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This is a follow up question to my previous question: Why define addition with successor?

In this one I'd like to ask about Russell's use of Peano's 5. Axiom to prove his definition of addition:

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.

I follow T. Tao's definition of the 5. Axiom as stated here: Fifth Peano axiom — Properties of the natural numbers

In Russell's case the property $P$ is the identity property of zero: $P(0) \equiv 0 + m = m$

Here Russell is not testing the relation P as true or false for zero but he defines the identity property of zero by using addition symbol "+".

But $P(n)$ and $P(n+1)$ are not true because neither $n$ nor $n+1$ has the identity property.

Since $P(n)$ and $P(n+1)$ are not true, the conditions for the 5. Axiom are not satisfied.

What's wrong?

zeynel
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    This is just induction: let's use $S(n)$ rather than $n++$ to avoid too many $+$s when discussing addition. For any $m \in \mathbb N$, $m+0=m$ so $m+n$ is defined for $n=0$, i.e. $P(0)$. If $m+n$ is defined for $n=k$, i.e. $P(n)$, then consider $m+S(k) = S(m+k)$, which is defined, so $m+n$ is defined for $n=S(k)$, i.e. $P(S(n))$. Using the axiom, $m+n$ is therefore defined for all $m$ and $n \in \mathbb N$. – Henry Nov 09 '23 at 10:55
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    A good companion to Russell's book can be Friedrich Waismann, Introduction to mathematical thinking: the formation of concepts in modern mathematics (1951) – Mauro ALLEGRANZA Nov 09 '23 at 11:17
  • @MauroALLEGRANZA Great book, very helpful, thanks. – zeynel Nov 09 '23 at 11:44
  • You are welcome :-) – Mauro ALLEGRANZA Nov 09 '23 at 11:48
  • @Henry What is the relation $P$ here? Looking at $P(0) = m+0$ I think $P$ refers to the identity property of zero. – zeynel Nov 09 '23 at 13:29
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    @zeynel $P(0)$ is "$m+0$ is defined" here for any given $m$, which we know since "we define $m+0$ as $m$". $P(n)$ is "$m+n$ is defined" for any given $m$, demonstrated using the axiom, i.e. induction over $n$. – Henry Nov 09 '23 at 13:49
  • @Henry Sorry, I still do not understand what $P$ is. What is the explicit definition of the property $P$ that we assume is true for zero, for $n$ and for $S(n)$? If $P$ is the identity property of zero as stated with $m+0=m$, then this property is not true for $n$ and for $S(n)$. Can you be more explicit? – zeynel Nov 12 '23 at 08:10
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    @zeynel $P(n)$ is saying that for a given natural number $m$ and any natural number $n$, $m+n$ is a natural number. $P(0)$ is true because the definition of $+$ says that $m+0=m$ which is a natural number, while $P(n)$ is true using the induction in the axiom. You can go further and show that $m+n$ changes when $n$ changes and that $0+n=n$. – Henry Nov 12 '23 at 10:46
  • @Henry Thanks. That was helpful. Do I understand correctly that $P(n)$ is a theorem that says $P(n) = m + n \in \mathbb{N}$ ? Can this be proved simply with the 2. Axiom: $P_2 = n+1 \in \mathbb{N}$ without using the 5. Axiom? Because, $(m+n) + 1 \in \mathbb{N}$ by $P_2$. Is this correct? Why do we have to use induction? – zeynel Nov 12 '23 at 19:16
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    @zeynel You want to show addition $m+n$ makes sense with this definition. Once you have done that, you can show it has other properties you want, such as $0+n=n$ and more generally $m+n=n+m$. – Henry Nov 13 '23 at 09:05
  • @Henry I guess by "this definition" you mean $m+0=m$ which you called "the definition of +". To me, $m+0=m$ cannot be a definition of "​+". A definition of "​+" should be something like: "​+ := {definition of the addition operator "​+​"} Russell is not defining addition, he is using addition to define the identity property of zero. – zeynel Nov 13 '23 at 14:19
  • @Henry And I don't understand how one can show that $m+n$ makes sense. Russell proposes to define $m+n$ then he rewrites the same expression $m+n$ for the special case $n=0$ as $m+0=m$. He never defined "+" but he uses it in his definitions. He proposes to define "​+" but he uses "​+" in his definitions. – zeynel Nov 13 '23 at 14:20
  • @Henry And his definitions are identities. I don't know what new information can be obtained from $m+n+1 = m+n+1$ . – zeynel Nov 13 '23 at 14:21
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    @zeynel $m+0=m$ is part of the definition you quote for $+$, while $0+n=n$ is a theorem which can be proved. Since $+$ is a binary operator, $m+n+1$ is not meaningful, at least until you have $m+(n+1)=(m+n)+1$ as an axiom or theorem – Henry Nov 13 '23 at 14:21
  • @Henry In your comment yesterday https://math.stackexchange.com/questions/4803539/is-russells-proof-of-addition-with-peanos-5-axiom-valid?noredirect=1#comment10221082_4803539 did we define or prove addition? No. We proved what $P(n)$ states, namely, $m+n \in \mathbb{N}$. We assumed addition to prove that $m+n \in \mathbb{N}$. – zeynel Nov 13 '23 at 14:43

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We have to start from the definition of sum in terms of $0$ and sucecssor:

$m+0=m$

and:

$m+s(n)=s(m+n)$,

and then we have to use induction to prove that the definition "works", i.e. "for every natural number $m$ there is natural number $n$ satisfying the definition and that this number is unique".

The relevant property $P(m)$ will be: $\forall y \exists z(m+y=z)$.

First step, to prove it for $0$, i.e. to prove that $\exists z(m+0=z)$. The result is immediate from the definition.

The next step, according to Russell's Axiom 5 (induction), will be: assume that the above holds for $m$ and prove that it holds also for $s(m)$.

See Addition of natural numbers for details. The same in Ch.2.2 Addition of Tao's Analysis I (page 24-on).

Mauro ALLEGRANZA
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  • "$\exists z(m+0=z)$" Should this be written with $z$ not $m$: $\exists z(z+0=z)$ ? – zeynel Nov 09 '23 at 18:06
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    @zeynel - no; for an $m$ whatever, we have to prove that there is a unique result of the add operation. This is by induction and the base step amounts to proving that the result of "add zero" exists, i.e. that "there is a $z$ such that $m+0=z$". – Mauro ALLEGRANZA Nov 10 '23 at 07:09
  • I understand that Russell uses Peano's 5. Axiom to prove that addition is valid for all numbers. Thanks for making this clear for me. But I don't see how this is a definition of addition. You write $m+0=m$ as the first part of the 5. Axiom. I read this as "the sum of $m$ and 0 is $m$." This means that Russell assumes addition. He uses addition to state the identity property of zero. Are you saying that $m+0=0$ is not an addition operation? – zeynel Nov 10 '23 at 08:15
  • @zeynel - see Russell's quote above: "we define m+0 as m, and m+(n+1) as the successor of m+n." This is the standard def of sum: Induction is needed to prove that the def is good. Compare with "round square=(def) a square that is circular". It is a definition, but this does not mean that round squares exist. – Mauro ALLEGRANZA Nov 10 '23 at 08:18
  • I still don't get it. The definition is, $m+0=m$". If you write "...that "there is a such that +0=" you are simply saying that $m=z$. What's the point? – zeynel Nov 24 '23 at 08:54