions. I'm not sure if I should open a new question. – zeynel Oct 31 '23 at 11:18
The first question is about your Succession function "succ". (I think this is the function written as $S$ mentioned in my other question.)
When you say $S(0)$ you are writing an instance of the function $S$, you are not defining it.

Can you write the definition of $S()$ explicitly?

To me this function can only be: $S(x) = x+1$. (Copying Russell who wrote "repeated addition of 1..."

I read this as: "The successor function adds the unit 1 to the last number $x$ in the natural order of the ordered list of numbers $\mathbb{N}$."

Is this correct? Do you define $S()$ differently?
– zeynel Oct 31 '23 at 11:19
As a clarification: Since (or if) $S(x) = x+1$ I still don't see how "addition" is not assumed. Clearly Russell assumes addition but wants to define it anyway. This is confusing. – zeynel Oct 31 '23 at 11:22
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It seems that your confusion is arising from the intuition that one should think of succ as a function $S:{\bf N}\rightarrow{\bf N}$. That intuition is understandable, but fallacious and misleading. After all, it is precisely the natural numbers, the set ${\bf N}={0, 1, 2,\ldots}$, that we are defining, along with addition. So here’s his (two-part recursive) definition of $\bf N$: “(1) there is a thing I’m calling 0, which is in $\bf N$, and (2) for any $n\in N$, there is another thing I’m calling the successor of $n$, which is also in $\bf N$.” – Paul Tanenbaum Oct 31 '23 at 15:38
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At that point, successor has no meaning whatsoever beyond serving as a mechanism for generating one so-called number from another. There does not (yet) exist any sense of order that we associate automatically with the everyday meaning of successor, no idea of one number’s following another. He could just as easily have used the label banana. – Paul Tanenbaum Oct 31 '23 at 15:43
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Once we have defined addition it becomes possible to define an order for these beasts we’ve defined. For any $n,m\in \bf N$—so each constructed from 0 and succ—we define the proposition $n\leq m$ to mean that there exists some $d\in\bf N$ such that $n+d=m$. – Paul Tanenbaum Oct 31 '23 at 15:50
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I fear I had furthered the confusion by referring to successor as a function (I did that several times). It’s better to think of it as a unary operator, or in terms perhaps more familiar to computer scientists, a constructor. – Paul Tanenbaum Oct 31 '23 at 16:18
I was thinking that Russell was "defining" addition. This doesn't make sense to me. Would it be better if I think about it as "deriving" addition from Peano's two initial assumptions? So, is it better to say that Russell is not defining a new type of addition but he is deriving Euclidean addition from "zero" and "successor"? His goal in this book is to show that mathematics is based on propositions of logic. – zeynel Nov 04 '23 at 05:31
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It’s not “a new type of addition” he’s defining. He’s defining the nonnegative integers and the addition operation on them. You could (probably) think of it as deriving them—depending on what you understand by the word derive. But what do you mean by “Euclidean addition”? Why drag him into it? Are you drawing an analogy to his axiomatic approach (to geometry, not arithmetic)? – Paul Tanenbaum Nov 04 '23 at 10:46
In any event, he’s responding to modern mathematicians’ expectations of logical rigor. We do not wish to try to build elaborate structures on unsound foundations. I don’t know if Isaac Newton ever stopped and seriously considered a precise definition of addition, let alone, say, seven. But if one doesn’t have such basics carefully established, one can wander off unwittingly into serious logical trouble. – Paul Tanenbaum Nov 04 '23 at 10:51
Thanks again for all your help. But I still don't see how Russell defines addition. I'm looking at this sentence from your comment above: "And he defines adding of and this S() to result in the successor of what you get from adding and ..." (Changed your succ(b) to S(b) for simplicity.) – zeynel Nov 05 '23 at 09:12
Here Russell states the associativity principle of addition. He does not define addition, he assumes it. Sum of $m$ and $S(b)$ = $m + b + 1$. And the successor of the sum of $m$ and $b = m + b + 1$. And the result is, $m+b+1 = m+b+1$. This is a tautogical statement. Where is the definition of addition in this statement? I don't get it. What am I missing? – zeynel Nov 05 '23 at 09:13
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Consider my comment above on 28 October at 11:47: “As one example, m+3 gets defined in terms of m+2, which in turn is defined in terms of m+1, which itself gets defined in terms of m+0, which is defined to be m.” So here’s exactly what’s going on. As he starts defining addition, the only things he has at hand are 0, S(0), S(S(0)), S(S(S(0))), and the like. There is nothing else at all in his universe but these beasts, which he calls numbers. – Paul Tanenbaum Nov 05 '23 at 10:35
