3

What is the skolemized form of (∀u∃v a(u,v)) ∧ (∀x∃y a(x,y)) ?

I am unsure, because there are different perenex normal forms possible:

  • ∀u∃v ∀x∃y (a(u,v) ∧ a(x,y))
  • ∀u∀x ∃v∃y (a(u,v) ∧ a(x,y))
  • …

There would different skolemized forms follow:

  • ∀u ∀x (a(u,f(u)) ∧ a(x,g(u,x)))
  • ∀u∀x (a(u,f(u,x)) ∧ a(x,g(u,x)))

In my mind, they are not equivalent to each other. Or am I here wrong?

Michael0x2a
  • 58,192
  • 30
  • 175
  • 224
Jojo
  • 51
  • 1
  • 5

1 Answers1

2

Yes, prenex normal forms are not unique for a given FO formula, and, correspondingly, Skolemizations are not unique. A simpler example for the same "scope escape" I think you are trying to show is the formula ∃xAx → ∃yBy, with prenex forms ∀x∃y (Ax → By) and ∃y∀x (Ax → By), and respective skolemizations ∀x (¬ Ax ∨ Bf(x)) and ∀x (¬ Ax ∨ B a), with a a constant.

Now, the pertinent question is whether the non-equivalence of those formulae matters for your particular problem. If it does, perhaps Skolemization is not the best tool for you: Skolemization is a process designed to preserve satisfiability of formulae, sometimes at the expense of equivalence.

(and in any case, it is a good exercise to see why distinct skolemizations of a single formula are equisatisfiable, if only on the examples above)

Francois G
  • 11,957
  • 54
  • 59
  • I'm thinking about this problem, because I have a bit more complex problem, that I try to solve. The intention is to use resolution to proof if the knowledge base can let to a given assumption. So the basic formula I try to disprove with resolution is "A ∧ B ∧ ¬C". I heared, that I can skolemize the parts of the formula one by one (1. "A", 2. "B" and 3. "¬C") and put them then together. But I don't know if this is true and if it is, why? What allows me to do this? I could figure that, if I do a part-skolemization then, rearange ∀s/∃s and skolemize the rest, but I don't know if I could do that… – Jojo Dec 27 '11 at 18:14