# 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 ?

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You're right that the sentences aren't equivalent. But since you've provided no justification for your assumption that they are, it's hard to pinpoint your error. – user92843 Jul 10 '11 at 19:07
Do you mean $\forall x: \lnot(x\in I \rightarrow P(x))$ or $\lnot (\forall x: x\in I \rightarrow P(x))$? They are not equivalent. The second needs a single $x \in I$ not to satisfy $P(x)$, while the first needs every $x \in I$ not to satisfy $P(x)$ – Ross Millikan Jul 10 '11 at 21:18

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.

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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.

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