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Note that these numbers have none of the behaviors or properties that you associate with your accumulated sense of what numbers are. Those properties and behaviors will eventually arise from the definitions he’s going to make. But for now a number is neither more nor less than either the meaningless beast 0 or a meaningless beast that has the form S(b), where $b$ is some other number. – Paul Tanenbaum Nov 05 '23 at 10:41
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Now, since every number is the result of applying one or the other of the two constructors, he defines addition in terms of them as well. As I wrote in my example, he’s going to define $m+k$ for any numbers $m$ and $k$. What do $m$ and $k$ look like? Each looks like S(S(0)) or S(0) or 0, or like S(S(S(S(S(S(S(S(0)))))))); you get the picture. And as for $m$, we needn’t concern ourselves with what we might call its “internal structure,” in other words with which number it is. – Paul Tanenbaum Nov 05 '23 at 10:51
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As for $k$, though, his approach does care. When he defines what it means to add any number $m$ and this $k$, he formulates that definition to handle each constructor explicitly. First, if $k$ was constructed using the constructor “0 is a number,” then he defines $m+k$ to be $m$. The only number $k$ for which this part of his definition is relevant is $k=0$. So what he’s just done is to define $m+0$ to be $m$. And for every other number $k$, for all the nonzero numbers, he relies on the fact that each of them has been constructed in terms of succession. – Paul Tanenbaum Nov 05 '23 at 11:00
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Any nonzero number $k$ is S(b) for some other number $b$. And he defines $m+k$, which is to say $m+\mbox{S}(b)$ to be $\mbox{S}(m+b)$. Thus, as one example, if $k$ is S(S(S(0)), then $m+k$ works out to be S(S(S(m))), which you can get by three successive applications of the second half of his definition of addition followed by one application of the first half of the definition. – Paul Tanenbaum Nov 05 '23 at 11:12
That’s recursion exactly as in computer science. – Paul Tanenbaum Nov 05 '23 at 11:13
Again, $m+\mbox{S}(\mbox{S}(\mbox{S}(0)))$ is defined as S($m+\mbox{S}(\mbox{S}(0)))$, which is defined as S(S($m+ \mbox{S}(0)))$, which is defined as S(S(S($m+0)))$, which is defined as S(S(S(m))). – Paul Tanenbaum Nov 05 '23 at 11:23
And now that he has the numbers and addition defined, it may at last be safe to introduce shorthand nicknames for all of those successors. So here goes: he defines “1” to mean S(0), and “2” to mean S(S(0)), and so on. So your quote all the way back in your OP, where he writes, “we define $m+0$ as $m$, and $m+(n+1)$ as the successor of $m+n$,” could be expressed as “we define $m+0$ as $m$, and $m+\mbox{S}(n)$ as S($m+n$).” Maybe if he had used the rewritten version you’d never have gotten confused to begin with. – Paul Tanenbaum Nov 05 '23 at 11:42
Thanks again. I read all your comments carefully I have many questions but it seems that this sentence is the sentence describing Russell's definition of addition: "And he defines +, which is to say +S() to be S(+)." But how can this be a definition because it simply says that $m+k=m+k$. Russell's definition of addition: "The sum of $m,k$ is defined as the sum of $m,k$". This is simple arithmetic: $k=b+1$ and $m+S(m+b) = m+b+1$, therefore $m+k=m+k$. First I thought he is defining addition as counting along the number line, but he is not even doing that. . – zeynel Nov 06 '23 at 13:39
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No, it doesn’t “simply [say] that $m+k=m+k$.” How about if you think of his defining an infinite set of unary operations we could call addition-of-$k$, one such operation for each of his numbers $k$. And recall that these $k$ are simply constructs of 0 and possibly some applications of S. Also for any of these beasts (any one of his numbers) $b$, let’s agree to call S(b) more complex than b, and conversely, let’s call b less complicated or simpler than S(b). These descriptions reflect nothing more than the syntactic relationship between the two beasts – Paul Tanenbaum Nov 06 '23 at 14:00
With those preliminaries taken care of, we can see that Russell is defining one unary operator, namely addition-of-$k$ for some particular $k$, as follows: If $k=0$, then addition-of-$k$ is the identity operation, in other words, for any $m$, $\mbox{addition-of-$k$}(m)=m$. And if $k\neq 0$, then $k$ must be S(b) for some $b$, and in that case, $\mbox{addition-of-$k$}(m)$ is defined to be S(addition-of-$b$($m$)). – Paul Tanenbaum Nov 06 '23 at 14:16
Thus once he’s past $k=0$, he is always defining addition-of-$k$ in terms of a bunch of addition-of-$t$’s where each of those $t$ is simpler than the one in whose definition it’s being used, and all of those $t$ are simpler than $k$. So there is no circularity. He defines each more complex addition-of in terms of simpler ones. – Paul Tanenbaum Nov 06 '23 at 14:18

