Ordering on Integers is Transitive

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


\(\ds \eqclass {a, b} {}\) \(\le\) \(\ds \eqclass {c, d} {}\)
\(\, \ds \land \, \) \(\ds \eqclass {c, d} {}\) \(\le\) \(\ds \eqclass {e, f} {}\)
\(\ds \implies \ \ \) \(\ds \eqclass {a, b} {}\) \(\le\) \(\ds \eqclass {e, f} {}\)

That is, ordering on the integers is transitive.


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:

\(\ds \eqclass {a, b} {}\) \(\le\) \(\ds \eqclass {c, d} {}\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds a + d\) \(\preccurlyeq\) \(\ds b + c\) Definition of Ordering on Integers
\(\ds \eqclass {c, d} {}\) \(<\) \(\ds \eqclass {e, f} {}\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds c + f\) \(\preccurlyeq\) \(\ds d + e\) Definition of Ordering on Integers
\(\ds \leadsto \ \ \) \(\ds a + d + f\) \(\preccurlyeq\) \(\ds b + c + f\) adding $f$ to both sides of $(1)$
\(\ds \leadsto \ \ \) \(\ds a + d + f\) \(\preccurlyeq\) \(\ds b + d + e\) from $(2)$: $b + \paren {c + f} \preccurlyeq b + \paren {d + e}$
\(\ds \leadsto \ \ \) \(\ds a + f\) \(\preccurlyeq\) \(\ds b + e\) subtracting $d$ from both sides
\(\ds \leadsto \ \ \) \(\ds \eqclass {a, b} {}\) \(\le\) \(\ds \eqclass {e, f} {}\) Definition of Ordering on Integers