# Strict Ordering on Integers is Well-Defined

## Theorem

Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.

Let:

 $\ds \eqclass {a, b} {}$ $=$ $\ds \eqclass {a', b'} {}$ $\ds \eqclass {c, d} {}$ $=$ $\ds \eqclass {c', d'} {}$

Then:

 $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {c, d} {}$ $\ds \leadstoandfrom \ \$ $\ds \eqclass {a', b'} {}$ $<$ $\ds \eqclass {c', d'} {}$

## Proof

This is a direct application of the Extension Theorem for Total Orderings.

$\blacksquare$