What will be the negation of this statement:

Question

Every street in the city has at least one house in which we can I find a person who is rich and beautiful or highly educated and kind.

Negation: 'There exists a street in the city where in every house we can find no person who is rich and beautiful or highly educated and kind.'

Is the negation correct? Please someone check...

Thank you..

Answer 1

Yes, that works.

In logic, the original is:

$\forall x (S(x) \rightarrow \exists y (H(y) \land I(y,x) \land \exists z (P(z) \land L(z,y) \land ((R(x) \land B(x)) \lor (E(x) \land K(x)))))$

If you negate this:

$\neg \forall x (S(x) \rightarrow \exists y (H(y) \land I(y,x) \land \exists z (P(z) \land L(z,y) \land ((R(x) \land B(x)) \lor (E(x) \land K(x))))) \Leftrightarrow$

$\exists x \neg (S(x) \rightarrow \exists y (H(y) \land I(y,x) \land \exists z (P(z) \land L(z,y) \land ((R(x) \land B(x)) \lor (E(x) \land K(x))))) \Leftrightarrow$

$\exists x (S(x) \land \neg \exists y (H(y) \land I(y,x) \land \exists z (P(z) \land L(z,y) \land ((R(x) \land B(x)) \lor (E(x) \land K(x))))) \Leftrightarrow$

$\exists x (S(x) \land \forall y \neg(H(y) \land I(y,x) \land \exists z (P(z) \land L(z,y) \land ((R(x) \land B(x)) \lor (E(x) \land K(x))))) \Leftrightarrow$

$\exists x (S(x) \land \forall y (H(y) \land I(y,x) \rightarrow \neg \exists z (P(z) \land L(z,y) \land ((R(x) \land B(x)) \lor (E(x) \land K(x)))))$

.. which is what your sentence says

If we push the negation further in, we get:

$\exists x (S(x) \land \forall y (H(y) \land I(y,x) \rightarrow \forall z \neg (P(z) \land L(z,y) \land ((R(x) \land B(x)) \lor (E(x) \land K(x)))))\Leftrightarrow$

$\exists x (S(x) \land \forall y (H(y) \land I(y,x) \rightarrow \forall z (P(z) \land L(z,y) \rightarrow \neg ((R(x) \land B(x)) \lor (E(x) \land K(x)))))\Leftrightarrow$

$\exists x (S(x) \land \forall y (H(y) \land I(y,x) \rightarrow \forall z (P(z) \land L(z,y) \rightarrow (\neg (R(x) \land B(x) \land \neg (E(x) \land K(x))))\Leftrightarrow$

$\exists x (S(x) \land \forall y (H(y) \land I(y,x) \rightarrow \forall z (P(z) \land L(z,y) \rightarrow ((\neg R(x) \lor \neg B(x) \land (\neg E(x) \lor \neg K(x))))$

which translates to:

"There is a street in the city where for every house in that street it is true that every person living in that house is not rich or not beautiful, and is also not highly educated or not kind"

Answer 2

The negation is NOT(Every street in the city has at least one house in which we can I find a person who is rich and beautiful or highly educated and kind.)

Which means --- NOT Every street in the city has at least one house in which we can I find a person who is rich and beautiful or highly educated and kind.

Which is equivalent to your statement.

Answer 3

Your answer is correct as far as it goes. Your instructor might want you to continue by changing

we can find no person ...

to

every person ...

and changing the ands and ors in the ... appropriately.

Answer 4

You said that: $$\forall S\in\mathcal{C},~\exists h\in S,~h\cap ((r\cap b)\cup(h\cap k))\neq\emptyset$$

where $S$ a street as a set of houses, $\mathcal{C}$ the city as a set of streets, $h$ an house as a set of persons, $r$ rich people, $b$ beautiful people, $h$ highly educated people, and $k$ kind people.

Its negation is $$\exists S\in\mathcal{C},~\forall h\in S,~h\subset (\bar r\cup \bar b)\cap(\bar h\cup \bar k)$$

Each person in every house of some street in the city has both the following qualities $(\bar r\cup \bar b)$ and $(\bar h\cup \bar k)$:

$~~(\bar r\cup \bar b)$: one between rich and beautiful, or neither of the two

$~~(\bar h\cup \bar k)$: one between highly educated and kind, or neither of the two

Answer 5

It's roughly the correct negation. Your negation has a few further implications than the direct logical inverse.

The original statement:

Every street in the city has at least one house in which we can find a person who is rich and beautiful or highly educated and kind.

The obvious exact negation:

Not every street in the city has at least one house in which we can find a person who is rich and beautiful or highly educated and kind.

Your negation:

There exists a street in the city where in every house we can find no person who is rich and beautiful or highly educated and kind.

There is a possibility with the exact negation that every street in the city with houses has at least one house in which (etc.), and that the only counterexample to the original claim is a street without any houses. Your negation precludes this possibility by implication.

If a street has no houses, it is not true to say "in every house on this street we can find no person who is rich, etc." It is true to say, "in no house on this street we can find a person who is rich, etc." The first implies the existence of the set "houses on this street." The latter does not.

---

Modifying the exact negation by progressive changes, retaining exact meaning:

Not every street in the city has at least one house in which we can find a person who is rich and beautiful or highly educated and kind.

There exists a street in the city which does not have at least one house in which we can find a person who is rich and beautiful or highly educated and kind.

There exists a street in the city where in no house can we find a person who is rich and beautiful or highly educated and kind.

This is really as far in as you can push the negation without any alteration of meaning. Any further and you imply the existence of sets that may or may not exist.

If we allow that, we may as well also allow "there is a" instead of "there exists a," despite the ambiguity regarding quantity (exactly one or at least one?). So then we get the following much sloppier sort-of-negation of the original, which implies existence of sets all over the place which were not required to exist by the ACTUAL negation:

There is a street in the city where in all the houses it has, we can't find a person who is rich and beautiful or highly educated and kind.

There is a city street with houses in all of which we can't find a person who is rich and beautiful or highly educated and kind.

Now we've implied houses, let's imply people. (There don't have to be ANY people per the original negation.)

There is a street in the city with its houses all full of people who are poor or ugly, and stupid or mean.