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# Exercise 47 - Chapter 2

edited June 2018

Start with a preorder $$(P, \le)$$, and use it to define a $$\textbf{Bool}$$-category as we did in Example 2.44. In the proof of Theorem 2.46 we showed how to turn that Bool-category back into a preorder. Show that doing so, you get the preorder you started with.

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edited May 2018

Bool $$:= (\mathbb{B}, \le, \tt{false}, \wedge)$$.

The following is not the complete proof, it only shows the mapping of data and not that the rules of the two concepts are honored.

Given a preorder $$\mathcal{P} : (P, \le)$$ use it to define, map to, a Bool-category $$\mathcal{Q}$$.

The objects of $$\mathcal{P}$$ are the objects of $$\mathcal{Q}$$.

For every pair of objects $$(x, y)$$ we assign an element of $$\mathbb{B} = \{false, true\}$$: simply assigning $$\tt{true}$$ if $$x \le y$$, and $$\tt{false}$$ otherwise.

Given a Bool-category $$\mathcal{Q}$$ use it to define, map to, a preorder $$\mathcal{P}$$.

The objects of $$\mathcal{Q}$$ are the objects of $$\mathcal{P}$$.

For every pair of objects $$(x, y)$$ we observe the value of its element, $$\mathbb{B} = \{false, true\}$$: simply add the arrow in $$\mathcal{P}$$ when $$\tt{true}$$ skipping the $$\tt{false}$$ pairs.

Comment Source:**Bool** \$$:= (\mathbb{B}, \le, \tt{false}, \wedge) \$$. The following is not the complete proof, it only shows the mapping of data and not that the rules of the two concepts are honored. Given a preorder \$$\mathcal{P} : (P, \le) \$$ use it to define, map to, a **Bool**-category \$$\mathcal{Q} \$$. The objects of \$$\mathcal{P} \$$ are the objects of \$$\mathcal{Q} \$$. For every pair of objects \$$(x, y) \$$ we assign an element of \$$\mathbb{B} = \\\{false, true\\\} \$$: simply assigning \$$\tt{true}\$$ if \$$x \le y \$$, and \$$\tt{false} \$$ otherwise. Given a **Bool**-category \$$\mathcal{Q} \$$ use it to define, map to, a preorder \$$\mathcal{P} \$$. The objects of \$$\mathcal{Q} \$$ are the objects of \$$\mathcal{P} \$$. For every pair of objects \$$(x, y) \$$ we observe the value of its element, \$$\mathbb{B} = \\\{false, true\\\} \$$: simply add the arrow in \$$\mathcal{P} \$$ when \$$\tt{true}\$$ skipping the \$$\tt{false}\$$ pairs.