Small doubt in quantifiers

Question

The answer given is only III).

I don't understand how can III) be equivalent to the expression given in the question.

In III), if we apply the negation starting from "there exists" then, "there exists becomes for all", "belongs to 'r' becomes not belongs to 'r' ", "~P(t) becomes P(t)". But in the question we just have "belongs to r" and not "not belongs to r". So I don't think they are equal.

Also, what is the scope of the negation operator here? Is it to the whole expression or only to quantifiers?

Please help me.

Thanks.

Answer 1

$¬∃¬$ is equivalent to $∀$.

We may use intuition: to say that "all Ravens are Black" is the same as to assert that "there are no Ravens that are not-Black.

We have to consider that $(\forall t \in r)P(t)$ is an abbreviation for:

$\forall t (t \in r \to P(t))$.

If we apply the above equivalence to rewrite it as:

$\lnot \exists t \ \lnot (t \in r \to P(t))$,

and then use the propositional equivaelnce between $\lnot (A \to B)$ and $A \land \lnot B)$, we get:

$\lnot \exists t (t \in r \land \lnot P(t))$.

And this in turn can be abbreviated to:

$\lnot (\exists t \in r)(\lnot P(t))$.

See Restricted quantifiers.

Answer 2

The negation is on the whole expression, however in order to make it easier to interpret we rewrite the $\in r$ in a nicer way. $\exists t \in r( P(t))$ can be rewritten as $\exists t (t\in r \wedge P(t))$ while $\forall t \in r (P(t))$ can be rewritten as $\forall t(t\in r \to P(t))$.

Thus $$\neg \exists t\in r (\neg P(t)) \equiv \neg \exists t (t\in r \wedge \neg P(t)) \equiv \forall t \neg (t\in r \wedge \neg P(t))\equiv \forall t (\neg t\in r \vee \neg\neg P(t))$$ $$\equiv \forall t (\neg t\in r \vee P(t)) \equiv \forall t (t\in r \to P(t)) \equiv \forall t\in r (P(t)). $$

Your solution is not correct since what you do is to apply the negation without taking care of how it actually applies to $\in$, like we did above.