Alternate definition of the integers Assuming we have already defined the natural numbers $\mathbb{N}$ and function iteration $f^n$ for a natural number $n$, a set $A$ and a function $f:A\rightarrow A$, do the following axioms (based off the Peano axioms) define the set of integers?


*

*There exist an element $0$ in $\mathbb{Z}$

*There exists a function $S:\mathbb{Z}\rightarrow\mathbb{Z}$. 

*There exists a function $P:\mathbb{Z}\rightarrow\mathbb{Z}$ such that for every element $x$ in $\mathbb{Z}$, $P(S(x)) = S(P(x)) = x$. 

*For all $n$ in $\mathbb{N}$ where $n\neq 0$ and for every element $x$ in $\mathbb{Z}$, $S^n(x) \neq x$ and $P^n(x) \neq x$. 

*If $A$ is a set such that $0$ is in $A$ and, for all $n$ in $\mathbb{N}$, $S^n(0)$ being in $A$ implies that $S(S^n(0))$ is in $A$ and $P^n(0)$ being in $A$ implies that $P(P^n(0))$ is in $A$, then $A$ contains every integer. 

 A: With these questions you want to make sure that you are constructing a unique object identical to the integers and not a class of objects. 
The parallels to the Peano axioms are clear: 


*

*Axiom 1 is the same as the equivalent for the natural numbers, and
rules out the empty set as a trivial object to the set of axioms.

*Axiom 2 is the same as the equivalent for the natural numbers, and
asserts that the function $S$ is closed over the set.

*Axiom 3 corresponds to its equivalent for the natural numbers, and
asserts that the function $S$ is bijective and has a unique inverse
function $P$, (while the original for the natural numbers only
asserts that the function $S$ is injective). It follows that the
function $P$ is also closed over the set.

*Axiom 4 asserts that while $S$ is a permutation, the cardinality of the set is not finite. This disallows finite cycles from being defined. 

*Axiom 5 is mathematical induction over the natural numbers on $S$ and
$P$ and corresponds to its equivalent over $S$ for the natural
numbers, and rules out any other elements other than those elements
accessible from $0$ through function composition of $S$ and $P$, just
as the original did for the natural numbers. Or equivalently, there does not exist a partition of the set into multiple mutually disjoint sets. This axiom also establishes that the set is countably infinite. 

*For $n = 2$, axiom 4 asserts that neither $S$ nor $P$ are
involutions, and with axiom 3, it follows that for every element $x$
in the set, $S(x) \neq P(x)$, and the two functions $S$ and $P$ are
not the same function. Therefore, the set is not identical to the
natural numbers.


These axioms together construct a structure with an initial element $0$ and two countably infinite arms branching off from $0$ in distinct directions, defined by function iteration of $S$ and $P$; this uniquely forms the integers. 
One can check indeed that this is isomorphic to the usual construction of the integers by equivalence classes of natural numbers (which does not need addition either if function iteration is defined):
We define the equivalence class $[a, b] \equiv [c, d]$ if $S^a(d) = S^c(b)$. We define $0 = [k, k]$ for all natural numbers $k$, which satisfies axiom 1. For the integer $x = [a, b]$ for natural numbers $a$ and $b$, we define $S(x) = [S(a), b]$ and $P(x) = [a, S(b)]$. Axiom two is satisfied because the successor function $S$ is closed over the natural numbers. Axiom three is satisfied because $P(S(x)) = S(P(x)) = [S(a), S(b)] \equiv x$. Axiom four is satisfied because the natural numbers are not a cycle. And axiom five is satisfied because the construction does not construct any elements inaccessible from $0$. 
