This is incorrect. You cannot assure that there are no bi-infinite chains disconnected from the starting chain, but you can assure that there no cycles disconn

This is incorrect. You cannot assure that there are no bi-infinite chains disconnected from the starting chain, but you can assure that there no cycles disconnected from the chain. To see this, note that Peano proves that, given x and y, if there exist z and w such that x+z=y and y+w=x, then x=y. Really, introducing natural numbers -- with addition and multiplication -- is overcomplicating this. There's absolutely no multiplication involved here; you could think of this as a purely additive thing, I suppose, but really directed graphs with outdegree 1 (i.e., functions on a set) are a better fit (if you want a simple model like that).