Logical squabbles I need some help with the follwing: Lets suppose that the sentence $\forall x: x\in I \rightarrow P(x)$ is false.
Now consider the sentence $$\forall x: x\in I \rightarrow (P(x) \ \& \ Q(x) ) \ \ \quad (1)$$ for any property $Q(x)$, which is also obviously false. 
But now I can define a set $I'=\left\{ x \in I| P(x) \right\}$.
So the sentence (1) should be equivalent to the sentence $$\forall x: x\in I' \rightarrow Q(x) $$
But since sentence is now true, since there aren't any $x$ such that $x \in I'$, because the first sentence was supposed to be false.
I'm sure the error is, that the two sentences are equivalent. But I can't pinpoint my error. Could someone tell me crystal-clear what I am doing wrong ?
 A: The sentence with the colon (first line) is not a sentence in any formal language that I am acquainted with.  Presumably it is meant to assert what would be ordinarily written as
$$\left(\forall x\right)\left(x\in I \implies P(x)\right).$$
We are told that this sentence is false.
So there is an element $a$ of $I$ such that $P(a)$ is false.  There may also be many elements $b$ of $I$ such that $P(b)$ is true.
Later, $I'$ is defined as the subset of $I$ consisting of the $x$ in $I$ such that $P(x)$ is true.  It is then asserted that $I'$ is empty.  But there is no reason to conclude that $I'$ is empty.  
Example: For example, let $I$ be the set of positive integers.  Let $P(x)$ be the assertion that $x$ is prime.  Then the assertion
$$\left(\forall x\right)\left(x\in I \implies P(x)\right)$$
is false, since there are non-primes.  But then $I'$ is the set of primes, which is demonstrably non-empty. 
Analysis:  Why the mistake?  The OP is experienced enough not to make an elementary error.  The problem is with the bad notation, which is a hybrid between two not unreasonable standard notations.  The first has already been used.  The second standard notation, in a corrected version, would read:
$$\left(\forall x: x\in I\right) (P(x)).$$
Note the absence of the implication symbol.  
A: It's a little neater to avoid $I'$ altogether and write your second displayed expression as $(\forall x \in I)[\lnot P(x) \to Q(x)]$; this captures the same idea.
I think that you've confused conjunction ('and') and disjunction ('or'). If your $(1)$ had been $(\forall x \in I)[P(x) \lor Q(x)]$ instead of $(\forall x \in I)[P(x) \land Q(x)]$, it would have been logically equivalent to $(\forall x \in I)[\lnot P(x) \to Q(x)]$, simply because $P(x) \lor Q(x)$ is logically equivalent to $\lnot P(x) \to Q(x)$. If at least one of $P(x)$ and $Q(x)$ has to be true for each $x \in I$, then it is indeed true that whenever $P(x)$ fails, $Q(x)$ must hold, and conversely. But your $(1)$ says that both $P(x)$ and $Q(x)$ have to be true for each $x \in I$, which is another matter altogether.
