User:KBlott:Questions\The first absorption law of the protointegers

Let $\mathbb N^2 =\mathbb N \times \mathbb N$ be the set of protointegers. Let $n = (n_0, n_1) \in \mathbb N^2$. Let $m = (m_0, m_1) \in \mathbb N^2$. Let $\wedge: \mathbb N^2 \times \mathbb N^2 \to \mathbb N^2$ such that $n \wedge m = (max(n_0,m_0), max(n_1,m_1))$. Let $\vee: \mathbb N^2 \times \mathbb N^2 \to \mathbb N^2$ such that $n \vee m = (min(n_0,m_0), min(n_1,m_1))$. Given that $min(x,max(x,y)) = x$ for every $x, y \in \mathbb N$,
 * we want to show that $n \vee (n \wedge m) = n$ for every $n, m \in \mathbb N^2$.

Proof
$n \vee (n \wedge m)$
 * $= n \vee (max(n_0,m_0),max(n_1,m_1))$
 * $ = ( min(n_0,max(n_0,m_0), min(n_1,max(n_1,m_1))$
 * $ = ( n_0, n_1)$
 * $ = n$

Therefore, Q.E.D.