The case of the ideal system d (on the multiplicative monoid of a commutative ring $A$) defined in the OP has a positive answer. In fact, the answer we get is rather strong and it is an immediate consequence of the following basic result.

**Lemma. $-$** Let $A$ be a commutative unital ring, $\mathfrak{a}_1$ and $\mathfrak{a}_2$ two ideals and $\mathfrak{p}$ a prime ideal of $A$. If $\mathfrak{p}$ contains $\mathfrak{a}_1\mathfrak{a}_2$, then the prime ideal $\mathfrak{p}$ contains one among $\mathfrak{a}_1$ or $\mathfrak{a}_2$.

*Proof.* Suppose $\mathfrak{p}$ does not contain $\mathfrak{a}_i$, $i=1, 2$. That is to say, there exist elements $x_i$ in $\mathfrak{a}_i$ such that $x_i$ does not belong to $\mathfrak{p}$, $i=1, 2$. Therefore $x_1x_2$ is not in $\mathfrak{p}$, as this is a prime ideal. Hence the prime ideal $\mathfrak{p}$ does not contain the product $\mathfrak{a}_1\mathfrak{a}_2$ and we get a contradiction.

**Corollary. $-$** Let $A$ be a commutative unital ring and $\mathfrak{a}_1$ and $\mathfrak{a}_2$ two ideals of $A$. Then
$$\text{coht}(\mathfrak{a}_1\mathfrak{a}_2) = \text{max}_{i=1}^2\{\text{coht}(\mathfrak{a}_i)\}.$$