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Exercise 60 - Chapter 1

edited June 2018

Show that a skeletal dagger preorder is just a discrete preorder, and hence just a set.

A dagger preorder is a preorder obeying the symmetry axiom:

$$x \le y \Leftrightarrow y \le x$$ Also recall the skeletal preorders (Remark 1.26) and discrete preorders (Example 1.27):

Remark 1.26 (Partial orders are skeletal preorders).

A preorder is a partial order if we additionally have that 3. $$x \cong y$$ implies $$x = y$$.

Example 1.27 (Discrete preorders).

Every set $$X$$ can be made into a discrete preorder. This means that the only order relations on $$X$$ are of the form $$x \le x$$; if $$x \neq y$$ then neither $$x \le y$$ or $$y \le x$$ hold.

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1.
edited April 2018

Recall that a skeletal dagger preorder is a preorder such that:

1. $$x \le y \wedge y \le x \Rightarrow x = y$$, and
2. $$x \le y \Leftrightarrow y \le x$$.

Combining these properties(*), we have that $$x \le y \Rightarrow x = y$$, which is the defining property of a discrete preorder.

(*) Since $$x \le y$$ and $$y \le x$$ are equivalent by (2), we can remove $$y \le x$$ from the precondition on (1). Formally, this involves substitution followed by idempotence of $$\wedge$$.

Comment Source:Recall that a _skeletal dagger preorder_ is a preorder such that: 1. \$$x \le y \wedge y \le x \Rightarrow x = y\$$, and 2. \$$x \le y \Leftrightarrow y \le x\$$. Combining these properties(\*), we have that \$$x \le y \Rightarrow x = y\$$, which is the defining property of a discrete preorder. (\*) Since \$$x \le y\$$ and \$$y \le x\$$ are equivalent by (2), we can remove \$$y \le x\$$ from the precondition on (1). Formally, this involves substitution followed by idempotence of \$$\wedge\$$.