# Prove that the ordering relation $<_Z$ on the integers is well defined.

I am in the beginning of learning set theory, and the author constructs $\mathbb{R}$, $\mathbb{Q}$ and lastly $\mathbb{Z}$, before $\mathbb{N}$.

Definition of $\sim$
The relation $\sim$ between ordered pairs of natural numbers is defined as: $a+d = c+b \implies (a,b)\sim (c,d)$

Definition of an Integer
The equivalence class under this equivalence relation for an ordered pair $(a,b)$, i.e. the set $\{(c,d)\in \mathbb{N}\times \mathbb{N}: (a,b)\sim (c,d)$}

Question
Assume $(a,b),(a',b'),(c,d),(c',d') \in \mathbb{N}\times \mathbb{N}$ and $(a,b)\sim (a',b')$ and $(c,d)\sim (c',d')$.
Use the standard properties of $\mathbb{N}$ to show that $a+_N d <_N c+_N b \iff a'+_N d'<_N c'+_N b'$.

(The reason for the title is that later $<_Z$ is defined as:
integer$(a,b) <_Z$integer$(c,d)$ if $a+_Nd<c+_Nb$)

Attempt
As far as I understand, since we work with $\mathbb{N}$ to define $\mathbb{Z}$, subtraction and division are not generally defined.

I know the following (all operations are those defined in $\mathbb{N}$):
(1.) $a+b'=a'+b$
(2.) $c+d'=c'+d$

I start by assuming $a+d < c+b$ and want to show that then $a'+d' < c'+b'$

From here I just feel stuck, because I would have wanted to maybe from (1.) and (2.) rewrite $a,b,c,d$ in terms of $a',b',c',d'$ but that requires subtraction or division which is not well defined. Overall I just feel very limited to do any algebra that I am used to when working with inequalities and equations because that usually requires subtraction or division.

I would appreciate a hint how to start, and possibly if you have time a solution underneath so that I can try to solve it myself.

There seems to be one way to solve this with only the operations and relations defined in $\mathbb{N}$, provided one knows of the following theorem:
T1: Let $a,b,c$ be natural numbers. Then $a<b \iff a+c<b+c$ The proof builds on some other theorems shown during the construction of the natural numbers that I won't prove here.
Showing that $a+d<c+b \iff a'+d'<c'+b'$ is then rather straightforward because we don't need to worry about adding terms to the inequality, since it is possible to "remove" them again by theorem T1:
By (1.) we can "create" $a'$ from $a$ by adding $b'$: $a+d < c+b \iff a+b'+d < c+b+b' \iff a'+b+d < c+b+b'$
In the same way we can create $d'$ from $d$ by (2.):
$a'+b+d < c+b+b' \iff a'+b+d+c' < c+b+b'+c' \iff a'+b+d'+c < c+b+b'+c' \iff a'+d'+(b+c)<c'+b'+(b+c)$
By theorem T1 we can then conclude that this is equivalent to $a'+d' < c'+b'$