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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, $\mathfrak{a}$ and $\mathfrak{b}$ two ideals and $\mathfrak{p}$ a prime ideal of $A$. If $\mathfrak{p}$ contains $\mathfrak{a}\cdot_d\mathfrak{b}$, then the prime ideal $\mathfrak{p}$ contains one among $\mathfrak{a}$ or $\mathfrak{b}$.

Proof. Suppose $\mathfrak{p}$ does not contain either $\mathfrak{a}$ or $\mathfrak{b}$. That is to say, there exist elements $x$ in $\mathfrak{a}$ and $y$ in $\mathfrak{b}$ such that $x$ and $y$ do not belong to $\mathfrak{p}$. Therefore $xy$ is not in $\mathfrak{p}$, as this is a prime ideal. Hence $\mathfrak{p}$ does not contain the product $\mathfrak{a}\cdot_d\mathfrak{b}$ and we get a contradiction.

Edit. Since I misread he definition of ideal in first place and it does not coincide with what I had in mind, I'd like to add some words about it and fix the corollary (see the comments).

Remarks. Notice that for any ideal $\mathfrak{a}$ of $A$, the value of $\text{coht}_d(\mathfrak{a})$ is precisely $\text{dim}(A/\mathfrak{a})$, the Krull dimension of $A/\mathfrak{a}$, when $\mathfrak{a}$ is a prime ideal and it is equal to $\text{dim}(A/\mathfrak{a}) + 1$ otherwise.

Corollary. $-$ Let $A$ be a commutative unital ring and $\mathfrak{a}$ and $\mathfrak{b}$ two ideals of $A$. Then $$\text{coht}_d(\mathfrak{a}\cdot_d\mathfrak{b}) \leq 1 + \text{max}\{\text{coht}_d(\mathfrak{a}), \text{coht}_d(\mathfrak{b})\}.$$

Proof. The previous lemma gives us the relation $$\text{dim}(A/\mathfrak{a}\cdot_d\mathfrak{b}) = \text{max}\{\text{dim}(A/\mathfrak{a}), \text{dim}(A/\mathfrak{b})\},$$ from which the inequality of the corollary immediately follows from the previous remark.

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, $\mathfrak{a}$ and $\mathfrak{b}$ two ideals and $\mathfrak{p}$ a prime ideal of $A$. If $\mathfrak{p}$ contains $\mathfrak{a}\cdot_d\mathfrak{b}$, then the prime ideal $\mathfrak{p}$ contains one among $\mathfrak{a}$ or $\mathfrak{b}$.

Proof. Suppose $\mathfrak{p}$ does not contain either $\mathfrak{a}$ or $\mathfrak{b}$. That is to say, there exist elements $x$ in $\mathfrak{a}$ and $y$ in $\mathfrak{b}$ such that $x$ and $y$ do not belong to $\mathfrak{p}$. Therefore $xy$ is not in $\mathfrak{p}$, as this is a prime ideal. Hence $\mathfrak{p}$ does not contain the product $\mathfrak{a}\cdot_d\mathfrak{b}$ and we get a contradiction.

Edit. Since I misread he definition of ideal in first place and it does not coincide with what I had in mind, I'd like to add some words about it and fix the proposition (see the comments).

Remarks. Notice that for any ideal $\mathfrak{a}$ of $A$, the value of $\text{coht}_d(\mathfrak{a})$ is precisely $\text{dim}(A/\mathfrak{a})$, the Krull dimension of $A/\mathfrak{a}$, when $\mathfrak{a}$ is a prime ideal and it is equal to $\text{dim}(A/\mathfrak{a}) + 1$ otherwise.

Proposition. $-$ Let $A$ be a commutative unital ring and $\mathfrak{a}$ and $\mathfrak{b}$ two ideals of $A$. Then $$\text{coht}_d(\mathfrak{a}\cdot_d\mathfrak{b}) \leq 1 + \text{max}\{\text{coht}_d(\mathfrak{a}), \text{coht}_d(\mathfrak{b})\}.$$

Proof. We can restrict to the case in which the $d$-coheights of the two ideals $\mathfrak{a}$ and $\mathfrak{b}$ are finite, the result being trivial otherwise. The previous lemma gives us the relation $$\text{dim}(A/\mathfrak{a}\cdot_d\mathfrak{b}) = \text{max}\{\text{dim}(A/\mathfrak{a}), \text{dim}(A/\mathfrak{b})\}.$$ Indeed, suppose wlog $n = \text{dim}(A/\mathfrak{a})\geq \text{dim}(A/\mathfrak{b})$ and consider a strict maximal chain of prime ideals $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \dots \subsetneq \mathfrak{p}_n$ of $A$, $n\geq 0$, such that $\mathfrak{a}\subseteq \mathfrak{p}_0$. If we had a prime ideal $\mathfrak{q}$ satisfying the relation $\mathfrak{a}\cdot_d \mathfrak{b} \subseteq \mathfrak{q} \subsetneq p_0$, the lemma would imply either $\mathfrak{a}\subseteq \mathfrak{q}$ or $\mathfrak{b}\subseteq \mathfrak{q}$, therefore producing a chain of primes over $\mathfrak{a}$ (resp. over $\mathfrak{b}$) of length $n+2$, which is impossible by assumption. The considerations of the above remark then immediately give the inequality regarding the $d$-coheights.

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, $\mathfrak{a}$ and $\mathfrak{b}$ two ideals and $\mathfrak{p}$ a prime ideal of $A$. If $\mathfrak{p}$ contains $\mathfrak{a}\cdot_d\mathfrak{b}$, then the prime ideal $\mathfrak{p}$ contains one among $\mathfrak{a}$ or $\mathfrak{b}$.

