Proving that the closure of a closed formula is the formula itself Closed Formula. A formula (in a first order language) is said to be closed if it has no free variables.
Closure of a Formula. Let $A[x_0, . . . , x_{n−1}]$ be a formula (in a first order language) whose free variables are among
$x_0, . . . , x_{n−1}$ and $x_{n−1}$ is free in $A$, where $x_0, . . . , x_{n−1}$ are the first $n$ variables in alphabetical order. We call
$\forall x_{n-1}\ldots\forall x_1\forall x_{0}A$
the closure of $A$.

Question.
Using these definitions, how can I show that if $A$ is closed, it is its own closure?

I thought that I will begin by assuming that $Q$ is the closure of $P$. But I don't understand what exactly it is that I need to show. Do I need to show that $P\leftrightarrow Q$ or that they are syntactically exactly same formula or something else? In any case I don't how to proceed. Can anyone help?
According to the definition, if the formula $A$ is closed, it has no free variables.

Thus, there is no $n$ such that:


  the free variables of $A$ are among $x_0, \ldots,x_{n−1}$ and $x_{n−1}$ is free in $A$.


Thus, the closure of $A$ has no "added" universal quantifiers upfront, which means that it is $A$ itself.



If we have a formula $A$, like $x_2=c_1$, the definition gives us a "recipe" to produce its closure (in this case we have $n=3$): 


  $∀x_2∀x_1∀x_0(x_2=c_1)$


because we have to prenex the universal quantifiers for the first $n$ variables in alphabetical order. 

If instead $A$ is $∃x_2(x_2=c_1)$, and we apply the definition:


  "let $A[x_0,\ldots,x_{n−1}]$ be a formula whose free variables are among $x_0,\ldots,x_{n−1}$ and $x_{n−1}$ is free", 


we get no $n$, because neither $x_2$ nor $x_1$ nor $x_0$ are free.

Thus, we have to "prenex" nothing and we get: $∃x_2(x_2=c_1)$ which is $A$ itself.

Note : in this case, it is not correct to say that $∀x_2∀x_1∀x_0(∃x_2(x_2=c_1))$ is the closure of $∃x_2(x_2=c_1)$.

Of course, the two are equivalent, because we can prove that:


  if $x$ does not occur free in $\alpha$, then $\vdash \alpha \leftrightarrow ∀x \alpha$.
You need to show that the formulae are syntactically equal.

The closure of a formula $\phi$ is the formula $\forall x_1 \ldots \forall x_n.\phi$ where $fv(\phi) = \{ x_1, \ldots, x_n \}$. If $\fv(\phi) = \emptyset$, then we of course will not prefix $\phi$ with any quantifiers, and the result is obvious just $\phi$.