Logic for Computer Scientists/Predicate Logic/Equivalence and Normal Forms

Equivalence and Normal Forms
Equivalence of formulae is defined as in the propositional case:

Definition 6
The formulae $$F$$ and $$G$$ are called (semantically) equivalent, if for all interpretations $$\mathcal{I}$$ for $$F$$ and $$G$$, $$\mathcal{I}(F)= \mathcal{I}(G)$$. We write $$F \equiv G$$.

The equivalences from the propositional case in theorem 4 equivalences hold and in addition we have the following cases for quantifiers.

Theorem 1
The following equivalences hold:

$$ \lnot \forall x F$$ $$ \equiv$$  $$ \exists x  \lnot F$$

$$ \lnot \exists x F$$ $$ \equiv$$ $$ \forall x \lnot F$$

If $$x$$ does not occur free in $$G$$:

$$ \forall x F \land G$$ $$ \equiv$$  $$ \forall x ( F \land G)$$

$$ \forall x F \lor G$$  $$\equiv$$  $$ \forall x ( F \lor G)$$

$$ \exists x F \land G$$ $$\equiv$$  $$ \exists x ( F \land G)$$

$$ \exists x F \lor G$$ $$\equiv$$  $$ \exists x ( F \lor G)$$

$$\forall x F \land \forall x G$$ $$ \equiv$$ $$ \forall x (F  \land G)$$

$$\exists x F \lor \exists x G$$ $$ \equiv$$  $$\exists x  (F \lor G)$$

$$\forall x \forall y \; F$$ $$\equiv$$ $$ \forall y \forall x F$$

$$\exists x \exists y \; F$$ $$ \equiv$$ $$\exists y \exists x F$$

Proof: We will prove only the equivalence $$ \forall x F \land G \equiv \forall x ( F \land G) $$ with $$x$$ has no free occurrence in $$G$$, as an example. Assume an interpretation $$\mathcal{I}= (U,\mathcal{A})$$ such that


 * $$ \mathcal{I}( \forall x F \land G) = true $$
 * $$\text{iff } \mathcal{I}(\forall x F)= true \text{ and } \mathcal{I}(G) = true $$


 * $$\text{iff } \text{ for all }  d \in U : \mathcal{I}_{[x/d]}(F) = true   \text{ and }

\mathcal{I}(G) = true $$


 * $$\text{iff } \text{ for all } d \in U : \mathcal{I}_{[x/d]}(F) = true   \text{ and }

\mathcal{I}_{[x/d]}(G) = true \text{ (x = does not occur free in = G)} $$


 * $$\text{iff } \text{ for all }  d \in U : \mathcal{I}_{[x/d]}((F \land G))= true $$


 * $$\text{iff } \mathcal{I}( \forall x ( F \land G)) = true.

$$ Note that the following symmetric cases do not hold:

$$\forall x F \lor \forall x G$$   is not equivalent to   $$\forall x  (F \lor G)$$

$$\exists x F \land \exists x G$$  is not equivalent to   $$\exists x  (F \land G)$$

The theorem for substituitivity holds as in the propositional case.

Example: Let us transform the following formulae by means of substituitivity and the equivalences from theorem 1:



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

\equiv ((\lnot \exists x P(x,y) \land \lnot \forall z Q(z)) \land \exists w  P(f(a,w))) $$

\equiv ((\forall x \lnot P(x,y) \land \exists z \lnot Q(z)) \land \exists w  P(f(a,w))) $$

\equiv ( \exists w P(f(a,w)) \land (\forall x \lnot  P(x,y) \land \exists z \lnot Q(z))) $$

\equiv \exists w ( P(f(a,w)) \land \forall x (\lnot  P(x,y) \land \exists z \lnot Q(z))) $$

\equiv \exists w ( \forall x ( \exists z \lnot Q(z) \land \lnot  P(x,y)) \land  P(f(a,w))) $$

\equiv \exists w ( \forall x   \exists z (\lnot Q(z) \land \lnot  P(x,y)) \land  P(f(a,w))) $$

\equiv \exists w  \forall x   \exists z ((\lnot Q(z) \land \lnot  P(x,y)) \land  P(f(a,w))) $$

Definition 7
Let $$F$$ be a formula, $$x$$ a variable and $$t$$ a term. $$F[x/t]$$ is obtained from $$F$$ by substituting every free occurrence of $$x$$ by $$t$$. Note, that this notion can be iterated: $$F[x/t_1][y/t_2]$$ and that $$t_1$$ may contain free occurrences of $$y$$.

