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Could anyone conceive of any predicates and Universe ( in mathematics, in the world, etc ) where we should use $\exists x ( P(x) \to Q(x) )$, and not necessarily $\forall x ( P(x) \to Q(x) )$ ?
I was thinking of some sittuation where the property of P implies property of Q for at least some individual on the domain, but not for all individuals on the domain. i think its non-existant sittuation ?
I tried to think , and the closest i got was to think that since we know that for some $n$, we have that $n$ is prime, plus $ 2^n-1$ is prime. So, i thought of using $\exists n$ ( $n$ is prime $\to$ $2^n -1$ is prime ) .
Would this be correct to use ?
Then i thought i could use $\exists n$ ( $n$ is prime $\land$ $2^n - 1 $ is prime ) instead, and those are not semantic equvialent, so now i'm kinda lost.

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up vote 1 down vote accepted

Consider $P(x) \equiv Cube(x)$, $Q(x) \equiv Small(x)$ in a world with a small cube $a$ and a big cube $b$.

Then $\exists x:P(x) \rightarrow Q(x)$ is true (due to $a$ being small), while $\forall x: P(x) \rightarrow Q(x)$ is false (because $b$ is a cube, but is not small).

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Hmmm i understand.But i'm thinking here we could use $\exists x : P(x) \wedge Q(x)$ . Since the $"\wedge"$ one is not equivalent to the $"->"$ one, which should we try to use on that case ?? – nerdy May 7 '14 at 9:57
Is it always the case that ( ∃x:P(x)→Q(x) ) -> ∃x:P(x)∧Q(x) ? Or would be the converse that is true ? – nerdy May 7 '14 at 10:06
@nerdy The first says: something is (either non-P or Q). The second says that something is (both P and Q). The second, of course, is stronger: if something is (both P and Q), then something is (simply Q), from which it follows that something is (Q or non-P), which is what the first says. The other direction isn't valid: consider a world with one thing that Qs but doesn't P. In that world, the first is true (because the object satisfies Q), but the second is false (because the object doesn't satisfy P). – Hunan Rostomyan May 7 '14 at 10:15
@nerdy - as explained by Peter and Hunan, the "interesting" couple is : $\forall x (P(x) \rightarrow Q(x))$ and $\exists x (P(x) \land Q(x))$; they are the so-called restricted (or bounded) quantifiers. We use them when we want to "restrict" the "search for" the objects such that $Q$ holds of them to the "domain" of objects such that $P$ holds. The canonical example is "all men are mortal", which we translate as $\forall x (Man(x) \rightarrow Mortal(x))$. The existential "counterpart" of it need $\land$ just because we want to avoid the "trivial" case ... 1/2 – Mauro ALLEGRANZA May 8 '14 at 10:24
... when $P$ holds of no objects (i.e. $\exists x P(x)$ is false). In order to avoid this we need $\land$, exactly because (as per Hunan's comment) $P \rightarrow Q$ and $P \land Q$ are not equivalent. Thus : "some man are wise" is translated as $\exists x (Man(x) \land Wise(x))$. 2/2 – Mauro ALLEGRANZA May 8 '14 at 10:27

It is worth remarking that a claim of the form $\exists x ( P(x) \to Q(x) )$ is typically unlikely to be interestingly informative and worth saying.

Why so?

Well, $\exists x ( P(x) \to Q(x) )$ is true so long as $P(a) \to Q(a)$ is true for some case where $a$ newly dubs an element of the domain. But that material conditional will be true so long as its antecedent is false.

Hence, so long as there is something $a$ in the domain which doesn't satisfy $P$, we have $\exists x ( P(x) \to Q(x) )$.

So, if you know already that $P$ is not universally true of everything in the domain (so there is something in the domain who can dub $a$ where $P(a)$ is false), and this is probably the typical case, then $\exists x ( P(x) \to Q(x) )$ gives you no new information.

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I think this is exactly the answer the OP was looking for. – Hunan Rostomyan May 7 '14 at 10:02
Thanks for the insight ! – nerdy May 7 '14 at 11:31

$$P(x)\equiv(x\le 1),\qquad Q(x)\equiv(x=1)$$

$$\exists x(x\le1\implies x=1)$$ is true, but

$$\forall x(x\le1\implies x=1)$$ is false.

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Thanks you two for the wto examples. Could you share a light on which of the two i should use for the mersenne primes ? since they are non-equivalent – nerdy May 7 '14 at 9:53
I'm thinking here we could use ∃x:P(x)∧Q(x) i think to describe roughly the same thing . Since the "∧" one is not equivalent to the "−>" one, which should we try to use on that case ?? – nerdy May 7 '14 at 9:59

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