Positive Integer is Well-Defined

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Theorem

"Positive" as applied to an integer is well-defined.


Proof

Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.

That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.

$\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.


In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested.


Thus, what we are trying to prove is:

$\eqclass {a, b} {} = \eqclass {c, d} {} \land b < a \implies d < c$


By definition:

$\eqclass {a, b} {} = \eqclass {c, d} {} \iff a + d = b + c$

So:

\(\ds b\) \(<\) \(\ds a\)
\(\ds \leadsto \ \ \) \(\ds \exists p \in \N: a\) \(=\) \(\ds b + p\)
\(\ds \leadsto \ \ \) \(\ds b + p + d\) \(=\) \(\ds b + c\)
\(\ds \leadsto \ \ \) \(\ds p + d\) \(=\) \(\ds c\)
\(\ds \leadsto \ \ \) \(\ds d\) \(<\) \(\ds c\)

$\blacksquare$