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# Exercise 26 - Chapter 2

edited June 2018

Let $$( \mathbb{B} , \le )$$ be as $$\mathbb{B} = \{ true, false \}$$ with $$false \le true$$, but now consider the monoidal product to be $$\vee$$ (OR).

$$\begin{array}{c|cc} \vee & false & true \\ \hline false & false & true \\ true & true & true \end{array}$$ $$\begin{array}{c|cc} max & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 1 \end{array}$$ What must the monoidal unit be in order to satisfy the conditions of Definition 2.1?

Does it satisfy the rest of the conditions?

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1.
edited May 2018

The monoidal unit, call it $$I$$, should be $$false$$ because

• $$false\vee I = I\vee false= false \vee false = false$$
• $$true\vee I = true \vee false =true = false\vee true = I\vee true$$

Moreover,

• $$(false\vee false)\vee false=false=false\vee(false\vee false)$$
• $$(false\vee false)\vee true=true=false\vee(false\vee true)$$
• $$(false\vee true)\vee false=true=false\vee(true\vee false)$$
• $$(true\vee false)\vee false=true=true\vee(false\vee false)$$
• $$(false\vee true)\vee true =true=false\vee(true\vee true)$$
• $$(true\vee false)\vee true=true=true\vee(false\vee true)$$
• $$(true\vee true)\vee false=true=true\vee(true\vee false)$$
• $$(true\vee true)\vee true=true=true\vee(true\vee true)$$,

which makes $$\vee$$ associative. Hence $$\mathbb{B}$$ is a monoid under $$\vee$$.

Finally, from the monoidal product table, we can see that if $$x, x', y\in \mathbb{B}$$ and $$x\le x'$$, then $$x\vee y \le x'\vee y$$, i.e. $$\vee$$ preserve the order. Hence, $$\mathbb{B}$$ is a monoidal preorder.

Comment Source:The monoidal unit, call it \$$I\$$, should be \$$false\$$ because - \$$false\vee I = I\vee false= false \vee false = false\$$ - \$$true\vee I = true \vee false =true = false\vee true = I\vee true\$$ Moreover, - \$$(false\vee false)\vee false=false=false\vee(false\vee false)\$$ - \$$(false\vee false)\vee true=true=false\vee(false\vee true)\$$ - \$$(false\vee true)\vee false=true=false\vee(true\vee false)\$$ - \$$(true\vee false)\vee false=true=true\vee(false\vee false)\$$ - \$$(false\vee true)\vee true =true=false\vee(true\vee true)\$$ - \$$(true\vee false)\vee true=true=true\vee(false\vee true)\$$ - \$$(true\vee true)\vee false=true=true\vee(true\vee false)\$$ - \$$(true\vee true)\vee true=true=true\vee(true\vee true)\$$, which makes \$$\vee\$$ associative. Hence \$$\mathbb{B}\$$ is a monoid under \$$\vee\$$. Finally, from the monoidal product table, we can see that if \$$x, x', y\in \mathbb{B}\$$ and \$$x\le x'\$$, then \$$x\vee y \le x'\vee y\$$, i.e. \$$\vee\$$ preserve the order. Hence, \$$\mathbb{B}\$$ is a monoidal preorder.