Proof of The Natural Numbers with Zero and Their Embedding into the Integers

The clause naturals. By The Set of Natural Numbers and the Number One §naturals, $\mathbb{N}=\mathbb{N}_{0}\setminus\{0\}$; so $0\notin\mathbb{N}$, $\mathbb{N}\subseteq\mathbb{N}_{0}$, and every element of $\mathbb{N}_{0}$ other than $0$ lies in $\mathbb{N}$. Since $0\in\mathbb{N}_{0}$ by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §inductive, it follows that $\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$.

Existence and uniqueness of $\iota_{0}$. By the clause just proved, every $n\in\mathbb{N}_{0}$ is either $0$ or an element of $\mathbb{N}$, and not both. Let $t(n)$ be $0_{\mathbb{Z}}$ if $n=0$ and $\iota(n)$ if $n\in\mathbb{N}$; then $t(n)\in\mathbb{Z}$ for every $n\in\mathbb{N}_{0}$, as $0_{\mathbb{Z}}\in\mathbb{Z}$ by The Integers §constants and $\iota$ maps $\mathbb{N}$ to $\mathbb{Z}$ by The Integers §embedding. Since $\mathbb{N}_{0}$ is a set by Omega Is the Least Inductive Class: It Is a Set, Induction from Zero, the Peano Properties, and Transitivity §set, Maps and Relations Given by Formulas §map gives exactly one map $\iota_{0}:\mathbb{N}_{0}\to\mathbb{Z}$ with $\iota_{0}(n)=t(n)$ for every $n\in\mathbb{N}_{0}$. A map $\mathbb{N}_{0}\to\mathbb{Z}$ satisfies $\iota_{0}(0)=0_{\mathbb{Z}}$ and $\iota_{0}(n)=\iota(n)$ for every $n\in\mathbb{N}$ exactly when its value at every $n\in\mathbb{N}_{0}$ is $t(n)$, so $\iota_{0}$ is the only map with the stated properties.

Facts about $\mathbb{Z}$. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ring, $\mathbb{Z}$ is a commutative ring with zero $0_{\mathbb{Z}}$, so $x+0_{\mathbb{Z}}=0_{\mathbb{Z}}+x=x$ and $xy=yx$ for all $x,y\in\mathbb{Z}$; by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §negation and commutativity, $0_{\mathbb{Z}}x=x\,0_{\mathbb{Z}}=0_{\mathbb{Z}}$. By The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §ordered-ring, $\le$ is a total order on $\mathbb{Z}$, hence reflexive and antisymmetric by Partial and Total Orders on a Set and the Associated Strict Relation §partial; and $x<y$ means $x\le y$ and $x\neq y$ by The Integers §operations, so $x\le y$ if and only if $x<y$ or $x=y$.

Positivity. For every $n\in\mathbb{N}_{0}$ we have $0_{\mathbb{Z}}\le\iota_{0}(n)$, and $\iota_{0}(n)=0_{\mathbb{Z}}$ if and only if $n=0$. Indeed, $\iota_{0}(0)=0_{\mathbb{Z}}$ and $0_{\mathbb{Z}}\le0_{\mathbb{Z}}$ by reflexivity; and for $n\in\mathbb{N}$, $0_{\mathbb{Z}}<\iota(n)$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §positive, that is, $0_{\mathbb{Z}}\le\iota_{0}(n)$ and $\iota_{0}(n)\neq0_{\mathbb{Z}}$.

Injectivity. Let $m,n\in\mathbb{N}_{0}$ with $\iota_{0}(m)=\iota_{0}(n)$. By positivity, $m=0$ if and only if $n=0$. If both are $0$, then $m=n$. Otherwise $m,n\in\mathbb{N}$ and $\iota(m)=\iota(n)$, so $m=n$ because $\iota$ is injective by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding.

Sums and products. Let $m,n\in\mathbb{N}_{0}$. If $n=0$, then $m+n=m$ by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero and $\iota_{0}(m)+\iota_{0}(n)=\iota_{0}(m)+0_{\mathbb{Z}}=\iota_{0}(m)$; if $m=0$, then likewise $m+n=n$ and $\iota_{0}(m)+\iota_{0}(n)=0_{\mathbb{Z}}+\iota_{0}(n)=\iota_{0}(n)$. If $m=0$ or $n=0$, then $mn=0$ by Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §zero, and $\iota_{0}(m)\,\iota_{0}(n)=0_{\mathbb{Z}}=\iota_{0}(mn)$ because one factor is $0_{\mathbb{Z}}$. If $m,n\in\mathbb{N}$, then $m+n,mn\in\mathbb{N}$ by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication §closed, and The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding gives $\iota_{0}(m+n)=\iota(m+n)=\iota(m)+\iota(n)=\iota_{0}(m)+\iota_{0}(n)$ and $\iota_{0}(mn)=\iota(m)\,\iota(n)=\iota_{0}(m)\,\iota_{0}(n)$.

Order. Let $m,n\in\mathbb{N}_{0}$. If $m=0$, both $m\le n$ and $\iota_{0}(m)\le\iota_{0}(n)$ hold, the first by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor §zero-least and the second by positivity, as $\iota_{0}(0)=0_{\mathbb{Z}}$. If $m\in\mathbb{N}$ and $n=0$, both fail: $m\le0$ would mean $m\in0$ or $m=0$ by The Order on Omega §order, which is impossible as $0=\emptyset$ by The Class Omega of Natural Numbers with Zero §zero and $m\neq0$; and $\iota_{0}(m)\le0_{\mathbb{Z}}$ together with $0_{\mathbb{Z}}\le\iota_{0}(m)$ would give $\iota_{0}(m)=0_{\mathbb{Z}}$ by antisymmetry, contrary to positivity. If $m,n\in\mathbb{N}$, then $m\le n$ if and only if $m<n$ or $m=n$, by Arithmetic and Order of the Natural Numbers §partial-order; $m<n$ if and only if $\iota(m)<\iota(n)$, by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §embedding; and $m=n$ if and only if $\iota(m)=\iota(n)$, by injectivity. Hence $m\le n$ if and only if $\iota(m)<\iota(n)$ or $\iota(m)=\iota(n)$, that is, if and only if $\iota_{0}(m)\le\iota_{0}(n)$.

Image. By positivity every value of $\iota_{0}$ lies in $\{x\in\mathbb{Z}:0_{\mathbb{Z}}\le x\}$. Conversely, let $x\in\mathbb{Z}$ with $0_{\mathbb{Z}}\le x$. If $x=0_{\mathbb{Z}}$, then $x=\iota_{0}(0)$. Otherwise $0_{\mathbb{Z}}<x$, so $x=\iota(n)=\iota_{0}(n)$ for some $n\in\mathbb{N}$ by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements §positive. Hence the image of $\iota_{0}$ is $\{x\in\mathbb{Z}:0_{\mathbb{Z}}\le x\}$.