Natural Sums
Why
We want to combine two groups.1
Defining result
For each natural number $m$, there exists a
function $s_m: \omega \to \omega $ which
satisfies
\[ s_m(0) = m \quad \text{ and } \quad s_m(\ssuc{n}) = \ssuc{(s_m(n))} \]
for every natural number $n$. The proof uses the recursion theorem (see Recursion Theorem).2
Let $m$ and $n$ be natural numbers. The value $s_m(n)$ is the sum of $m$ with $n$.
Notation
We denote the sum $s_m(n)$ by $m + n$.Properties
The properties of sums are direct applications of the principle of mathematical induction (see Natural Induction).3
Let $k$, $m$, and $n$ be natural numbers.
Then
\[ (k + m) + n = k + (m + n). \]
Let $m$ and $n$ be natural numbers.
Then
\[ m + n = n + m. \]
Relation to addition
Let $k$, $m$, and $n$ be natural numbers.
Then
\[ k \cdot (m + n) = (k \cdot m) + (k \cdot n). \]