Derivation of Null Quantification in Logic? I was reading page 10-8 of this: https://faculty.washington.edu/smcohen/120/Chapter10.pdf
and I was wondering if the distributive qualities could be derived, e.g. $\forall x (P \lor Q(x)) \Leftrightarrow P \lor \forall x Q(x)$, and the equivalent one for $\exists , \land$.
 A: The following proof uses the rules of Chapter 12: Methods of Proof for Quantifiers.
I'll consider only the case :  $∀x(P∨Q(x)) \Rightarrow P∨∀xQ(x)$; the other one is similar.
(1)  $∀x(P∨Q(x))$ --- assumed
(2)  $P∨Q(a)$ --- from (1) by universal instantiation, or $\forall$-elim
Now we need some "propositional" transformation : we use the equivalence between $A \lor B$ and $\lnot A \Rightarrow B$
(3) $\lnot P \Rightarrow Q(a)$ --- from (2) by tautological equivalence
(4) $\lnot P$ --- assumed
(5) $Q(a)$ --- by $\Rightarrow$-elim (or modus ponens)
(6) $\forall xQ(x)$ --- by universal introduction (or $\forall$-intro) : the constant $a$ is not free in the assumptions
(7) $\lnot P \Rightarrow \forall xQ(x)$ --- from (4) and (6) by $\Rightarrow$-intro
(8) $P \lor \forall xQ(x)$ --- from (7) by tautological equivalence

(9) $∀x(P∨Q(x)) \Rightarrow P \lor \forall xQ(x)$ --- from (1) by $\Rightarrow$-intro.


Note
If we don't want to use the tautological equivalence in steps (3) and (8), we can use a "proof by cases".
From step :
(2) $P \lor Q(a)$
(3) $P$ --- assumed for $\lor$-elim
(4) $P \lor \forall x Q(x)$ --- from (3) by $\lor$-intro
(5) $Q(a)$ --- assumed for $\lor$-elim
(6) $\forall xQ(x)$ --- by universal introduction (or $\forall$-intro) : the constant $a$ is not free in the assumptions
(7) $P \lor \forall x Q(x)$ --- from (6) by $\lor$-intro
(8) $P \lor \forall x Q(x)$ --- from (2), (4) and (7) by $\lor$-elim, "discharging" the temporary assumptions (3) and (5). 
Theorem follows as above, by $\Rightarrow$-intro.

Added
The other "direction" is easier.
From the assumption : $P \lor \forall x Q(x)$, we derive from both $P$ and $\forall xQ(x)$ separately : $P \lor Q(a)$ [in one case by $\lor$-intro; in the other case by $\forall$-elim followed by $\lor$-intro].
Then we apply again proof by cases ($\lor$-elim) to conclude from : $P \lor \forall x Q(x)$ to : $P \lor Q(a)$.
Finally we apply $\forall$-intro to derive : $\forall x(P \lor Q(x))$ followed by a final $\Rightarrow$-intro.

Comment
Please note that, in general : $\forall xP(x) \lor \forall xQ(x) \vdash \forall x(P(x) \lor Q(x))$, but not vice versa, i.e. :


$\forall x(P(x) \lor Q(x)) \nvdash \forall xP(x) \lor \forall xQ(x)$.


In other words, the above proof does not works if $x$ is free in $P$.
Why so ?
Because (consider the first proof above) in step (6) we have to apply $\forall$-intro to $Q(a)$ to get $\forall xQ(x)$.
If $x$ is free in $P$ [and this is so if we try to apply the proof to the assumption $\forall x(P(x) \lor Q(x))$, because in step (2) we have instantiated it to $P(a) \lor Q(a)$], if we try to apply $\forall$-intro we have to violate the proviso that the constant $a$ must not be free in the assumptions, and at that step of the proof we still have an "undischarged" assumption containing $a$ : exactly $P(a)$.
A: It doesn't seem like the notes you're reading define any deductive system for FOL at all, and you can't derive anything before you have a deductive system to do it in.
However, once you have a deductive system for FOL (of which there are several to choose among), you will be able to derive these laws in it. Otherwise the system you have is not FOL!
