> **Puzzle 195.** Show that in any closed monoidal poset we have \\( I \multimap x = x \\) for every element \\(x\\).

By the anti-symmetry property of posets we need to show:

1. \\( x \le I \multimap x \\)

2. \\( I \multimap x \le x\\).

For the first statement, by reflexivity we have that

\[ x \le x \]

and we apply the monoidal unit \\(I\\) to the left side

\[ I \otimes x \le x \]

and by the definition of closed we get

\[ x \le I \multimap x .\]

For the second statement, by reflexivity we have that

\[ I \multimap x \le I \multimap x \]

and we apply the definition of closed

\[ I \otimes (I \multimap x) \le x \]

and use the fact that \\(I\\) is the monoidal unit to get

\[ I \multimap x \le x .\]

By the anti-symmetry property of posets we need to show:

1. \\( x \le I \multimap x \\)

2. \\( I \multimap x \le x\\).

For the first statement, by reflexivity we have that

\[ x \le x \]

and we apply the monoidal unit \\(I\\) to the left side

\[ I \otimes x \le x \]

and by the definition of closed we get

\[ x \le I \multimap x .\]

For the second statement, by reflexivity we have that

\[ I \multimap x \le I \multimap x \]

and we apply the definition of closed

\[ I \otimes (I \multimap x) \le x \]

and use the fact that \\(I\\) is the monoidal unit to get

\[ I \multimap x \le x .\]