# Peano arithmetic - Why is adding n to m the same as incrementing m n times?

Addition(+) is defined using the successor function(++) in Peano arithmetic as:

1. 0 + m = m
2. (n++) + m = (n + m)++

While these are intuitive axioms that are consistent with my previous, elementary, understanding of addition, I don't understand how it follows from these axioms that n + m is the same thing as incrementing m n times.

Although I can see that it is true for specific cases:

1. 1 + m = (0++) + m = (0 + m)++ = m++
2. 2 + m = (1++) + m = (1 + m)++ = (m++)++
3. etc.

Thanks very much.

• You continue the pattern and prove with induction. – Gareth Ma Apr 3 at 15:53
• I think I would have defined it as $m+0=m$ and $m+(n++)=(m+n)++$. – Angina Seng Apr 3 at 16:02
• @AnginaSeng Isn't that equivalent to "0 + m = 0" and "(n++) + m = (n + m)++"? i.e. You could just do induction to get the other. – Haziq Muhammad Apr 3 at 16:44

We can prove it by induction on $$m$$. If $$m=0$$, $$n+m=n+0=n$$ is the result of not incrementing $$n$$ at all, i.e. doing it $$0$$ times. Suppose $$n+k$$ is the result of $$k$$ increments starting from $$n$$. For the inductive step, associativity of $$+$$ gives $$n+(k+1)=(n+k)+1$$, i.e. we increment $$k$$ times, then once more.
• @HaziqMuhammad You would define it recursively: $0$ increments leaves $n$ unchanged, $1$ changes it to $n$'s successor $n+1$, & $k+1$ increments applies $k$ then $1$. – J.G. Apr 3 at 16:32