# What is the absolute minimum that must be accepted/defined in order to prove 1+1=2?

As I know the "1 + 1 = 2" question has been asked before, my question (for now anyway) is more specific: Assuming as little as possible and accepting as few definitions as possible, what are the very basic axioms and definitions that you absolutely must accept in order to prove 1 + 1 = 2?

Whenever this question gets asked, the immediate responses are things like, "well you have to prove what '1', '2', '+', and '=' are." And that's fine, but in order to prove all four of those things, and then ultimately what they mean together, what is the absolute bare minimum that you simply must accept for any approach to this problem with a completely blank slate? e.g. Peano's axioms?

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I think the Peano axioms would do perfectly. And of course a definition of the decimal system, since peano axioms are written in essentially a unary system. –  user54609 Aug 2 '13 at 3:12
Why did you delete this question only to ask it again? A comment on the last version indicated correctly that the bare minimum is an axiom that says $1+1=2$. That's just not an interesting way to do it. –  dfeuer Aug 2 '13 at 3:12
It was a silly answer, which is why it was a comment. There are certainly more interesting answers. I did not think that my comment made the question less interesting. For instance, en.wikipedia.org/wiki/Robinson_arithmetic –  Baby Dragon Aug 2 '13 at 3:38
@BabyDragon: it was a silly answer that pointed out the inherent weakness in the question. –  dfeuer Aug 2 '13 at 4:22
@dfeuer Of course as someone who will never be as skilled as you likely are in math, it might be out of place for me to say this, but I think you know what I'm asking for (at least, something more than just making 1+1=2 an axiom itself) and are instead just trying to immediately say the question is weak, without actually saying why it is weak, how it could be phrased better, and so on. –  CptSupermrkt Aug 2 '13 at 13:28
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We do not look for a minimal set of axioms, since as has been pointed out in comments, excessively trivial answers can be given. Instead we operate within the traditional framework. The answer below is still essentially trivial, but it introduces some of the language and axioms that one actually uses.

Consider the first-order language $L$ whose non-logical symbols consist of (i) a constant symbol $0$; (ii) a unary function symbol $S$, intended to be interpreted as the successor function; (iii) binary function symbols $A$ and $M$, intended to be interpreted as addition and multiplication respectively. (For your question, we do not need $M$.)

First we define $1$ and $2$. These are abbreviations for the formal terms $S(0)$ and $S(S(0))$.

Then we give the basic axioms for addition. They are (a) $\forall x(A(x,0)=x)$ and (b) $\forall x \forall y(A(x,S(y))=S(A(x,y))$. (These could be made into a single axiom if we wished.)

If we wish to be semi-minimalist, the above is enough. The axioms we have written down are part of the traditional list. Very little can be proved from them except for the desired $1+1=2$. We cannot even prove that $\lnot(0=S(0))$.

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"First-order language" appears to be the complicated bit there. –  dfeuer Aug 2 '13 at 4:26
I could, and perhaps should, have omitted the "first-order" part. Professional reflex! Was distancing myself from the "logicist" tradition. –  André Nicolas Aug 2 '13 at 4:33
It's good, it gives me something else to learn about. –  CptSupermrkt Aug 2 '13 at 13:23
Awesome answer though, I just re-read it twice. I wonder, does the "traditional list" of axioms have a name? –  CptSupermrkt Aug 2 '13 at 13:35
The traditional axioms I was amputating are called the Peano Axioms. –  André Nicolas Aug 2 '13 at 14:26

To make it interesting, you'll want to define $2$ as something other than $1+1$. It's reasonable to define $2$ to be the successor of $1$, and $1$ to be the successor of $0$. Then you just need these two axioms:

• $x+0=x$
• $x+S(y)=S(x+y)$

We have $2=S(1)=S(1+0)=1+S(0)=1+1$.

Edit: André Nicolas' answer is much better than mine. :) Let me try to add something useful by showing how weak these axioms are. The following statements are independent of the above axioms:

• Only $0$ has an additive inverse.
• Every number has an additive inverse.
• $2\neq0$
• $S(x)\neq x$
In fact, many of those statements are independent of each other. You can have fun playing the the following models: the natural numbers $\mathbb N$, the integers $\mathbb Z$ , the ordinal number $\omega^2$, the cyclic group $\mathbb Z/2\mathbb Z$, and the trivial group $\{0\}$.