Dan wrote:

> I don't believe that the set of all legal chess positions, with \\(\leq\\) meaning "can produce by legal moves", is a preorder set, since it does not satisfy the reflexivity condition. For example, if the current position \\(x\\) is a check in which the only legal response is a pawn move, then \\(x\nleq x\\). (Maybe this can be fixed by allowing the number of intervening legal moves to be 0, and perhaps that's what you meant anyway.)

I hope that's what he meant. Here's what a category theorist would say if they weren't trying to be polite:

> The most reasonable interpretation of "can produce by legal moves" is "can produce by some natural number of legal moves" - and any mathematician who doesn't count 0 as a natural number ought to be taken out and shot, along with those who don't think the empty set is a set.

We take nothing seriously. _Very_ seriously. Math works much more smoothly if you treat degenerate cases, like zero and the empty set, with the same respect as nondegenerate ones.

> I don't believe that the set of all legal chess positions, with \\(\leq\\) meaning "can produce by legal moves", is a preorder set, since it does not satisfy the reflexivity condition. For example, if the current position \\(x\\) is a check in which the only legal response is a pawn move, then \\(x\nleq x\\). (Maybe this can be fixed by allowing the number of intervening legal moves to be 0, and perhaps that's what you meant anyway.)

I hope that's what he meant. Here's what a category theorist would say if they weren't trying to be polite:

> The most reasonable interpretation of "can produce by legal moves" is "can produce by some natural number of legal moves" - and any mathematician who doesn't count 0 as a natural number ought to be taken out and shot, along with those who don't think the empty set is a set.

We take nothing seriously. _Very_ seriously. Math works much more smoothly if you treat degenerate cases, like zero and the empty set, with the same respect as nondegenerate ones.