Negative vs Positive Free Logic Witness

Question

Free logic comes in different flavors(*). In negative free logic a predicate with non-existent argument is supposed to fail. In positive free logic there is no such restriction

which leads to a dual universe semantics. What would be a closed free logic formula, that is a theorem of negative free logic, but not a theorem of positive free logic.

(*) Free Logic - Stanford Encyclopedia of Philosophy
https://plato.stanford.edu/entries/logic-free/

Answer 1

Ok, gotcha, a solution, is relatively simple. Take this FOL formula:

   FOL |- ~∃xp(x) => ~p(a)

It is provable in FOL. It is not provable in positive free logic.
At least this translation is not provable in FOL:

   /* shows ~∃xp(x) => ~p(a) is not provable
      in positive free logic */
   FOL |/- ~∃x(u(x) & p(x)) => ~p(a)

But it is provable in negative free logic.
At least this translation is provable in FOL:

   /* shows ~∃xp(x) => ~p(a) is provable
      in negative free logic */
   FOL |- ∀x(p(x) => u(x)) => (~∃x(u(x) & p(x)) => ~p(a))