Seon-Ho Park's Blog (sunoblog) Study on Computer Science, Security, Model

13Jul/106

Quantifier Equivalences in Predicate Logic

Theorem for Quantifier Equivalences

$\neg \forall x \phi \dashv \vdash \exists x \neg \phi$
$\neg \exists x \phi \dashv \vdash \forall x \neg \phi$
$\forall x \phi \wedge \psi \dashv \vdash \forall x (\phi \wedge \psi)$
$\forall x \phi \vee \psi \dashv \vdash \forall x (\phi \vee \psi)$
$\exists x \phi \wedge \psi \dashv \vdash \exists x (\phi \wedge \psi)$
$\exists x \phi \vee \psi \dashv \vdash \exists x (\phi \vee \psi)$
$\forall x (\psi \rightarrow \phi) \dashv \vdash \psi \rightarrow \forall x \phi$
$\exists x (\phi \rightarrow \psi) \dashv \vdash \forall x \phi \rightarrow \psi$
$\forall x(\phi \rightarrow \psi) \dashv \vdash \exists x \phi \rightarrow \psi$
$\exists x(\psi \rightarrow \phi) \dashv \vdash \psi \rightarrow \exists x \phi$
$\forall x \phi \wedge \forall x \psi \dashv \vdash \forall x (\phi \wedge \psi)$
$\exists x \phi \vee \exists x \psi \dashv \vdash \exists x (phi \vee \psi)$
$\forall x \forall y \phi \dashv \vdash \forall y \forall x \phi$
$\exists x \exists y \phi \dashv \vdash \exists y \exists x \phi$

$Assuming~that~x~is~not~free~in~\psi,~for~theorem3-10.$

Tagged as: first order logic, Predicate Logic Leave a comment

Comments (6) Trackbacks (0) ( subscribe to comments on this post )

Seon-Ho Park
July 13th, 2010 - 14:55

Actually, this post is for testing WP Latex, but that theorem is also important to me.

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