• Se x non occorre libera in P allora

$$ \forall x P\Leftrightarrow_{FO}P\Leftrightarrow_{FO}\exist x P $$

• Dunque

$$ \exist x\forall x P(x) \Leftrightarrow_{FO} \forall x \forall xP(x) \Leftrightarrow_{FO}\forall x P(x) \\ \exist x\exist xP(x)\Leftrightarrow_{FO}\forall x\exist x P(x) \Leftrightarrow_{FO} \exist x P(x) $$

$$ \forall x (P \vee Q(x))\Leftrightarrow_{FO}P\vee\forall x Q(x) \\ \exist x (P\wedge Q(x))\Leftrightarrow_{FO}P\wedge\exist x Q(x) $$