Integer Multiplication Identity is One

Theorem
The identity of integer multiplication is $$1$$.

Proof
Let us define $$\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$$ as in the formal definition of integers.

That is, $$\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$$ 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 $$\left({x_1, y_1}\right) \boxminus \left({x_2, y_2}\right) \iff x_1 + y_2 = x_2 + y_1$$.

In order to streamline the notation, we will use $$\left[\!\left[{a, b}\right]\!\right]$$ to mean $$\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$$, as suggested.

We need to show that $$\forall a, b, c \in \N: \left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{c+1, c}\right]\!\right] = \left[\!\left[{a, b}\right]\!\right] = \left[\!\left[{c+1, c}\right]\!\right] \times \left[\!\left[{a, b}\right]\!\right]$$.

From Natural Numbers form Semiring, we take it for granted that:
 * addition and multiplication are commutative and associative on the natural numbers $$\N$$;
 * natural number multiplication is distributive over natural number addition.

So:

$$ $$ $$ $$

So $$\left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{c+1, c}\right]\!\right] = \left[\!\left[{a, b}\right]\!\right]$$.

The identity $$\left[\!\left[{a, b}\right]\!\right] = \left[\!\left[{c+1, c}\right]\!\right] \times \left[\!\left[{a, b}\right]\!\right]$$ is demonstrated similarly.