QED
Predicate Logic · step 13 of 13

Skolemisation & clausal form

Skolemisation removes existential quantifiers by naming the witnesses. An ∃ with no universals in front becomes a fresh constant; an ∃ inside universals becomes a function of exactly those universals, recording the dependence that quantifier order encodes. The result is not logically equivalent to the original — it is equisatisfiable, which is all resolution needs.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

  1. Convert to prenex normal formRename apart and pull every quantifier to the front, preserving the alternation order.
  2. Replace each ∃ in turn, left to rightIf ∃y is preceded by universals x₁ … xₖ, replace every y with f(x₁, …, xₖ) for a fresh function symbol f. With no preceding universals, use a fresh constant.
  3. Drop the universal quantifiersAll remaining variables are implicitly universally quantified in clausal form.
  4. Convert the matrix to CNF and splitDistribute to CNF and write each conjunct as a separate clause. That set is the input to resolution.

Worked example

Skolemise ∀x ∃y ∀z ∃w R(x, y, z, w).

  1. The first ∃y is preceded by the universal x, so y becomes f(x).
  2. Substituting: ∀x ∀z ∃w R(x, f(x), z, w).
  3. The ∃w is preceded by universals x and z, so w becomes g(x, z).
  4. Substituting: ∀x ∀z R(x, f(x), z, g(x, z)); drop the universals.

Answer. R(x, f(x), z, g(x, z)), with f unary and g binary — the arities record exactly which universals each witness depended on.

Where marks get dropped

These are the specific errors that cost credit on skolemisation & clausal form questions — QED's rubric penalises each of them separately.

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Skolemisation & clausal form — frequently asked questions

Why is Skolemisation only equisatisfiable?

Because it commits to a specific choice function for the witnesses. The original may have models where no definable choice exists, but if the original has a model, a Skolem model can be built from it — enough for refutation.

What if the formula has a leading ∃?

With no universals in scope, the witness depends on nothing, so it becomes a fresh constant — a Skolem constant, which is just a zero-arity Skolem function.

Do I Skolemise before or after negating the goal?

After. Negate the conclusion first, then convert the whole set — otherwise the negation flips the quantifiers and your Skolem functions are for the wrong ones.

The rest of Predicate Logic

Quantifiers, predicates, binding, and validity. Each subtopic below has its own method, worked example and mark-losing traps.

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