[proofplan] The proof uses the convention $\mathbb{N} := \{n \in \mathbb{Z} : n > 0\}$ for the positive integers, together with the [well-ordering principle](/theorems/721) for $\mathbb{N}$. For the bounded-below subset statement, choose an integer lower bound $b$ for $S$, translate $S$ upward from $b$, and obtain a nonempty subset of $\mathbb{N}$. The well-ordering principle for $\mathbb{N}$ gives a least translated element, and translating it back gives the least element of $S$. [/proofplan] [step:Identify the least positive integer] Let $\mathbb{N} := \{k \in \mathbb{Z} : k > 0\}$ denote the positive integers inside $\mathbb{Z}$. By the well-ordering principle for the natural numbers, the set $\mathbb{N}$ has least element $1$. Hence, if $n \in \mathbb{Z}$ and $n > 0$, then $n \in \mathbb{N}$, so $1 \le n$. Therefore $1$ is the least positive integer. [/step] [step:Translate a bounded-below subset of $\mathbb{Z}$ into a subset of $\mathbb{N}$] Let $S \subset \mathbb{Z}$ be nonempty and bounded below. Choose a lower bound $b \in \mathbb{Z}$, so that $b \le s$ for every $s \in S$. Define a set $A \subset \mathbb{N}$ by \begin{align*} A := \{s - b + 1 : s \in S\}. \end{align*} Since $S$ is nonempty, the set $A$ is nonempty. If $s \in S$, then $b \le s$, so $0 \le s-b$ by compatibility of addition with the order [citetheorem:9705]. Hence $1 \le s-b+1$, and therefore $s-b+1 \in \mathbb{N}$. Thus $A$ is a nonempty subset of $\mathbb{N}$. [guided] The reason for introducing $A$ is that subsets of $\mathbb{Z}$ need not have least elements, while nonempty subsets of $\mathbb{N}$ do. The lower bound $b$ lets us shift every element of $S$ to a nonnegative integer. Choose $b \in \mathbb{Z}$ such that $b \le s$ for every $s \in S$. Define \begin{align*} A := \{s - b + 1 : s \in S\}. \end{align*} This set is nonempty because $S$ is nonempty: choosing any $s_0 \in S$ gives $s_0-b+1 \in A$. We also need to check that $A \subset \mathbb{N}$. If $s \in S$, then $b \le s$. Adding $-b$ to both sides is valid by the compatibility of integer order with addition [citetheorem:9705], giving \begin{align*} 0 \le s-b. \end{align*} Adding $1$ gives \begin{align*} 1 \le s-b+1. \end{align*} Therefore $s-b+1$ is a positive integer, so it lies in $\mathbb{N}$. Hence $A$ is a nonempty subset of $\mathbb{N}$. [/guided] [/step] [step:Apply well-ordering and translate the least element back] By the well-ordering principle for the natural numbers, there exists $m \in A$ such that $m \le a$ for every $a \in A$. Since $m \in A$, there exists $\ell \in S$ such that \begin{align*} m = \ell - b + 1. \end{align*} We claim that $\ell$ is the least element of $S$. Let $s \in S$. Then $s-b+1 \in A$, so the minimality of $m$ gives \begin{align*} \ell-b+1 \le s-b+1. \end{align*} Adding $b-1$ to both sides is valid by compatibility of order with addition [citetheorem:9705], and yields \begin{align*} \ell \le s. \end{align*} Since $s \in S$ was arbitrary, $\ell \le s$ for every $s \in S$. Also $\ell \in S$ by construction. Therefore $\ell$ is the least element of $S$. [guided] The set $A$ is a nonempty subset of $\mathbb{N}$, so the well-ordering principle for the natural numbers gives an element $m \in A$ such that $m \le a$ for every $a \in A$. Because $m \in A$, the definition of $A$ gives an element $\ell \in S$ satisfying \begin{align*} m = \ell - b + 1. \end{align*} We now translate the minimality of $m$ back to the original subset $S$. Let $s \in S$. By the definition of $A$, the integer $s-b+1$ lies in $A$. The minimality of $m$ therefore gives \begin{align*} \ell-b+1 \le s-b+1. \end{align*} Adding $b-1$ to both sides is valid by the compatibility of integer order with addition [citetheorem:9705], and it gives \begin{align*} \ell \le s. \end{align*} Since $s \in S$ was arbitrary, $\ell \le s$ for every $s \in S$. Also $\ell \in S$ by construction from $m \in A$. Hence $\ell$ is the least element of $S$. [/guided] [/step]