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The question is composed of 2 sub-questions, I've already answered the first one but I'm gonna write it since it's important for the rest of the question.

We will state that formula A is of CNF formation if there exists $A_{i,j}$ elementary formulas, or the negation of elementary formulas, so that:

$A = (A_{1,1} \vee ... \vee A_{1,n1}) \land ... \land (A_{k,1} \vee ... \vee A_{k,nk})$

Prove that for each quantifier-less formula A there exists a quantifier less formula B in CNF formation, so that $ \vdash A \leftrightarrow B$

This I proved, but this is necessary for the next question that I found some trouble in:

Show that if $\vdash A \leftrightarrow B$ then $\vdash QxA\leftrightarrow QxB $

For $ Q = \exists, \forall$

Thanks!

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    It sounds like for the second part you need to show that if formulas $A$ and $B$ are equivalent, then $\forall x A$ is equivalent to $\forall x B$ and $\exists x A$ is equivalent to $\exists x B$. OK, I can show you how to do something like that, but that would be completely independent of the first part. That is, this result is true whether $A$ and/or $B$ are in CNF or not. But you said the CNF part was important ... making me worry that maybe I misunderstand your question. Can you please clarify a bit more? – 2017-01-12
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    Also, do you need to show equivalence, or do you need to show provability? .. since you use the $\vdash$ symbol, which is for provability in some kind of proof system ... but I don't know what proof system you have. Again, please clarify. – 2017-01-12
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    First, okay, I thought that it may be related to the first question but if you state that it doesn't matter then go ahead and write your intended answer. About the second question, I forgot to add something. I fixed it, by using the symbol $\vdash$ also in $A \leftrightarrow B $. This means that if a proof system provides the first statement, it will also provide the following. – 2017-01-12
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    OK. But like I said, I don't know what proof system you are using with, so either you will have to tell me what rules it has, or I will assume that it is sound and complete, and outline the proof for logical equivalence, i.e that if $\vDash A \leftrightarrow B$ then $\vDash \forall x A \leftrightarrow \forall x B$ – 2017-01-12
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    You can assume it is sound and complete, it is the usual proof system with the known axioms. There is no restrictions nor modifications – 2017-01-12
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    There are a *lot* of different proof systems. In fact, there are already different *types* of proofs systems, like axiom systems, sequent systems, Fitch systems, etc. And in some of these systems I cannot even prove $A \leftrightarrow B$ if $A$ and $B$ contain free variables. So, maybe the best thing to do is for me to using formal semantics to prove equivalence? And by the way, how did you the first part? That might tell me something more about what kind of proof you want/need. – 2017-01-12
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    I have a sneaking feeling that my (rather complicated!) Answer below is not the way you were supposed to answer this question, and that you did have to do this by considering dormal proofs (which most likely would make the answer a lot simpler) ... Is that right? But if so, again, We really need to know the rules of the particular proof system you have to work with in order to help you. – 2017-01-14

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OK, not knowing the proof system but assuming it is sound and complete, i.e. for any $\varphi$ it holds that:

$\vdash \varphi$ iff $\vDash \varphi$

I'll show that:

(*) If $\vdash A \leftrightarrow B$ then $\vdash \forall x A \leftrightarrow \forall x B$

By showing that

If $\vDash A \leftrightarrow B$ then $\vDash \forall x A \leftrightarrow \forall x B$

For then we can simply do:

If

$\vdash A \leftrightarrow B$

then (sound)

$\vDash A \leftrightarrow B$

so (*)

$\vDash \forall x A \leftrightarrow \forall x B$

so (complete)

$\vdash \forall x A \leftrightarrow \forall x B$

OK, but how do we show (*)?

First, let's assume that $A$ and $B$ do not contain $x$ as a free variable. Then, by Null Quantification:

$\forall x A \Leftrightarrow A$ and $\forall x B \Leftrightarrow B$

Since for any $\varphi$ and $\psi$:

$\vDash \varphi \leftrightarrow \psi$ iff $\varphi \Leftrightarrow \psi$

We thus get that if

$\vDash A \leftrightarrow B$

then:

$A \Leftrightarrow B$

and so:

$\forall x A \Leftrightarrow \forall x B$

and thus:

$\vDash \forall x A \leftrightarrow \forall x B$

OK, but what if $A$ and $B$ do contain $x$ as a free variable?

Well, let's first assume they have the same free variables ... which actually doesn't have to be, e.g. if $A = P(x) \lor \neg P(x)$ and $B = P(y) \lor \neg P(y)$, then $\vDash A \leftrightarrow B$, but let's ignore those variables that 'do no interesting work'.

OK, so let $x,x_1,x_2,...$ be all the free variables in both $A$ and $B$.

Then by $A \Leftrightarrow B$ we mean that for any interpretation (structure) $I$ with domain $D$, it holds that the exact same tuples $$ with $d,d_1,d_2, ... \in D$ will satisfy the formulas $A$ as well as $B$, defined as follows:

$$ satisfies formula $A(x,x_1,x_2,...)$ in $I$ iff $I \vDash A(x,x_1,x_2,...)[d,d_1,d_2,...]$ and where we mean (in general):

$I \vDash A(x)[d/x]$ iff by extending our language with a new constant symbol $c$, and extending our interpretation $I$ to become a new interpretation $I[d/c]$ which is just like $I$, but in addition interprets $c$ as $d$, we have that $I[d/c] \vDash A(c/x)$ where $A(c/x)$ is the result of replacing all free variables $x$ in $a$ with constant symbol $c$.

OK, so if we assume that

$\vDash A(x,x_1,x_2,...) \leftrightarrow B(x,x_1,x_2,...)$

then:

$A(x,x_1,x_2,...) \Leftrightarrow B(x,x_1,x_2,...)$

and thus for any interpretation (structure) $I$ with domain $D$ it holds that:

For all tuples $$ with $d,d_1,d_2, ... \in D$: $I \vDash A(x,x_1,x_2,...)[d/x,d_1/x_1,d_2/x_2,...]$ iff $I \vDash B(x,x_1,x_2,...)[d/x,d_1/x_1,d_2/x_2,...]$

So now we have that for any interpretation $I$ and all tuples $$ with $d_1,d_2, ... \in D$:

$I \vDash \forall x A(x,x_1,x_2,...)[d_1/x_1,d_2/x_2,...]$ iff

for all $d \in D$: $I \vDash A(x,x_1,x_2,...)[d/x,d_1/x_1,d_2/x_2,...]$ iff

for all $d \in D$: $I \vDash B(x,x_1,x_2,...)[d/x,d_1/x_1,d_2/x_2,...]$ iff

$I \vDash \forall x B(x,x_1,x_2,...)[d_1/x_1,d_2/x_2,...]$ iff

So:

$\forall x A(x,x_1,x_2,...) \Leftrightarrow \forall x B(x,x_1,x_2,...)$

and thus:

$\vDash \forall x A(x,x_1,x_2,...) \leftrightarrow \forall x B(x,x_1,x_2,...)$ or simply put:

$\vDash \forall x A \leftrightarrow \forall x B$