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Exercise 26 - Chapter 2

edited June 2018 in Exercises

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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Comments

  • 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.
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