# How to construct a partial ordering from Peano's 5 Axioms?

I am trying to formally construct the usual partial ordering LTE from Peano's 5 Axioms. Would the following construction work?

$$\forall a,b: [(a,b)\in LTE \iff(a,b)\in N^2$$

$$\land ~ \forall c\subset N^2:[\forall d\in N: [(d,d) \in c]$$

$$\land ~ \forall d,e:[(d,e)\in c \implies(d, S(e))\in c]$$

$$\implies (a,b) \in c]]$$

where $$N$$ is the set of natural numbers and $$S$$ is the usual successor function on $$N$$.

EDIT:

Informally, for all $$x,y\in N$$, we can recursively define $$\le$$ on $$N$$ as the "smallest" set of ordered pairs of natural numbers that satisfies:

$$x \le x$$

$$x\le y \implies x\le S(y)$$

• In the second line, you switch from defining $LTE$ to making a statement about all relations, which seems like it's not what you want. Apr 20, 2021 at 22:14
• LTE means.... what? Apr 20, 2021 at 22:28
• @LeeMosher LTE (less than or equal) is a set of ordered pairs of natural numbers. Apr 20, 2021 at 22:29
• I think your definition is fine: You are saying that $a \le b$ holds if whenever $R$ (which you call $c$) is a relation on $N$ that is reflexive and is such that $d\mathrel{R}S(e)$ whenever $d\mathrel{R}e$, then $a\mathrel{R}b$.That says that $(\le)$ is the intersection of all reflexive relations $R$ such that $(S \circ R) \subseteq R$, which is true. Apr 20, 2021 at 22:46
• Do note that the definition involves some second-order logic, though. Unless you're considering $N$ as being a set within some first-order set theory such as ZFC, or something along those lines. Apr 20, 2021 at 23:23

Thanks all for your help, but I didn't find my construction here to be very workable. I was trying to avoid using an addition function to "simplify things." That was probably a mistake.

Using Peano's Axioms, some basic set theory and a form of natural deduction, I formally constructed (in order) the function $$+$$ and the relations $$\leq$$ and $$\lt$$ as sets of ordered n-tuples such that, for all $$x, y \in N$$ we have:

$$x+0 = x$$

$$x+S(y) = S(x+y)$$

$$x\leq y \iff \exists z\in N: x+z=y$$

$$x \lt y \iff x\leq y ~\land ~ x\neq y$$

Everything then fell into place. I was just now able to derive the usual fundamental properties of $$+$$, $$\leq$$ and $$\lt$$ for a larger project that I am working on.