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Let $X$ be a set, regarded as discrete category. if $X$ has structure of closed monoidal category $(X,\cdot,e)$, then it is easy to show that $X$ is a group: since all the morphisms are identities, the monoidal structure makes $X$ a monoid. the evaluation morphism $[x,e]\cdot x\overset{=}{\to}e$ for every $x\in X$ says that every element has left inverse, so $X$ is a group.

Now, if $X$ is instead a poset $(X,\preceq)$ and closed monoidal, $X$ is still a monoid (since every isomorphism is identity), but the evaluation morphism is read as $[x,e]\cdot x\preceq e$.
Question: is $X$ necessarily a preordered group? it seems not, but I don't have any counterexamples.

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No, there are closed monoidal posets that are not groups. For example, take any boolean algebra (or even any Heyting algebra) – then it has the structure of a symmetric monoidal category where the monoidal product is $\land$ and the internal hom is $\to$.

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