> With regard to the monoidal product distributing over the joins, note here how the two languages do not mesh together well. In the lattice theoretic world, you impose that the monoidal product distributes over joins, whereas in the category theoretic world you impose that the monoidal product preserves the order. Are these equivalent?

No, they are not equivalent.

I believe if \\((X,\leq, \vee, \otimes)\\) has a monoidal product which distributes over joins, then it must have a monoidal product which preserves order.

**Proof.**

Let \\(x \leq y\\). We must show \\(z \otimes x \leq z \otimes y\\).

The assumption \\(x \leq y\\) is equivalent to \\(x \vee y = y\\), which means

\[z \otimes (x \vee y) = z \otimes y . \tag{★}\]

Since the monoidal product here distributes, we know \\(z \otimes (x \vee y) = (z \otimes x) \vee (z \otimes y)\\), which together with (★) means \\((z \otimes x) \vee (z \otimes y) = z \otimes y\\). But this is equivalent to \\(z \otimes x \leq z \otimes y\\), which is what we wanted to show. \\( \qquad \blacksquare \\)

The converse is not true, as you hint below:

> Is there a monoidal preorder in which the monoidal product does not distribute over joins?

Yes. John Baez asked a similar question in [Lecture 61](https://forum.azimuthproject.org/discussion/2291/lecture-61-chapter-4-closed-monoidal-posets)

> **Puzzle 194.** From [Lecture 11](https://forum.azimuthproject.org/discussion/1991/lecture-11-chapter-1-the-poset-of-partitions/p1) we know that for any set \\(X\\) the set of partitions of \\(X\\), \\(\mathcal{E}(X)\\), becomes a poset with \\(P \le Q\\) meaning that \\(P\\) is finer than \\(Q\\). It's a monoidal poset with product given by the meet \\(P \wedge Q\\). Is this monoidal poset closed?

I tried to show in [comment #13](https://forum.azimuthproject.org/discussion/comment/20139/#Comment_20139) that this monoidal poset did not always have a product that distributes over joins. I used this and the adjoint functor theorem to argue \\(\mathcal{E}(X)\\) is not closed in general.

No, they are not equivalent.

I believe if \\((X,\leq, \vee, \otimes)\\) has a monoidal product which distributes over joins, then it must have a monoidal product which preserves order.

**Proof.**

Let \\(x \leq y\\). We must show \\(z \otimes x \leq z \otimes y\\).

The assumption \\(x \leq y\\) is equivalent to \\(x \vee y = y\\), which means

\[z \otimes (x \vee y) = z \otimes y . \tag{★}\]

Since the monoidal product here distributes, we know \\(z \otimes (x \vee y) = (z \otimes x) \vee (z \otimes y)\\), which together with (★) means \\((z \otimes x) \vee (z \otimes y) = z \otimes y\\). But this is equivalent to \\(z \otimes x \leq z \otimes y\\), which is what we wanted to show. \\( \qquad \blacksquare \\)

The converse is not true, as you hint below:

> Is there a monoidal preorder in which the monoidal product does not distribute over joins?

Yes. John Baez asked a similar question in [Lecture 61](https://forum.azimuthproject.org/discussion/2291/lecture-61-chapter-4-closed-monoidal-posets)

> **Puzzle 194.** From [Lecture 11](https://forum.azimuthproject.org/discussion/1991/lecture-11-chapter-1-the-poset-of-partitions/p1) we know that for any set \\(X\\) the set of partitions of \\(X\\), \\(\mathcal{E}(X)\\), becomes a poset with \\(P \le Q\\) meaning that \\(P\\) is finer than \\(Q\\). It's a monoidal poset with product given by the meet \\(P \wedge Q\\). Is this monoidal poset closed?

I tried to show in [comment #13](https://forum.azimuthproject.org/discussion/comment/20139/#Comment_20139) that this monoidal poset did not always have a product that distributes over joins. I used this and the adjoint functor theorem to argue \\(\mathcal{E}(X)\\) is not closed in general.