MJD · Answer 2 · 2023-10-30T09:59:34.207

The point here is that we don't have to assume that the numbers come to us already imbued with an additive structure. All we need is for the numbers to have the much simpler notion of “successor”:

There is a first number.
Each number has a following number.

From this tiny amount of structure, we can construct the whole idea of arithmetic addition. Any group of things with those two properties must include a thing that behaves like $2$, and a thing that behaves like $4$, and that do $2+2 =4$. It must also have $197 + 313 = 510$. You get all this just from those two properties.

Russell is confusing you by using the standard notation with “$0$” and “$n+1$” for these abstract ideas (“first number” and “number following $n$”) without showing that they really match up. He's combined two steps that should be separate.

First, any system with a “first number” and “each number has a following number” can be shown to have a notion something like addition, based just on those two simpler ideas
And if we do that with the ordinary numbers, taking the “first number” to be $0$, and “the number following $n$” to be $n+1$, then the addition-like thing we get is actually regular addition

Why do we do this at all? Because it allow us to understand addition as a result of simpler properties. For example, we expect that addition should the property that $n+m $ always equals $m+n$. We know this is true for ordinary addition of numbers. But is is a consequence of same the tiny bit of structure? Or does it come from somewhere else? We can show that actually no more properties are needed: any group of things with those two properties not only has something like addition, but that something is commutative.

In contrast, though: ordinary numbers have a “$<$” relation and it's always true that $a < a+b+1$. Can we get that from the two simple properties above? It turns out we can't, we need something extra! Isn't that interesting?

Another way this turns out to be useful is that it lets us model addition in systems that are very unlike ordinary numbers. For example, suppose we are studying computation, and we have a simplified model of a computer. We would like to show that this model can perform all the computations that real computers can, such as additions. If we can show that the model can represent zero, and successors, and manipulate them in a few simple ways, Russell's construction gives a recipe for how the model can do addition.

In the comment you asked:

All I see is that Russell assumes addition when he uses n+1. Calling n+1 “successor” or “number following n” does change the fact that addition is assumed. Can you explain how addition follows from your two initial assumption?

Let's undo the confusion by using different notation. Let's write $Z$ for the first number and $S(n)$ for the number after $n$. Notice that $S$ is not addition, it is simpler and more limited than addition. Addition combines any two numbers. $S$ just tells you the number that comes after any single number.

We define $a+b$ this way: Either $b$ is the first number, or it comes after some other number.

If $b$ is the first number, $Z$, then $a+b = a$.
If $b$ comes after some other number $p$, so that $b=S(p)$, then $a+b = S(a+p)$.

If we define “$+$” this way, we can prove that it has all the properties we expect addition to have. For example: $0+n = n+0$; $p+(q+r) = (p+q) +r$; $2+2=4$; $197+313=510$.

