Negating a statement with an arbitrary variable Consider the statement $P(a)$, where $a$ is an arbitrary element from $D$.
Which of these is its negation?

*

*$\neg P(a)$, where $a$ is an arbitrary element from $D$


*$\neg P(a)$, where $a$ is some element from $D$


Consider the statement $P(a)$, where $a$ is an arbitrary element from $D$.

Although we can apply Universal Generalisation to $P(a),$ $P(a)$ itself remains an open formula (i.e., propositional function, i.e., predicate), not a statement.

Which of these is its negation?

*

*$\neg P(a)$, where $a$ is an arbitrary element from $D$


*$\neg P(a)$, where $a$ is some element from $D$

The negation of $P(a)$ is simply $\,¬P(a)\,;$ the denotation of $a$ as an arbitrary element of $D$ is unaffected.

Based on previous answers given, I think I have come to a conclusion:
If I negate the proposition like this "For an arbitrary integer $a$, $\neg P(a)$",

No. Here, you've negated $P(a)$ and applied Universal Generalisation after.

If I negate the proposition like this "$\neg$ (For an arbitrary integer $a$, $P(a)$)",

No. Here, you've derived $\forall a\,P(a)$ and then negated the result.
Now, remember, negation is a logical operation, not an inferential rule. So, there is no reason to expect these four formula from above to be equivalent to one another. Indeed, no pair is equivalent to each other:

*

*$P(a)$

*$\lnot P(a)$

*$\forall\lnot P(a)$

*$\lnot\forall P(a)$