Lemma 1
Let $$F$$ be a formula, $$x$$ a variable and $$t$$ a term. $$\mathcal{I}(F[x/t]) = \mathcal{I}_{[x/\mathcal{I}(t)]}(F)$$

Lemma 2 (Bounded Renaming)
Let $$F= Q x G$$ be a formula, where $$Q \in \{\forall, \exists\}$$ and $$y$$ a variable without an occurrence in $$G$$, then $$F \equiv Q y G[x/y]$$

Definition 8
A formula is called proper if there is no variable which occurs bound and free and after every quantifier there is a distinct variable.

Lemma 3 (Proper Formula)
For every formula $$F$$ there is a formula $$G$$ which is proper and equivalent to $$F$$. Proof: Follows immediately by bounded renaming. Example: $$ F = \forall x \exists y\; p(x, f(y)) \land \forall x (q(x, y) \lor r(x)) $$ has the equivalent and proper formula $$ G= \forall x \exists y\;( p(x, f(y))) \land \forall z(q(z, u) \lor r(z) $$

Definition 9
A formula is called in prenex form if it has the form $$ Q_1 \cdots Q_n F$$, where $$Q_i \in \{\forall, \exists\}$$ with no occurrences of a quantifier in $$F$$

Theorem 2
For every formula there is a proper formula in prenex form, which is equivalent. Example: $$\forall x \exists y\;( p(x, g(y, f(x))) \lor \lnot q(z)) \lor \lnot \forall x \;r(x,z)$$

$$\forall x \exists y\;( p(x, g(y, f(x)) \lor \lnot q(z)) \lor \exists x \; \lnot  r(x,z)$$

$$\forall x \exists y\;( p(x, g(y, f(x)) \lor \lnot q(z)) \lor \exists v \lnot  r(v,z)$$

$$\forall x \exists y \exists v\; ( p(x, g(y, f(x)) \lor \lnot q(z)) \lor \lnot   r(v,z)$$

Proof: Induction over the structure of the formula gives us the theorem for an atomic formula immediately.
 * $$F = \lnot F_0$$: There is a $$G_0 = Q_1 y_1 \cdots Q_n y_n G'$$ with  $$Q_i \in \{ \forall, \exists\} $$, which is equivalent to $$F_0$$. Hence we have  $$F \equiv  \overline{Q_1} y_1 \cdots \overline{Q_n} y_n \lnot G' $$

where $$\overline{Q_i} = \begin{cases} \;\;\,\exists & if Q_i = \forall \\ \;\;\, \forall & if Q_i = \exists \end{cases} $$ $$ G_1 =  Q_1 y_1 \cdots Q_k y_k G_1' $$ $$G_2 =  Q_1' z_1 \cdots Q_l' z_l G_2' $$ where $$\{y_1, \cdots, y_n\} \cap \{z_1, \cdots , z_l\} = \emptyset $$ and hence $$ F \equiv Q_1 y_1 \cdots Q_k y_k Q_1' z_1 \cdots Q_l' z_l ( G_1' \circ G_2') $$ In the following we call proper formulae in prenex form PP-formulae or PPF’s.
 * $$F = F_1 \circ F_2$$ with $$\circ \in \{\land, \lor\} $$. There exists $$G_1, G_2$$ which are proper and in prenex form and $$G_1 \equiv F_1$$  and $$G_2 \equiv F_2$$. With bounded renaming we can construct

Definition 10
Let $$F$$ be a PPF. While $$F$$ contains a $$\exists$$-Quantifier, do the following transformation: $$F$$ has the form $$ \forall y_1, \cdots \forall y_n \exists z\; G$$ where $$G$$ is a PPF and $$f$$ is a $$n$$-ary function symbol, which does not occur in $$G$$.

Let $$F$$ be $$ \forall y_1, \cdots \forall y_n\;  G[z/f(y_1, \cdots ,y_n)] $$

If there exists no more $$\exists$$-quantifier, $$F$$ is in Skolem form.

Theorem 3
Let $$F$$ be a PPF. $$F$$ is satisfiable iff the Skolem form of $$F$$ is satisfiable.