Proof. Suppose $\mathfrak{p}$ does not contain either $\mathfrak{a}$ or $\mathfrak{b}$. That is to say, there exist elements $x$ in $\mathfrak{a}$ and $y$ in $\mathfrak{b}$ such that $x$ and $y$ do not belong to $\mathfrak{p}$. Therefore $xy$ is not in $\mathfrak{p}$, as this is a prime ideal. Hence $\mathfrak{p}$ does not contain the product $\mathfrak{a}\cdot_d\mathfrak{b}$ and we get a contradiction.

Corollary. $-$ Let $A$ be a commutative unital ring and $\mathfrak{a}$ and $\mathfrak{b}$ two ideals of $A$. Then $$\text{coht}_d(\mathfrak{a}\cdot_d\mathfrak{b}) = \text{max}\{\text{coht}_d(\mathfrak{a}), \text{coht}_d(\mathfrak{b})\}.$$

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, $\mathfrak{a}$ and $\mathfrak{b}$ two ideals and $\mathfrak{p}$ a prime ideal of $A$. If $\mathfrak{p}$ contains $\mathfrak{a}\cdot_d\mathfrak{b}$, then the prime ideal $\mathfrak{p}$ contains one among $\mathfrak{a}$ or $\mathfrak{b}$.

Proof. Suppose $\mathfrak{p}$ does not contain either $\mathfrak{a}$ or $\mathfrak{b}$. That is to say, there exist elements $x$ in $\mathfrak{a}$ and $y$ in $\mathfrak{b}$ such that $x$ and $y$ do not belong to $\mathfrak{p}$. Therefore $xy$ is not in $\mathfrak{p}$, as this is a prime ideal. Hence $\mathfrak{p}$ does not contain the product $\mathfrak{a}\cdot_d\mathfrak{b}$ and we get a contradiction.

Edit. Since I misread he definition of ideal in first place and it does not coincide with what I had in mind, I'd like to add some words about it and fix the corollary (see the comments).

Remarks. Notice that for any ideal $\mathfrak{a}$ of $A$, the value of $\text{coht}_d(\mathfrak{a})$ is precisely $\text{dim}(A/\mathfrak{a})$, the Krull dimension of $A/\mathfrak{a}$, when $\mathfrak{a}$ is a prime ideal and it is equal to $\text{dim}(A/\mathfrak{a}) + 1$ otherwise.

Corollary. $-$ Let $A$ be a commutative unital ring and $\mathfrak{a}$ and $\mathfrak{b}$ two ideals of $A$. Then $$\text{coht}_d(\mathfrak{a}\cdot_d\mathfrak{b}) \leq 1 + \text{max}\{\text{coht}_d(\mathfrak{a}), \text{coht}_d(\mathfrak{b})\}.$$

Proof. The previous lemma gives us the relation $$\text{dim}(A/\mathfrak{a}\cdot_d\mathfrak{b}) = \text{max}\{\text{dim}(A/\mathfrak{a}), \text{dim}(A/\mathfrak{b})\},$$ from which the inequality of the corollary immediately follows from the previous remark.

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, $\mathfrak{a}$ and $\mathfrak{b}$ two ideals and $\mathfrak{p}$ a prime ideal of $A$. If $\mathfrak{p}$ contains $\mathfrak{a}\mathfrak{b}$, then the prime ideal $\mathfrak{p}$ contains one among $\mathfrak{a}$ or $\mathfrak{b}$.

Proof. Suppose $\mathfrak{p}$ does not contain neither $\mathfrak{a}$ nor $\mathfrak{b}$. That is to say, there exist elements $x$ in $\mathfrak{a}$ and $y$ in $\mathfrak{b}$ such that $x$ and $y$ do not belong to $\mathfrak{p}$. Therefore $xy$ is not in $\mathfrak{p}$, as this is a prime ideal. Hence $\mathfrak{p}$ does not contain the product $\mathfrak{a}\cdot_d\mathfrak{b}$ and we get a contradiction.

Corollary. $-$ Let $A$ be a commutative unital ring and $\mathfrak{a}$ and $\mathfrak{b}$ two ideals of $A$. Then $$\text{coht}_d(\mathfrak{a}\cdot_d\mathfrak{b}) = \text{max}\{\text{coht}_d(\mathfrak{a}), \text{coht}_d(\mathfrak{b})\}.$$

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, $\mathfrak{a}$ and $\mathfrak{b}$ two ideals and $\mathfrak{p}$ a prime ideal of $A$. If $\mathfrak{p}$ contains $\mathfrak{a}\cdot_d\mathfrak{b}$, then the prime ideal $\mathfrak{p}$ contains one among $\mathfrak{a}$ or $\mathfrak{b}$.

Proof. Suppose $\mathfrak{p}$ does not contain either $\mathfrak{a}$ or $\mathfrak{b}$. That is to say, there exist elements $x$ in $\mathfrak{a}$ and $y$ in $\mathfrak{b}$ such that $x$ and $y$ do not belong to $\mathfrak{p}$. Therefore $xy$ is not in $\mathfrak{p}$, as this is a prime ideal. Hence $\mathfrak{p}$ does not contain the product $\mathfrak{a}\cdot_d\mathfrak{b}$ and we get a contradiction.

Corollary. $-$ Let $A$ be a commutative unital ring and $\mathfrak{a}$ and $\mathfrak{b}$ two ideals of $A$. Then $$\text{coht}_d(\mathfrak{a}\cdot_d\mathfrak{b}) = \text{max}\{\text{coht}_d(\mathfrak{a}), \text{coht}_d(\mathfrak{b})\}.$$