Which of these statements are always true? this problem has been asked on mathexchange already, but i fail to understand why only the second option is always true:
1.$((\forall x(P(x) \lor Q(x)))) \implies ((\forall xP(x)) \lor (\forall xQ(x)))$
2.$(\forall x(P(x) \implies Q(x))) \implies ((\forall xP(x)) \implies (\forall xQ(x)))$
3.$(\forall x(P(x)) \implies \forall x(Q(x))) \implies (\forall x(P(x) \implies Q(x)))$
4.$(\forall x(P(x)) \iff (\forall x(Q(x)))) \implies (\forall x(P(x) \iff Q(x)))$
To make the formulas easier to read, because each predicate has the same subject $(x)$, let $P(x) = A$ and $Q(x) = B$. Because in each formula the predicates are expressed in such a way that only universal quantifiers are used, though I'm not sure, my guess is they can be discarded (by mentally 'factoring out' to the beginnings of the expressions). In that case I get the following:

*

*$(((A \lor B))) \implies ((A) \lor (B)) = \lnot (A \lor B) \lor (A \lor B) = (\lnot A \lor \lnot B) \lor (A \lor B) = (A \land B) \lor (A \lor B)$
this is not tautology.


*$((A \implies B)) \implies ((A) \implies (B)) = \lnot (\lnot A \lor B) \lor (\lnot A \lor B) = (A \lor \lnot B) \lor (\lnot A \lor B)$
this is tautology.


*$((A) \implies (B)) \implies ((A \implies B)) = \lnot (\lnot (A) \lor (B)) \lor ((\lnot A \lor B)) = \lnot (\lnot A \lor B) \lor (\lnot A \lor B) = (A \lor \lnot B) \lor (\lnot A \lor B)$
this is tautology.


*$((A) \iff ((B))) \implies ((A \iff B))$
this is tautology (implicated identity).
 A: Using your procedure to “factorise” out quantifiers from a formula does change its meaning. For example, assuming a non-empty domain of discourse,
$$\bigg(\forall x \big[\,P(x)\to Q(x)\,\big]\bigg) \kern.8em\not\kern-.8em\iff \bigg(\forall x P(x)\to\forall x Q(x)\bigg) \\ \iff \bigg(\forall y \exists x \big[\,P(x)\to Q(y)\,\big] \bigg) \\ \iff \bigg(\exists x \forall y \big[\,P(x)\to Q(y)\,\big] \bigg).$$
A: When you "factor out" the quantifier, you are applying something like a "law of distribution", which however does not exist for this.
To show you why for example equation 1) is not true, I'll let the domain over which $x$ varies be the positive integers, and $P(x)$ means "$x$ is the square of an integer" and $Q(x)$ means "x is not the square of an integer".
Then
$$\forall x (P(x) \lor Q(x))$$
means "for all positive integers $x$, $x$ is the square of an integer or not the square of an integer", which is true: Any positive integer is either a square or not.
However, $(\forall x P(x)) \lor (\forall x Q(x))$ means "all positive integers are squares of integers or all positive square integers are not squares of integers", which is wrong: It's an "or"-Statement, and both parts are false, as some positve integers are not squares (so $\forall x P(x)$ is false) and some positve integers are squares (so $\forall x Q(x)$ is false).
