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Recursion Theorem
Needed by:
Integer Order
Integer Partitions
Integer Sums
Natural Equations
Natural Products
Natural Summation
Number of Disjoint Unions
Number Partitions
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Natural Sums


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 $n$.


We denote the sum $s_m(n)$ by $m + n$.


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

\[ k \cdot (m + n) = (k \cdot m) + (k \cdot n). \]

  1. Future editions will change this section. ↩︎
  2. Future editions will give the entire account. ↩︎
  3. Future editions will include the accounts. ↩︎
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