# Strict Ordering on Integers is Trichotomy

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## Theorem

Let $\eqclass {a, b} {}$ and $\eqclass {c, d} {}$ be integers, as defined by the formal definition of integers.

Then exactly one of the following holds:

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

That is, strict ordering is a trichotomy.

## Proof

By the formal definition of integers, we have that $a, b, c, d, e, f$ are all natural numbers.

To eliminate confusion between integer ordering and the ordering on the natural numbers, let $a \preccurlyeq b$ denote that the natural number $a$ is less than or equal to the natural number $b$.

We have:

 $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {c, d} {}$ $\ds \leadsto \ \$ $\ds \eqclass {c, d} {}$ $\nless$ $\ds \eqclass {a, b} {}$ Strict Ordering on Integers is Asymmetric

Then:

 $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {c, d} {}$ $\ds \leadsto \ \$ $\ds a + d$ $\prec$ $\ds b + c$ Definition of Strict Ordering on Integers $\ds \leadsto \ \$ $\ds a + d$ $\ne$ $\ds b + c$ Definition of Ordering on Natural Numbers $\ds \leadsto \ \$ $\ds \eqclass {a, b} {}$ $\ne$ $\ds \eqclass {c, d} {}$ Definition of Strict Ordering on Integers

Similarly:

 $\ds \eqclass {a, b} {}$ $>$ $\ds \eqclass {c, d} {}$ $\ds \leadsto \ \$ $\ds \eqclass {c, d} {}$ $\ngtr$ $\ds \eqclass {a, b} {}$ Strict Ordering on Integers is Asymmetric

Then:

 $\ds \eqclass {a, b} {}$ $>$ $\ds \eqclass {c, d} {}$ $\ds \leadsto \ \$ $\ds a + d$ $\succ$ $\ds b + c$ Definition of Strict Ordering on Integers $\ds \leadsto \ \$ $\ds a + d$ $\ne$ $\ds b + c$ Definition of Ordering on Natural Numbers $\ds \leadsto \ \$ $\ds \eqclass {a, b} {}$ $\ne$ $\ds \eqclass {c, d} {}$ Definition of Strict Ordering on Integers

and:

 $\ds \eqclass {a, b} {}$ $=$ $\ds \eqclass {c, d} {}$ $\ds \leadsto \ \$ $\ds a + d$ $=$ $\ds b + c$ Definition of Strict Ordering on Integers $\ds \leadsto \ \$ $\ds a + d$ $\nprec$ $\ds b + c$ Definition of Ordering on Natural Numbers $\, \ds \land \,$ $\ds a + d$ $\nsucc$ $\ds b + c$ Definition of Ordering on Natural Numbers $\ds \leadsto \ \$ $\ds \eqclass {a, b} {}$ $\nless$ $\ds \eqclass {c, d} {}$ Definition of Strict Ordering on Integers $\, \ds \land \,$ $\ds \eqclass {a, b} {}$ $\ngtr$ $\ds \eqclass {c, d} {}$ Definition of Strict Ordering on Integers

This demonstrates that $<$, $=$ and $>$ are mutually exclusive.

Now:

 $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {c, d} {}$ $\ds \leadsto \ \$ $\ds a + d$ $\prec$ $\ds b + c$ Definition of Strict Ordering on Integers $\ds \leadsto \ \$ $\ds b + c$ $\preccurlyeq$ $\ds a + d$ Definition of Ordering on Natural Numbers $\ds \leadsto \ \$ $\ds \eqclass {c, d} {}$ $\le$ $\ds \eqclass {a, b} {}$ Definition of Strict Ordering on Integers

Similarly:

 $\ds \eqclass {a, b} {}$ $>$ $\ds \eqclass {c, d} {}$ $\ds \leadsto \ \$ $\ds a + d$ $\succ$ $\ds b + c$ Definition of Strict Ordering on Integers $\ds \leadsto \ \$ $\ds b + c$ $\succcurlyeq$ $\ds a + d$ Definition of Ordering on Natural Numbers $\ds \leadsto \ \$ $\ds \eqclass {c, d} {}$ $\ge$ $\ds \eqclass {a, b} {}$ Definition of Strict Ordering on Integers

and:

 $\ds \eqclass {a, b} {}$ $\ne$ $\ds \eqclass {c, d} {}$ $\ds \leadsto \ \$ $\ds a + d$ $\ne$ $\ds b + c$ Definition of Strict Ordering on Integers $\ds \leadsto \ \$ $\ds a + d$ $\prec$ $\ds b + c$ Definition of Ordering on Natural Numbers $\, \ds \lor \,$ $\ds a + d$ $\succ$ $\ds b + c$ Definition of Ordering on Natural Numbers $\ds \leadsto \ \$ $\ds \eqclass {a, b} {}$ $<$ $\ds \eqclass {c, d} {}$ Definition of Strict Ordering on Integers $\, \ds \land \,$ $\ds \eqclass {a, b} {}$ $\>$ $\ds \eqclass {c, d} {}$ Definition of Strict Ordering on Integers

demonstrating that either $<$, $=$ or $>$ must hold.

$\blacksquare$