Consider $\N$, defined as a naturally ordered semigroup $\struct {S, \circ, \preceq}$. For simplicity, let $S' = p^\ge$. First an '''admissible mapping''' is defined. Let $n \in S'$. The mapping $h: \closedint p n \to T$ is defined as an '''admissible mapping''' for $n$ {{iff}}: :$\forall r \in \closedint p n: \map h r = \begin{cases} a & : r = p \\ \map g {\map h y} & : r = y \circ 1 \end{cases}$ where $\closedint p n$ denotes an integer interval of $S$: :$\closedint p n = \set {r \in S: p \le r \le n}$ Define: :$A = \set {x \in S': x \text { has exactly one admissible mapping} }$ We now use the Principle of Mathematical Induction for a Naturally Ordered Semigroup to prove that $A = S'$. === Basis for the Induction === Consider the integer interval $\closedint p p = \set p$. Let the mapping $h_p: \set p \to T$ be defined as: :$\map {h_p} p = a$ By definition, $h_p$ is the unique '''admissible mapping''' for $p$. Thus $p \in A$. This is the basis for the induction. === Induction Hypothesis === Let $n \in A$ such that $p \le n$. That is, there exists a unique '''admissible mapping''' for $n$. This is the induction hypothesis. === Induction Step === From the induction hypothesis: :Let $h$ be the unique '''admissible mapping''' for $n$. By Sum with One is Immediate Successor in Naturally Ordered Semigroup: :$n \circ 1 \notin \closedint p n$ So by Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor: :$\closedint p {n \circ 1} = \closedint p n \cup \set {n \circ 1}$ So we can define a mapping $\phi: \closedint p {n \circ 1} \to T$ by: :$\forall r \in \closedint p {n \circ 1}: \map \phi r = \begin{cases} \map h r & : r \in \closedint p n \\ \map g {\map h n} & : r = n \circ 1 \end{cases}$ By definition, $\phi$ is an '''admissible mapping''' for $n \circ 1$. Let $\phi'$ be an '''admissible mapping''' for $n \circ 1$. Uniqueness of $\phi$ is then proved by showing that $\phi' = \phi$, as follows. The restriction of $\phi'$ to $\closedint p n $ is an '''admissible mapping''' for $n$ Thus by the induction hypothesis: :$\phi' {\restriction_{\closedint p n} } = \phi {\restriction_{\closedint p n} }$ It follows that: {{begin-eqn}} {{eqn | l = \map {\phi'} {n \circ 1}       | r = \map g {\map {\phi'} n}       | c =  }} {{eqn | r = \map g {\map h n}       | c =  }} {{eqn | r = \map \phi {n \circ 1}       | c =  }} {{end-eqn}} Thus as: :$\phi' {\restriction_{\closedint p n} } = \phi {\restriction_{\closedint p n} }$ and: :$\map {\phi'} {n \circ 1} = \map \phi {n \circ 1}$ it follows by Equality of Mappings that: :$\phi' = \phi$ Thus: :$n \circ 1 \in A$ Thus by the Principle of Mathematical Induction for a Naturally Ordered Semigroup it follows that: :$A = S'$ For all $n \in S'$, let $h_n$ be the unique '''admissible mapping''' for $n$. Let $n \ge p$. The restriction of $h_{n \circ 1}$ to $\closedint p n$ is an '''admissible mapping''' for $n$. Therefore: :$h_{n \circ 1} {\restriction_{\closedint p n} } = h_n$ So: :$\forall r \in \closedint p n: \map {h_{n \circ 1} } r = \map {h_n} r$ and in particular: :$\map {h_{n \circ 1} } n = \map {h_n} n$ Let $f: S' \to T$ be the mapping defined as: :$\forall n \in S': \map f n = \map {h_n} n$ Then: :$\map f p = \map {h_p} p = a$ and, for all $n \in S'$: {{begin-eqn}} {{eqn | l = \map f {n \circ 1}       | r = \map {h_{n \circ 1} } {n \circ 1} }} {{eqn | r = \map g {\map {h_{n \circ 1} } n} }} {{eqn | r = \map g {\map {h_n} n} }} {{eqn | r = \map g {\map f n} }} {{end-eqn}} Thus $f: S' \to T$ is a mapping with the desired properties.   Let $f'$ be any mapping satisfying the desired properties.  It remains to be proved that: :$f' = f$ that is, that $f$ is the only mapping with the desired properties. By the foregoing, for all $n \in S'$, the restriction of $f'$ to $\closedint p n$ is an '''admissible mapping''' for $n$. Therefore: :$f' = h_n$ So: :$\forall n \in S': \map {f'} n = \map {h_n} n = \map f n$ and the proof is complete. {{qed}}