The problem I am working on is:
Express each of these statements using quantifiers. Then
form the negation of the statement, so that no negation
is to the left of a quantifier. Next, express the negation in
simple English. (Do not simply use the phrase “It is not
the case that.”)a) Some old dogs can learn new tricks.
b)No rabbit knows calculus.c) Every bird can fly.
d)There is no dog that can talk.
e) There is no one in this class who knows French and
Russian.
I am having trouble with only two parts–namely, d) and e)
For d): $P(x)= x$ cannot talk
$\exists xP(x)$ Negating this, $\neg \exists xP(x) \rightarrow \forall x \neg P(x)$
This would read in English, "Every dog can talk". However, the answer is, "There is a dog that talks."
For e): $F(x)= x$ doesn't know French.$\qquad R(x)= x$ doesn't know Russian.
$\forall x(F(x) \wedge R(x)$ Negating this, $\neg \forall x(F(x) \wedge R(x) \rightarrow \exists x(\neg F(x) \vee \neg R(x))$ Translating this back into English: There is a person in class that knows French or Russian."
However the answer is, "There is someone in this class who knows French and Russian"
Best Answer
In both cases you’re starting in the wrong place, translating the original statement into symbols incorrectly.
For (d) the original statement is essentially ‘There does not exist a dog that can talk’, i.e., $\neg\exists xP(x)$, where $P(x)$ is ‘$x$ is a dog that can talk’. Negating that gives you simply $\exists xP(x)$, ‘There is a dog that can talk’.
Similarly, assuming that the universe of discourse is this class, (e) is $\neg\exists\big(F(x)\land R(x)\big)$, where $F(x)$ is ‘$x$ does know French’ and $R(x)$ is ‘$x$ does know Russian’, so its negation is $\exists\big(F(x)\land R(x)\big)$ ‘There is someone in this class who knows French and Russian’.