John: Thanks for the kind words. It's been 12 years since I finished my CS and math degrees so I'm a bit rusty on things. Glad to see I'm not making any obvious errors yet.

Regarding latest **puzzle**:

My first thought is that a preorder that is *both* symmetric *and* antisymmetric would be a discrete preorder.

Symmetry requires that (using \\(\leq\\)): if \\(x \leq y\\) then \\(y \leq x\\). However, it permits *either* \\(x = y\\) *or* \\(x \neq y\\).

However, antisymmetry gives us: \\(x \leq y \wedge y \leq x \implies x = y\\).

Having both, then, forces every pair that is related by \\(\leq\\) to be equal to each other. This leaves us with only discrete preorders.

***

I'm expanding my last paragraph a bit. On rereading it I felt it wasn't as clear as could be.

Having both symmetry and antisymmetry, if we have \\(x,y \in X\\) and \\(x \leq y\\) then by symmetry we get \\(y \leq x\\).

With both \\(x \leq y \wedge y \leq x\\) and antisymmetry we get \\(x = y\\). This is equivalent to the text's definition for a discrete preorder on page 11.

Regarding latest **puzzle**:

My first thought is that a preorder that is *both* symmetric *and* antisymmetric would be a discrete preorder.

Symmetry requires that (using \\(\leq\\)): if \\(x \leq y\\) then \\(y \leq x\\). However, it permits *either* \\(x = y\\) *or* \\(x \neq y\\).

However, antisymmetry gives us: \\(x \leq y \wedge y \leq x \implies x = y\\).

Having both, then, forces every pair that is related by \\(\leq\\) to be equal to each other. This leaves us with only discrete preorders.

***

I'm expanding my last paragraph a bit. On rereading it I felt it wasn't as clear as could be.

Having both symmetry and antisymmetry, if we have \\(x,y \in X\\) and \\(x \leq y\\) then by symmetry we get \\(y \leq x\\).

With both \\(x \leq y \wedge y \leq x\\) and antisymmetry we get \\(x = y\\). This is equivalent to the text's definition for a discrete preorder on page 11.