3-SAT mixed with 2-SAT formulas Context: Refering to the  question: Complexity of the $(3,2)_s$ SAT problem? and since the paper by Porshen and Speckenmayer : Satisfiability of mixed Horn formulas, we know that even when $F_3$ is Horn, the problem of deciding the satisfiability of $F_3 \wedge F_2$ is NP-complete - where $F_3$ and $F_2$ are respectively 3-CNF and 2-CNF formulas.
I am wondering if there exist some cases where $F_3 \wedge F_2$ is easy to decide.
Hence my question:
Let $F_3$ a 3-CNF containing only clauses with exactly 3 different literals and $F_2$ a 2-CNF defined on the same variables as $F_3$.
What is the complexity of deciding the satisfiability of $F_3 \wedge F_2$ when $F_3$ and $F_2$ are both monotone?
Thank you.
If $F_3$ and $F_2$ are both monotone, satisfiability can be checked in polynomial time (or even in coNLOGTIME), as $F_3\land F_2$, which is also monotone, is satisfiable iff it is satisfied by the $\vec1$ assignment, that is, iff it does not contain the empty clause.
If one of the formulas is allowed to be monotone and the other one negated monotone (i.e., having only negative literals), then satisfiability of $F_3\land F_2$ is NP-complete: given a 3-CNF $F$ in variables $x_1,\dots,x_n$, let $F_3$ be the monotone 3-CNF obtained from $F$ by replacing all negative literals $\neg x_i$ with new variables $y_i$, and let $F_2=\bigwedge_i(\neg x_i\lor\neg y_i)$. Then $F$ is equisatisfiable with $F_3\land F_2$.
In particular, if $(\vec x,\vec y)$ is a satisfying assignment for $F_3\land F_2$, then for each $i$, at most one of $x_i$ or $y_i$ gets value $1$, that is, $y_i\le\neg x_i$. Thus, if we modify the assignment so that $y_i:=\neg x_i$, it will still satisfy $F_3$ as it is monotone. It follows that $\vec x$ satisfies the original formula $F$.