Proof: Let $$F = \forall y_1 \cdots \forall y_n \exists z\; G$$; after one transformation according to the while-loop we have $$ F' = \forall y_1 \cdots \forall y_n \; G [z/ f(y_1, \cdots, y_n) $$ where $$f$$ is a new function symbol. We have to prove that this transformation is satisfiability preserving: Assume $$F'$$ is satisfiable. than there exists a model $$\mathcal{I}$$ for $$F'$$ $$\mathcal{I} $$ is an interpretation for $$F$$. From the model property we have for all $$u_1, \cdots, u_n \in U_\mathcal{I}$$ $$ \mathcal{I}_{[y_1/u_1] \cdots [y_n/u_n]}(G[z/f(y_1, \cdots, y_n)]) = true $$ From Lemma 1 we conclude $$ \mathcal{I}_{[y_1/u_1] \cdots [y_n/u_n] [z/v]} (G) = true $$ where $$v = \mathcal{I}(f(u_1, \cdots, u_n)$$. Hence we have, that for all $$u_1, \cdots , u_n \in U_\mathcal{I}$$ there is  a $$ v \in U_\mathcal{I}$$, where  $$ \mathcal{I}_{[y_1/u_1] \cdots [y_n/u_n] [z/v]} (G) = true $$ and hence we have, that $$\mathcal{I}(\forall y_1 \cdots \forall y_n \exists z\; G) = true$$, which means, that $$\mathcal{I}$$ is a model for $$F$$.

For the opposite direction of the theorem, assume that $$F$$ has a model $$\mathcal{I}$$. Then we have, that for all $$u_1, \cdots, u_n \in U_\mathcal{I}$$, there is a $$ v \in U_\mathcal{I}$$, where $$ \mathcal{I}_{[y_1/u_1] \cdots [y_n/u_n] [z/v]} (G) = true $$ Let $$\mathcal{I'}$$ be an interpretation, which deviates from $$\mathcal{I}$$ only, by the fact that it is defined for the function symbol $$f$$, where $$\mathcal{I}$$ is not defined. We assume that $$f^{\mathcal{I}'}(u_1, \cdots, u_n) = v$$, where $$v$$ is chosen according to the above equation.

Hence we have that for all $$u_1, \cdots, u_n \in U_\mathcal{I}'$$ $$ \mathcal{I}'_{[y_1/u_1] \cdots [y_n/u_n][z/f^{\mathcal{I}'}(u_1, \cdots, u_n)]}(G) = true $$ and from Lemma 1 we conclude that for all $$u_1, \cdots, u_n \in U_\mathcal{I}$$ $$ \mathcal{I}'_{[y_1/u_1] \cdots [y_n/u_n]}(G[z/f^(y_1, \cdots, y_n)]) = true $$ which means, that $$\mathcal{I}'(\forall y_1 \cdots \forall y_n \; G[z/f(y_1, \cdots, y_n)]) = true$$, and hence $$\mathcal{I}'$$ is a model for $$F'$$. The above results can be used to transform a Formula into a set of clauses, its clause normal form:

Problem 6 (Predicate)
Let $$F$$ be a formula, $$x$$ a variable and $$t$$ a term. Then $$F[x/t]$$ denotes the formula which results from $$F$$ by replacing every free occurrence of $$x$$ by $$t$$. Give a formal definition of this three argument function $$F[x/t]$$,  by induction over  the structure of the formula $$F$$. $$\Box$$

Problem 7 (Predicate)
Show the following semantic equivalences: $$\Box$$
 * 1) $$\forall x (p(x) \to (q(x) \land r(x))) \equiv \forall x (p(x) \to q(x)) \land \forall x (p(x) \to r(x))$$
 * 2) $$\forall x (p(x) \to (q(x) \lor r(x))) \not\equiv \forall x (p(x) \to q(x)) \lor \forall x (p(x) \to r(x))$$

Problem 8 (Predicate)
Show the following semantic equivalences:
 * 1) $$(\forall x p(x)) \to q(b) \equiv \exists x (p(x) \to q(b))$$
 * 2) $$(\forall x r(x)) \lor (\exists y \lnot r(y)) \equiv (\forall x r(x)) \to (\exists y r(y))$$

$$\Box$$

Problem 9 (Predicate)
Show that for arbitrary formulae $$F$$ and $$G$$, the following holds: $$\Box$$
 * 1) $$\forall x (F \lor G) \not\equiv \forall x F \lor \forall x G$$
 * 2) If $$G = F[x/t]$$, than $$G \models \exists x F$$.