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Exercise 81 - Chapter 2

edited June 2018 in Exercises

Show that \( \textbf{Bool} = ( \mathbb{B} , \le, true, \wedge ) \) is monoidal closed.

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  • 1.

    Definining \((true \multimap false) = false\) and \((true \multimap true) = (false \multimap true) = (false \multimap false) = true\) should work.

    1. the falsity of \( (true \wedge true) \leq false\) requires the falsity of \(true \leq (true \multimap false)\), hence the falsity of \( (true \multimap false)\).
    2. the truth of \( (true \wedge true) \leq true\) requires the truth of \(true \leq (true \multimap true)\), hence the truth of \( (true \multimap true)\).
    3. the truth of \( (true \wedge false) \leq x\) for all \(x \in \mathbb{B}\) requires the truth of \(true \leq (false \multimap x)\), hence the truth of \( (false \multimap x)\) for all \(x \in \mathbb{B}\).
    4. the truth of all expressions of the form \( (false \wedge x) \leq y\) and \( false \leq z\) for any x,y,z in \(\mathbb{B}\) implies that they give no further restrictions on \(\multimap\).
    Comment Source:Definining \\((true \multimap false) = false\\) and \\((true \multimap true) = (false \multimap true) = (false \multimap false) = true\\) should work. 1. the falsity of \\( (true \wedge true) \leq false\\) requires the falsity of \\(true \leq (true \multimap false)\\), hence the falsity of \\( (true \multimap false)\\). 2. the truth of \\( (true \wedge true) \leq true\\) requires the truth of \\(true \leq (true \multimap true)\\), hence the truth of \\( (true \multimap true)\\). 3. the truth of \\( (true \wedge false) \leq x\\) for all \\(x \in \mathbb{B}\\) requires the truth of \\(true \leq (false \multimap x)\\), hence the truth of \\( (false \multimap x)\\) for all \\(x \in \mathbb{B}\\). 4. the truth of all expressions of the form \\( (false \wedge x) \leq y\\) and \\( false \leq z\\) for any x,y,z in \\(\mathbb{B}\\) implies that they give no further restrictions on \\(\multimap\\).
  • 2.

    More simply, \((v \multimap w) :=\neg (v \land \neg w) = \neg v \lor w \)

    Comment Source:More simply, \\((v \multimap w) :=\neg (v \land \neg w) = \neg v \lor w \\)
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