Enderton (set). Theorem 4H.
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\subsection{\sorry{Theorem 4H}}%
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\subsection{\pending{Theorem 4H}}%
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\hyperlabel{sub:theorem-4h}
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\hyperlabel{sub:theorem-4h}
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\begin{theorem}[4H]
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\begin{theorem}[4H]
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\end{theorem}
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\end{theorem}
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\begin{proof}
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\begin{proof}
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TODO
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Let $\langle N, S, e \rangle$ be a \nameref{ref:peano-system}.
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By the \nameref{sub:recursion-theorem-natural-numbers}, there exists a
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unique function $h \colon \omega \rightarrow N$ such that $h(0) = e$ and
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for every $n \in \omega$, $h(n^+) = h(\sigma(n)) = S(h(n))$.
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All that remains is proving $h$ is one-to-one and onto.
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\suitdivider
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\noindent
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We first show $h$ is one-to-one by induction.
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Define
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$$S = \{n \in \omega \mid \forall m, h(m) = h(n) \Rightarrow m = n\}.$$
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We show that (i) $0 \in S$ and (ii) if $n \in S$, then $n^+ \in S$.
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Afterward we show (iii) that $h$ is one-to-one.
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\paragraph{(i)}%
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\hyperlabel{par:theorem-4h-i}
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Let $m \in \omega$ such that $h(m) = h(0) = e$.
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Suppose $m \neq 0$.
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Then \nameref{sub:theorem-4c} indicates there exists some natural number
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$p$ such that $p^+ = m$.
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But then $h(m) = h(p^+) = S(h(p)) = e$.
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By definition of a Peano system, $e \not\in \ran{S}$.
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This is a contradiction.
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Hence $m = 0$, i.e. $0 \in S$.
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\paragraph{(ii)}%
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\hyperlabel{par:theorem-4h-ii}
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Suppose $n \in S$.
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Let $m \in \omega$ such that $h(m) = h(n^+) = S(h(n))$.
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There are two cases to consider:
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\subparagraph{Case 1}%
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Suppose $m = 0$.
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Then $h(0) = e = S(h(n))$.
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But, by definition of a Peano system, $e \not\in \ran{S}$.
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Thus we have a contradiction.
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\subparagraph{Case 2}%
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Suppose $m \neq 0$.
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Then \nameref{sub:theorem-4c} indicates there exists some natural number
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$p$ such that $p^+ = m$.
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Then $$S(h(n)) = h(m) = h(p^+) = S(h(p)).$$
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By definition of a Peano system, $S$ is one-to-one.
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Therefore $h(n) = h(p)$.
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Since $n \in S$, it follows $n = p$.
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Therefore $n^+ = p^+ = m$.
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\subparagraph{Subconclusion}%
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The above two cases are exhaustive.
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Hence $n^+ \in S$.
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\paragraph{(iii)}%
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By \nameref{par:theorem-4h-i} and \nameref{par:theorem-4h-ii}, $S$ is an
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\nameref{ref:inductive-set}.
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By \nameref{sub:theorem-4b}, $S = \omega$.
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Thus for all natural numbers $m, n \in \omega$, if $h(m) = h(n)$, then
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$m = n$.
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In other words, $f$ is one-to-one.
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\suitdivider
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\noindent
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We next show that $\ran{h} = N$.
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By the Peano posulate, every subset $A$ of $N$ containing $e$ and closed under $S$ is $N$ itself.
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Thus it suffices to prove that (i) $e \in \ran{h}$ and (ii) $\ran{h}$ is closed under $S$.
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\paragraph{(i)}%
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That is, $h(0) = e$ by definition.
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Thus $e \in \ran{h}$.
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\paragraph{(ii)}%
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Let $y \in \ran{h}$.
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Then there exists some $n \in \omega$ such that $h(n) = y$.
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By definition of $h$, $h(n^+) = S(h(n)) = S(y)$.
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Therefore $S(y) \in \ran{h}$.
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Since this holds for any $y \in \ran{h}$, $\ran{h}$ is closed under $S$.
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\end{proof}
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\end{proof}
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\section{Arithmetic}%
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\section{Arithmetic}%
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