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# Exercise 90 - Chapter 2

edited June 2018

Show that $$\textbf{Bool} = ( \mathbb{B} , \le, true, \wedge )$$ is a quantale.

A unital commutative quantale is a symmetric monoidal closed preorder $$\mathcal{V}=(V,\leq,I,\otimes)$$ that has all joins: $$\vee A$$ exists for every $$A\subseteq V$$. In particular, we denote the empty join $$0:=\vee\emptyset$$.
In exercise 81 it was shown that Bool is monoidal closed. Ex. 2.24 shows that it is symmetric monoidal (and thus symmetric). So, it remains to show that Bool has all joins. Let $$A\subseteq\mathbb{B}$$, then $$A\in\{\emptyset,\{\mathrm{true}\},\{\mathrm{false}\},\{\mathrm{true},\mathrm{false}\}\}$$. The joins of these sets are $$\mathrm{false},\mathrm{true},\mathrm{false}$$, and $$\mathrm{true}$$, respectively. Since these joins all exist and lie in $$\mathbb{B}$$, Bool is a quantale.
Comment Source:**Definition 2.87** A _unital commutative quantale_ is a symmetric monoidal closed preorder \$$\mathcal{V}=(V,\leq,I,\otimes)\$$ that has all joins: \$$\vee A\$$ exists for every \$$A\subseteq V\$$. In particular, we denote the empty join \$$0:=\vee\emptyset\$$. In exercise 81 it was shown that **Bool** is monoidal closed. Ex. 2.24 shows that it is symmetric monoidal (and thus symmetric). So, it remains to show that **Bool** has all joins. Let \$$A\subseteq\mathbb{B}\$$, then \$$A\in\\{\emptyset,\\{\mathrm{true}\\},\\{\mathrm{false}\\},\\{\mathrm{true},\mathrm{false}\\}\\}\$$. The joins of these sets are \$$\mathrm{false},\mathrm{true},\mathrm{false}\$$, and \$$\mathrm{true}\$$, respectively. Since these joins all exist and lie in \$$\mathbb{B}\$$, **Bool** is a quantale.