Notice that Russell has actually given this definition, but has confused you by using “$0$” to mean the first number and “$n+1$” to mean the number after $n$. By writing $n+1$ for the number after $n$, Russell has made it look like he is using addition to define addition. But he isn't, he is using the simpler and more limited operation of finding the next number after $n$.

“ First, any system with a “first number” and “each number has a following number” can be shown to have a notion something like addition, based just on those two simpler ideas.” Where is this shown? All I see is that Russell assumes addition when he uses $n+1$. Calling $n+1$ “successor” or “number following $n$” does change the fact that addition is assumed. Can you explain how addition follows from your two initial assumption? — zeynel, Oct 30 '23 at 06:37
“ Russell is confusing you by using the standard notation with “0” and “n+1” for these abstract ideas (“first number” and “number following n”)…” Yes, this is true. I didn’t realize the first number can be any number. — zeynel, Oct 30 '23 at 06:39
Thanks for this answer. I'm still studying it but I still don't see how "addition" follows from "first number" and "successor". — zeynel, Nov 02 '23 at 07:46
My question is very simple: Is $n+1$ the arithmetic operation we call "addition". Yes. Does "successor" mean $n+1$? Yes. Then, Russell already knows addition. His "successor" is nothing but another word for "addition." — zeynel, Nov 02 '23 at 07:47
Russell's so-called definition of addition (as quoted in my question) is nothing but a tautology: $(m+n)+1 = m+(n+1)$. Since addition is associative the parens are unnecessary, so Russell simply defines addition as $m+n+1 = m+n+1$. I don't see a definition here. If this defines anything it is adding two numbers by walking along the number line. That's nothing new, it's counting with fingers. Russell describes counting and calls it a new definition of addition. — zeynel, Nov 02 '23 at 07:47
No, it isn't a tautology. Russell isn't stupid. If you think he is it's because you don't understand the purpose of what is being done here. — MJD, Nov 02 '23 at 11:44
Yes, you may be right. I think there is a misunderstanding on my part about the terminology used. I thought Russell was "defining" addiditon. But in fact, what he is doing is to "derive" addition from Peano's two initial assumptions. Is this correct? — zeynel, Nov 03 '23 at 06:01
@MJD "any group of things with those two properties has commutative addition" is not true: ordinals have those two properties but not commutative addition. Try to find the property that you missed. ;) — Veky, Dec 31 '23 at 18:26

score 0 · Answer 3 · answered Oct 27 '23 at 14:41

It would be beneficial for you to delve into the field of Set Theory, particularly focusing on Transfinite Induction and Transfinite Recursion. In Set Theory, we can define $\alpha+1=\alpha\cup\{\alpha\}$, and arithmetic operations are defined on a unique class of sets known as Ordinals. We only have the definition of $\alpha+1$, and addition is represented as a mapping from $\mathrm{Ord}^2$ to $\mathrm{Ord}$, or more simply, from $\mathbb{N}^2$ to $\mathbb{N}$. Transfinite Recursion provides a systematic and comprehensive way to define such mappings.

Transfinite Recursion is particularly tailored for Ordinals. If your focus is solely on arithmetic within $\mathbb{N}$, then Mathematical Induction is sufficient, and you might refer to it as Mathematical Recursion.

In summary, what we require is a rigorous and general method for defining a mapping through recursion. For instance, a strict definition of a function on $\mathbb{N}$ (or on $\mathrm{Ord}$) with the following holds: $$f(0)=a_0, \text{ and } f(n+1)=F(f(n)).$$ (For the $\mathrm{Ord}$ case, you can refer to any standard Set Theory textbook.)

We are so accustomed to arithmetic within $\mathbb{N}$ that we seldom question the existence of such mappings. However, their existence is guaranteed through certain forms of induction. This is the mechanism of recursion. To gain a better understanding of induction and recursion, in my opinion, a better way is to study ordinal arithmetic, a fundamental and essential concept in Set Theory. Because Ordinals are abstract enough to necessitate a departure from our intuition and prior experiences, compelling us to contemplate the underlying general principles.

Why define addition with successor?

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