Logic and Proof MT24, Skolemization


Flashcards

@Define what it means for a formula to be in Skolem form.


It is in the form

\[\forall x_1 \cdots \forall x_n \varphi^\ast\]

where $\varphi$ is quantifier-free.

@State and @prove a proposition that lets you convert any logical formula into an equivalent one in Skolem form.


Suppose

\[\forall y_1 \forall y_1 \cdots \forall y_n \exists y_{n+1} \varphi^\ast\]

is a sentence where $\varphi^\ast$ is quantifier free. Given a function symbol $f$ of arity $n$ that doesn’t occur in $\varphi$, write

\[\psi = \forall y_1 \forall y_2 \cdots \forall y_n \varphi^\ast [f(y_1, \ldots, y_n)/y_{n+1}]\]

Then $\varphi$ and $\psi$ are equisatisfiable.

Proof: Given a model $\mathcal A$ for $\varphi$, obtain a model $\mathcal B$ for $\psi$ by interpreting the Skolem function $f$ by emulating the existential quantifier (given $y _ 1, \ldots, y _ n$ set $f$ to whichever value of $y _ {n+1}$ satisfied the formula)

Likewise given a model $\mathcal B$ for $\psi$, obtain a model $\mathcal A$ for $\varphi$ by setting $y _ {n+1} = f(y _ 1, \ldots, y _ n)$.

@Justify why for every first-order formula $\varphi$, there is an equisatisfiable formula $\psi$ in Skolem form.


  • Translate the formula into rectified prenex form (a formula is rectified if no variable occurs both bound and free and all quantifiers in formula refer to different variables)
  • Repeatedly eliminate the outermost existential quantifier until an equisatisfiable formula in Skolem form is obtained (skolemization).

Examples

Convert

\[\forall x \exists y \forall z \exists w (\lnot P(a, w) \lor Q(f(x), y))\]

into an equisatisfiable formula in Skolem form.


\[\begin{aligned} &\forall x \exists y \forall z \exists w (\lnot P(a, w) \lor Q(f(x), y)) \text{ is satisfiable} \\\\ \text{iff }&\forall x \forall z \exists w (\lnot P(a, w) \lor Q(f(x), \pmb{g(x)})) \text{ is satisfiable} \\\\ \text{iff }&\forall x \forall z (\lnot P(a, \pmb{h(x, z)}) \lor Q(f(x), g(x))) \text{ is satisfiable} \end{aligned}\]

@example~

How would you put the following formula

\[\forall x (P(x) \lor \forall x (Q(x)))\]

into rectified prenex form?


Deeper quantifiers override shallow ones, so we may rewrite as

\[\forall x (P(x) \lor \forall y Q(y))\]

and then this becomes

\[\forall x \forall y (P(x) \lor Q(y))\]

@example~

Convert

\[(\exists x(P(x) \lor Q(x))) \to (\exists x(P(x) \lor \exists x Q(x)))\]

into an equisatisfiable formula in Skolem form.


Rename variables to rectify the formula:

\[(\exists x (P(x) \lor Q(x))) \to (\exists y (P(y) \lor \exists z Q(z)))\]

Expand the definition of $\to$:

\[\lnot (\exists x (P(x) \lor Q(x))) \lor (\exists y (P(y) \lor \exists z Q(z)))\]

Move negation inwards:

\[(\forall x(\lnot(P(x) \lor Q(x)))) \lor (\exists y (P(y) \lor \exists z Q(z)))\]

Take quantifiers out:

\[\forall x \exists y \exists z (\lnot (P(x) \lor Q(x)) \lor P(y) \lor Q(z))\]

Replace existential quantifiers with functions:

\[\forall x(\lnot (P(x) \lor Q(x)) \lor P(f(x))\lor Q(g(x,y)))\]

@example~ @exam~




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