Htykuut

I know this is valid. But from LHS we can infer only for same $x$ for both $P$ and $Q$. So, I want to know how we can directly arrive at $P(a)→Q(b)$, when only information we have is $P(a)→Q(a)$ and $P(b)→Q(b)$.

If P holds for all x and for each x, P(x) implies Q(x), then Q holds for all x.
You never need P(a) implies Q(b), but you have guaranteed, on RHS that P(x) for all x, and by LHS, P(x) implies Q(x).

As an example, let the domain be {a, b}.
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))". Note that we are NOT proving "for all x (P(x) implies for all x Q(x))".

Do you mean valid in a semantic or deductive way?

Deductively it can be shown as follows:

Assume $forall x( P(x) rightarrow Q(x))$

Suppose $ forall x P(x)$,
Therefore $P(x)$ and as such $Q(x)$ by the assumption. As such $forall x Q(x)$.

That completes the proof. Since natural deduction using the usual rules is semantically valid, then so is the formula. It's valid even constructively.

For an example, let the domain be $A={a,b}$. Now assume:

$$forall x( P(x) rightarrow Q(x))$$

As such:

$$ P(a) rightarrow Q(a),P(b) rightarrow Q(b)$$

Now assume:
$P(a),P(b)$

Therefore by the above: $Q(a),Q(b)$. Since $a,b$ is the whole domain we are allowed to bring in the "for all" quantifier.

And that concludes the proof. Note that model theory would be inconsistent if this was not the case. By the above proof the sentence should hold for all possible worlds.

Finally, there is small error in the question. $P(a)rightarrow Q(b)$ is not the case.