If the statement $∀x (P (x) → Q (x))$ is FALSE then it's FALSE also $∀x P (x)$ Consider the proposition $∀x (P (x) → Q (x))$, where $P (x)$ and $Q (x)$ are predicates on a domain $U$. Prove or disprove the following statement, justifying the answer.
If the statement $∀x (P (x) → Q (x))$ is FALSE then it's FALSE also $∀x P (x)$
How can I solve this exercise?
 A: Consider the statement, 
For all $x$, if $x^2\geq 0$, then $x\geq 0$.
with the domain consisting of real numbers. This is false. What about 
For all $x$, $x^2\geq 0$?
A: If $∀x (P (x) → Q (x))$ is false then it just means there exists some $x$ such that $P(x)$ is true and $Q(x)$ is false. It does not tell us anything about weather $P(x)$ is true for every $x$ or not ... can you take it from here?
A: It's typically a good strategy to try to come up with a counterexample to the given statement: if you can find a counterexample, then the statement is false, and if you can't, then you probably get some idea why you can;t, and that might translate into a proof why the statement is in fact true.
Now, to find a counterexample to your given statement:

If the statement $∀x (P (x) → Q (x))$ is FALSE then it's FALSE also $∀x P (x)$

we should try to set  $∀x (P (x) → Q (x))$ to FALSE, but $∀x P (x)$ *not to FALSE, i.e. to TRUE.
The latter means that everything is a $P$, and the former means that not every $P$ is a $Q$.
So, can you think of a domain of objects where every object in the domain has some property (which will be $P$), but not everything has some othr property (which will be $Q$)?
