Answer
a) $ P(-2) \lor P(-1) \lor P(0) \lor P(1) \lor P(2)$
b) $ P(-2) \land P(-1) \land P(0) \land P(1) \land P(2)$
c) $ \neg P(-2) \lor \neg P(-1) \lor \neg P(0) \lor \neg P(1) \lor \neg P(2)$
d) $ \neg P(-2) \land \neg P(-1) \land \neg P(0) \land \neg P(1) \land \neg P(2)$
e) $ \neg (P(-2) \lor P(-1) \lor P(0) \lor P(1) \lor P(2))$
f) $\neg ( P(-2) \land P(-1) \land P(0) \land P(1) \land P(2))$
Work Step by Step
INTERPRETATION SYMBOLS
Negation : $\neg p$
Disjunction : $p \lor q$
Conjunction : $ p \land q$
Existential quantification : $\exists$ $ x$ $P(x)$
Universal quantification : $\forall$ $x$ $P(x)$
Solution-
a) $\exists$ $ x$ $P(x)$ means that there exists a value of x for which P(x) is true, thus P(-2) is true or P(-1) is true or P(0) is true or P(1) is true or P(2) is true.
So, we can rewrite the proposition as:
$$ P(-2) \lor P(-1) \lor P(0) \lor P(1) \lor P(2)$$
b) $\forall$ $ x$ $P(x)$ means that there exists a value of x for which P(x) is true, thus P(-2) is true and P(-1) is true and P(0) is true and P(1) is true and P(2) is true.
So, we can rewrite the proposition as:
$$ P(-2) \land P(-1) \land P(0) \land P(1) \land P(2)$$
c)$\exists$ $ x$ $\neg P(x)$ means that there exists a value of x for which $\neg P(x)$ is true, thus $\neg P(-2)$ is true or $\neg P(-1)$ is true or $\neg P(0)$ is true or $\neg P(1)$ is true or $\neg P(2)$ is true.
So, we can rewrite the proposition as:
$$ \neg P(-2) \lor \neg P(-1) \lor \neg P(0) \lor \neg P(1) \lor \neg P(2)$$
d)$\forall$ $ x$ $\neg P(x)$ means that there exists a value of x for which P(x) is true, thus $\neg P(-2)$ is true and $\neg P(-1)$ is true and $\neg P(0)$ is true and $\neg P(1)$ is true and $\neg P(2)$ is true.
So, we can rewrite the proposition as:
$$ \neg P(-2) \land \neg P(-1) \land \neg P(0) \land \neg P(1) \land \neg P(2)$$
e) $\neg$ $ \exists x$ $P(x)$ is the negation of $ \exists x$ $P(x)$ in part a) , thus we take the negation of part (a):
$$ \neg (P(-2) \lor P(-1) \lor P(0) \lor P(1) \lor P(2))$$
f) $\neg \forall x$ $P(x)$ is the negation of $\forall$$ x$ $P(x)$ in part b) , thus we take the negation of the result in part b):
$$\neg ( P(-2) \land P(-1) \land P(0) \land P(1) \land P(2))$$
