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.
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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.