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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.

## Comments

Recall that a

skeletal dagger preorderis a preorder such that: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 \(\wedge\).

`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\\).`