# On an alternative definition of addition, $S(a)\ +'\ S(b) = S(a\ +'\ (b\ +'\ 1))$, assuming the first nine Peano axioms.

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?

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

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: $$a\ +'\ 0=0\quad \forall a.$$

• 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. Sep 16, 2022 at 11:32
• 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... Sep 16, 2022 at 11:47
• 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? Sep 16, 2022 at 11:52
• Yes. $3\ +'\ 2 = S(2)\ +'\ S(1) = S(2\ +'\ (1\ +'\ 1)) = S(2\ +'\ 2) = S(4) = 5.$ Sep 16, 2022 at 11:57
• Ah, ok. So the reason you need an extra axiom for $a+'0$ is because $0$ is not a successor of any number. Sep 16, 2022 at 12:01