Integer Multiplication Identity is One

Theorem
The identity of integer multiplication is $$1$$:
 * $$\exists 1 \in \Z: \forall a \in \Z: a \times 1 = a = 1 \times a$$

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.

From the method of construction, $$\left[\!\left[{c+1, c}\right]\!\right]$$, where $$c$$ is any element of the natural numbers $$\N$$, is the isomorphic copy of $$1 \in \N$$.

To ease the algebra, we will take $$\left[\!\left[{1, 0}\right]\!\right]$$ as a canonical instance of this equivalence class.

So, we need to show that $$\forall a, b \in \N: \left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{1, 0}\right]\!\right] = \left[\!\left[{a, b}\right]\!\right] = \left[\!\left[{1, 0}\right]\!\right] \times \left[\!\left[{a, b}\right]\!\right]$$.

From Natural Numbers form Commutative 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 $$\left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{1, 0}\right]\!\right] = \left[\!\left[{a, b}\right]\!\right]$$.

The identity $$\left[\!\left[{a, b}\right]\!\right] = \left[\!\left[{1, 0}\right]\!\right] \times \left[\!\left[{a, b}\right]\!\right]$$ is demonstrated similarly.