The three-element generated subalgebra conjecture for clone-minimal dispersive algebras

Let $\mathbb{A}$ be a clone-minimal dispersive algebra, and let $a,b\in\mathbb{A}$ with $\operatorname{Sg}\{a,b\}$ denoting the subalgebra they generate. Suppose there is an element $c\in\operatorname{Sg}\{a,b\}$ such that, for every $d\in\operatorname{Sg}\{a,b\}$, the set $\{c,d\}$ is a two-element projection subalgebra. In the direct square $\mathbb{A}^2$, let $\operatorname{Sg}_{\mathbb{A}^2}\{(a,b),(b,a)\}$ denote the subalgebra generated by $(a,b)$ and $(b,a)$.

Three-element generated subalgebra conjecture. Under these hypotheses,

$$\operatorname{Sg}\{a,b\}=\{a,b,c\}.$$

It would suffice to prove that at least one of $(a,c)$ and $(b,c)$ belongs to

$$\operatorname{Sg}_{\mathbb{A}^2}\{(a,b),(b,a)\}.$$

This is presented among the paper's difficult conjectures on clone-minimal dispersive algebras, and the source gives no evidence of a resolution.