Counting reduction from #SAT to #HornSAT? Is it possible to find a counting reduction from #SAT to #HornSAT? I haven't found this question posted here, so decided to check if anyone has any answer to this. Let me explain what do I mean by counting reduction. 

Suppose $f,g : \{0,1\}^* \to \mathbb N$ are two counting problems. For example, #SAT asks how many satisfiable assignments are there for a specific instance $\phi$, and $f,g$ are similar counting problems finding total number of witnesses. A weakly parsimonious counting reduction from $f$ to $g$ consists of a pair of polynomial time computable functions $\sigma : \{0,1\}^* \to \{0,1\}^*$ and $\tau : \{0,1\}^* \times \mathbb N \to \mathbb N$ such that $f(x) = \tau (x, g(\sigma(x)))$. In the case that $f(x) = g(\sigma(x))$, this is known as strongly parsimonious counting reduction.

I can see that if there is any such counting reduction from #SAT to #HornSAT, it must be weakly parsimonious reduction: a strong reduction would imply that the #SAT and #HornSAT instances will have zero or non-zero number of solutions together, and assuming that $\mathsf P \ne \mathsf{NP}$, this is impossible (as HornSAT $\in \mathsf P$ while SAT is $\mathsf{NP}$-complete).

So my question is: is there any weakly parsimonious counting reduction from #SAT to #HornSAT? If so, can anyone please give me some reference ?
This post discusses #P-completeness of #Monotone-2SAT under weakly parsimonious reductions. If you negate all literals in a monotone 2-CNF formula $\phi$, you obtain a Horn 2-CNF formula $\psi$ with the same number of satisfying assignments.