A question about Roger Penrose's spin networks and mathematical formalization?

Question

Let $a,b,c$ be "units" in the spin network. Then there are there are the following three requirements to fulfill (according to the relevant Wikipedia entry):

0. $a,b,c \in \mathbb{N}$
1. Triangle inequality: $a \le b+c, b \le a+c, c \le a+b$
2. $a+b+c \equiv 0 \mod (2) $

Why are these requirements not modeled as a metric space over the natural numbers? For example, I'd expect something like: for all natural numbers $A,B,C$ we have

0. $d(A,B) \in \mathbb{N}$
1. Triangle inequality: $d(A,B) \le d(A,C) + d(C,B)$
2. $d(A,B)+d(A,C)+d(B,C) \equiv 0 \mod (2)$

"Units" $a,b,c$ are interpreted as distances in some metric space with (0-2): $$a = d(B,C), b = d(A,C) , c = d(A,B).$$

I have been playing around with a metric space induced by the shortest paths in a graph, and it seems to have this property (Related question: Why is this bipartite graph a partial cube, if it is?) $$ d_s(A,B) = \Omega\left(\frac{AB}{\gcd(A,B)^2}\right) $$ Here $\Omega$ is the prime counting function with multiplicities.
I have three questions sorted descending by vagueness:

1. Is there a reason, why this mathematical formalism is not applied in physics?

2. Are there any metric spaces on the natural numbers which satisfiy (0-2)?

3. Does the metric $d_s$ as defined above, always satisfy the requirement $\equiv 0 \mod (2)$ ?

Edit: The third question can be answered positively - I think - since every shortest path between $A$ and $B$ goes through $\gcd(A,B)$, and this means that:

$$d_s(A,B) = d_s\big(A,\gcd(A,B)\big) + d_s\big(\gcd(A,B),B\big)$$

But then: $$d_s(A,B) = \Omega\left(\frac{A}{\gcd(A,B)}\right)+\Omega\left(\frac{B}{\gcd(A,B)}\right)$$

So: $$ \begin{split} d_s(A,B) +d_s(A,C)+d_s(B,C) &= \Omega\left(\frac{A}{\gcd(A,B)}\right)+\Omega\left(\frac{B}{\gcd(A,B)}\right)\\ &\qquad+\Omega\left(\frac{A}{\gcd(A,C)}\right)+\Omega\left(\frac{C}{\gcd(A,C)}\right)\\ &\qquad\qquad+\Omega\left(\frac{B}{\gcd(B,C)}\right)+\Omega\left(\frac{C}{\gcd(B,C)}\right) \end{split} $$

But how to proceed from here?

Second edit: The third question can be answered positively, but for this reason: Since $\Omega$ is completely additive, we get: $$d_s(A,B) +d_s(A,C)+d_s(B,C) = \Omega \left( \frac{AB}{\gcd(A,B)^2} \frac{AC}{\gcd(A,C)^2} \frac{BC}{\gcd(B,C)^2} \right) = \Omega \left( \left( \frac{ABC}{\gcd(A,B)\gcd(A,C)\gcd(B,C)} \right)^2 \right) \equiv 0 \mod(2)$$

since $\Omega(x^2)\equiv 0 \mod(2)$ for a square $x^2$.