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# Exercise 81 - Chapter 2

edited June 2018

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

• Options
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\$$.
• Options
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 \$$