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Full disclaimer: I have edited the question to make it simpler, and therefore some of the comments may no longer make much sense.

Assume we have defined the set $\mathbb{N}$ using the first nine Peano axioms. For brevity, $1$ is defined as $S(0),\ 2\ $ is defined as $S(1),\ $ etc.

Define the function, $+': (\mathbb{N}\cup\{0\}) \times (\mathbb{N}\cup\{0\})\to (\mathbb{N}\cup\{0\}) $ recursively by:

  • $0\ +'\ a=a\qquad (1)$
  • $S(a)\ +'\ S(b) = S(a\ +'(\ b\ +'\ 1)\ )\qquad (2)$

So for example,

$1\ +'\ 1\ \overset{def}{=}\ S(0)\ +'\ S(0) \overset{(2)}{=}\ S( 0\ +'\ (0\ +'\ 1)\ ) \overset{(1)}{=}\ S(0\ +'\ 1) \overset{(1)}{=}\ S(1) \overset{def}{=} 2. $

And I believe we can prove all other additions match our expectations using similar reasoning.

Edit: I don't see how we show that $\ 1\ +'\ 0 = 1.$

I think this is an alternative way of defining addition.

Is this correct? Does my definition of addition give us the expected results and therefore also match wikipedia's definition?

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  • $\begingroup$ Where have you defined the term $a+''b+''1$ that is used in ($2\ast$)? $\endgroup$
    – Dan Rust
    Sep 12, 2022 at 12:42
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    $\begingroup$ Actually, don't you need to define $1$ in order to use it for both $+'$ and $+''$? $\endgroup$
    – Dan Rust
    Sep 12, 2022 at 12:45
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    $\begingroup$ You haven't defined $+''1.$ $\endgroup$
    – Thomas Andrews
    Sep 12, 2022 at 12:53
  • $\begingroup$ I'd try to prove $0+'' a =a$ in your definition. $\endgroup$
    – Thomas Andrews
    Sep 12, 2022 at 12:55
  • $\begingroup$ Isn’t $1$ defined as $S(0)$ in the Peano axioms ? $\endgroup$
    – Adam Rubinson
    Sep 12, 2022 at 12:57

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Given the two axioms in the question, we can deduce that $a\ +'\ b = a+b\ $ for all $a$ and all $b\geq 1.$

However, I don't see a reason why $1\ +'\ 0$ cannot equal any natural number (or zero).

Same for $a\ +'\ 0$ for any $a:\ $ it can be equal to any natural number and this doesn't lead to contradictions, e.g. by the pigeonhole principle.

Obviously there are many ways a third axiom would give us $a\ +'\ 0 = a\quad \forall a,\ $ for example associativity:

$ a\ +'\ (b\ +'\ c) = (a\ +'\ b)\ +'\ c\quad \forall\ a,b,c $

This would work because, for example:

$S(1)\ +'\ S(0) = S(1\ +'\ (0\ +'\ 1)) = S(2) = S((1\ +'\ 0)\ +'\ 1)\implies\ 1\ +'\ 0 = 1,\ $ because $+': (\mathbb{N}\cup\{0\}) \times (\mathbb{N}\cup\{0\})\to (\mathbb{N}\cup\{0\}) $ is a function and the only value of $k$ for which $k\ +'\ 1 = 2\ $ is $\ k=1.$

Commutativity is an alternative third axiom that would make the usual addition work for all $a,b$ because along with axiom $(1)$ we would have: $a\ +'\ 0=0\quad \forall a.$

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  • $\begingroup$ The first sentence still is not quite correct. "All $a$ and all $b\geq1$" makes the claim that $3+'2=3+2=5,$ but $3+'2$ only has to be $(1+'0)+4,$ and as you say, $1+'0$ could be anything. $\endgroup$
    – David K
    Sep 16, 2022 at 11:32
  • $\begingroup$ With regards to my first sentence and speaking about the two axioms in the question: "But $3\ +'\ 2\ $ only has to be $(1\ +'\ 0)\ +'\ 4.$" No, I don't think my claim that $ a\ +'\ b = a+b\ $ for all $a$ and all $b\geq 1$ implies that $(1\ +' 0)\ +'\ 4 = 5.$ And yes, I agree that $1\ +'\ 0$ could be anything and therefore $(1\ +' 0)\ +'\ 4$ could be anything. So I'm not quite sure what you're saying... $\endgroup$
    – Adam Rubinson
    Sep 16, 2022 at 11:47
  • $\begingroup$ You said you could deduce that $a+'b = a+b$ for all $a$ and all $b\geq 1.$ That includes for $a=3$ and $b=2.$ Do you still think you can deduce that $a+'b = a+b$ for $a=3$ and $b=2$? Have you tried proving it? $\endgroup$
    – David K
    Sep 16, 2022 at 11:52
  • $\begingroup$ Yes. $3\ +'\ 2 = S(2)\ +'\ S(1) = S(2\ +'\ (1\ +'\ 1)) = S(2\ +'\ 2) = S(4) = 5.$ $\endgroup$
    – Adam Rubinson
    Sep 16, 2022 at 11:57
  • $\begingroup$ Ah, ok. So the reason you need an extra axiom for $a+'0$ is because $0$ is not a successor of any number. $\endgroup$
    – David K
    Sep 16, 2022 at 12:01

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