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 $D$
 A: Good question. I think I see what you mean.
You could put it this way: in order to be a proposition, the expression $P(a)$ has to have all of its arguments filled in.
For example, if $a$ already has a specific value such as "the sky" then $P(a)$ is a proposition like "The sky is blue." It is a proposition because all of the arguments have been filled in. Its negation is $\neg P(a)$, "The sky is not blue."
In addition to propositions, you can also talk about the generic expression $P(\cdot)$, which means something like "______ is blue."  This is not a proposition, because we have not yet filled in all of the arguments. It is a function with arguments that you can plug in; by plugging in different values,  you get different propositions $P(a)$, $P(b)$, $P(c)$.
Sometimes instead of writing the function like $P(\cdot)$, we name its argument and call it, say, $P(x)$, where $x$ is just a placeholder for something you could fill in later. This way of writing it is potentially confusing, because just by glancing at the page, you are not sure if $x$ refers to a specific thing, or if it is just the name we've given to a function's argument.
If $P(x)$ is just a function and its arguments have not yet been filled in, then $P(x)$ is not a proposition. It does not mean "A specific thing $x$ is blue", or "Every $x$ is blue." It is a form with a blank in it; it is "____ is blue." You can tell because inside those parentheses, $x$ means "the name of $P$'s function argument" and outside of those parentheses, $x$ has not been defined.
Some people write functions like $(x \mapsto P(x))$ so they have a different way of writing it than $P(a)$.

*

*When you already have defined $a$ ("the sky"), you can negate a proposition like $P(a)$ ("The sky is blue"). When you negate it, you get $\neg P(a)$ ("The sky is not blue.").

*You can negate a proposition like $\forall a. P(a)$ ("Every $a$ in the universe is blue.") When you negate it, you get $\neg \forall a. P(a)$ ("Not every $a$ in the universe is blue").

*You can negate a proposition like $\exists a. P(a)$ ("Some $a$ in the universe is blue."). When you negate it, you get $\neg \exists a. P(a)$. ("No $a$ in the universe is blue.")

*When you have not already defined $x$, then $P(x)$ is a function whose argument is named $x$. You technically cannot negate a function, because it is not a proposition. But if you have one function $(x \mapsto P(x))$ you can consider the related function $(x\mapsto \neg P(x))$, which corresponds to "_____ is not blue".


Extra note: In lambda calculus, this sort of difference between functions and propositions is made precise and concrete as the difference between $\lambda x. fx$ and $(\lambda x. fx)(a)$.
A: Based on previous answers given, I think I have come to a conclusion:
Let $P(x)$ be the propositional function "$x$ is an even number".
Then the proposition "For an arbitrary integer $a$, $P(a)$" states "For an arbitrary integer $a$, $a$ is an even number.
The negation of this proposition can be interpreted in two ways:
If I negate the proposition like this "For an arbitrary integer $a$, $\neg P(a)$", I am stating "For an arbitrary integer $a$, it is not the case that $a$ is an even number."
If I negate the proposition like this "$\neg$ (For an arbitrary integer $a$, $P(a)$)", I am stating "It is not the case that for an arbitrary integer $a$, $a$ is an even number." Or in other words, there exist integers that are not even numbers.
A: 
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:

*

*$P(a)$

*$\lnot P(a)$

*$\forall\lnot P(a)$

*$\lnot\forall P(a)$
