Distributive Property of Quantifiers I know that given $$ \forall  x \;\; P(x) \wedge  \forall  x \;\; Q(x)  $$ can be simplified to $$ \forall  x \;\;  (P(x) \vee Q(x)) $$  but does the same apply if its  $ \neg \forall  x   P(x) \wedge   \neg \forall  x \;\; Q(x) $ ? Can you simplify that to $ \neg\forall  x \;\;  (P(x) \wedge Q(x)) $ ?
My main question is whether the distributive property applies when all the quantifiers are negated? 
Also can the same work with the existential quantifier?
does $$ \neg \exists x P(x) \vee \neg \exists x   Q(x) =  \neg \exists x(P(x)\vee Q(x))$$ 
 A: No, not as you have it.
$$\neg \forall x\;P(x)\;\wedge\; \neg \forall x\;Q(x) \iff \neg(\forall x\;P(x)\vee\forall x\;Q(x))$$
$$\neg \forall x\;P(x)\;\wedge\; \neg \forall x\;Q(x) \iff \exists x\;\neg P(x)\wedge\exists x\;\neg Q(x))$$
However:
$$\neg \forall x\;P(x)\;\vee\; \neg \forall x\;Q(x) \iff \neg(\forall x\;P(x)\wedge\forall x\;Q(x)) \iff \neg \forall x\;(P(x)\wedge Q(x))$$
$$\neg \forall x\;P(x)\;\vee\; \neg \forall x\;Q(x) \iff \exists x\;\neg P(x)\vee\exists x\;\neg Q(x))\iff \exists x\;(\neg P(x)\vee\neg Q(x))$$
In short: remember your Dual Negation rules.
A: here are some basic distributive properties in quantifiers, hope it might help someone.
∀x(P(x) ∧ Q(x)) ≡   (∀xP(x) ∧ ∀xQ(x))
∃x(P(x) ∧ Q(x)) →   (∃xP(x) ∧ ∃xQ(x))
∀x(P(x) ∨ Q(x)) ←   (∀xP(x) ∨ ∀xQ(x))
∃x(P(x) ∨ Q(x)) ≡   (∃xP(x) ∨ ∃xQ(x))
∀x(P(x) → Q(x)) ←   (∃xP(x) → ∀xQ(x))
∃x(P(x) → Q(x)) ≡   (∀xP(x) → ∃xQ(x))
∀x¬P(x) ≡   ¬∃xP(x)
∃x¬P(x) ≡   ¬∀xP(x)
∀x∃yT(x,y)  ←   ∃y∀xT(x,y)
∀x∀yT(x,y)  ≡   ∀y∀xT(x,y)
∃x∃yT(x,y)  ≡   ∃y∃xT(x,y)
∀x(P(x) ∨ R)    ≡   (∀xP(x) ∨ R)
∃x(P(x) ∧ R)    ≡   (∃xP(x) ∧ R)
∀x(P(x) → R)    ≡   (∃xP(x) → R)
∃x(P(x) → R)    →   (∀xP(x) → R)
∀x(R → Q(x))    ≡   (R → ∀xQ(x))
∃x(R → Q(x))    →   (R → ∃xQ(x))
∀xR ←   R
∃xR →   R
The following formulas are not valid.
   A                B           counterexample

∃x(P(x) ∧ Q(x)) ←   (∃xP(x) ∧ ∃xQ(x))   D = {a, b}, M = {P(a), Q(b)}
∀x(P(x) ∨ Q(x)) →   (∀xP(x) ∨ ∀xQ(x))   D = {a, b}, M = {P(a), Q(b)}
∀x(P(x) → Q(x)) →   (∃xP(x) → ∀xQ(x))   D = {a, b}, M = {P(a), Q(a)}
∀x∃yT(x,y)  →   ∃y∀xT(x,y)  D = {a, b}, M = {T(a,b), T(b,a)}
∃x(P(x) → R)    ←   (∀xP(x) → R)    D = Ø, M = {R}
∃x(R → Q(x))    ←   (R → ∃xQ(x))    D = Ø, M = Ø
∀xR →   R   D = Ø, M = Ø
∃xR ←   R   D = Ø, M = {R}
Note: if empty domains are not allowed, then the last four implications are in fact valid.
