If a formula $\varphi$ is independent of $p$, there is some $\varphi' \sim \varphi$ where $p$ does not occur in $\varphi'$ 
Show that if a formula $\varphi$ is independent of a propositional symbol $p$, then there exists a formula $\varphi'$ such that $\varphi' \sim \varphi$ and 
  $$VOC(\varphi') = VOC(\varphi)\setminus \{ p\}$$
  where $VOC(\psi)$ is the set of all propositional symbols that occur in $\psi$.

We say that a formula $\varphi$ is independent of a symbol $p$ is for every valuation $v$ we have that:
$$[[\varphi]]^{v[p/\bot]} = [[\varphi]]^{v[p/\top]} = [[\varphi]]^v$$
that is, $[[\varphi]]^v$ does not depend on whether $v(p)=0$ or $v(p)=1$.
I am having some trouble trying prove this but I came up with a "possible proof" by induction. 
For the base case, if $\varphi = \top$, then $VOC(\varphi) = VOC(\top) = 0$, and we can take $\varphi' = \varphi$ since $\varphi$ does not depend on any $p$. If $\varphi = \bot$ we can do just the same thing. Again, if $\varphi = p$, $\varphi$ depends on $p$ the hypothesis $\varphi$ does not depend on $p$ doesn't apply. Finally, if $\varphi = q$ and $q \neq p$, we can take $\varphi' = \varphi$.
Now, assume that $\varphi = \neg \psi$ and that $\varphi$ is independent of $p$. Then we must have that $\psi$ is independent of $p$ as well, and by induction hypothesis there is some $\psi' \sim \psi$ such that $VOC(\psi')=VOC(\psi) \setminus \{ p\}$. Then we take $\varphi ' = \neg \psi '$. This works, since 
$$VOC(\varphi) = VOC(\psi)$$
and 
$$VOC(\varphi') = VOC(\psi') = VOC(\psi) \setminus \{ p \} = VOC(\varphi) \setminus \{ p \}$$ 
Now, if $\varphi = \psi_2 \square \psi_2$ where $\square \in \{ \land, \lor, \rightarrow, \leftrightarrow \}$ and $\varphi$ is independent of $p$, then $\psi_1$ and $\psi_2$ must be independent of $p$. By induction hypothesis, there are some $\psi_1' \sim \psi_1$ and $\psi_2' \sim \psi_2$ such that
$$VOC(\psi_1') = VOC(\psi_1)\setminus \{p\}$$
and 
$$VOC(\psi_2') = VOC(\psi_2) \setminus \{ p \}$$
In that case, we take $\varphi' = \psi_1' \square \psi_2'$ and check that it works. Since
$$VOC(\varphi) = VOC(\psi_1) \cup VOC(\psi_2)$$
we have that 
$$\begin{align*}
VOC(\varphi') &= VOC(\psi_1') \cup VOC(\psi_2') = \left( VOC(\psi_1) \setminus \{p \} \right) \cup \left( VOC(\psi_2) \setminus \{p \}\right)\\
& = \left( VOC(\psi_1) \cup VOC(\psi_2) \right) \setminus \{ p \}\\
& = VOC(\varphi) \setminus \{ p \}
\end{align*}$$
which ends the proof.
I would really thank someone who could point me out if I made any mistakes here and help me with a correct proof of the statement. Thank you in advance!
 A: Any propositional formula is equivalent to the disjunction of the conjunction of propositional variables or their negation. That is just look at the truth table and take the disjunction of the assignments which make the formula true. In case it is independent of $p$ they will now cancel out, for example 
$$(q\wedge p)\vee(q\wedge \neg p)\equiv q.$$  
A: It is not clear why, if $\varphi = \neg \psi$ is independent of $p$, $\psi$ is independent of $p$ as well.  This is not trivial.
More pressingly: for the case where $\varphi = \varphi_1 * \varphi_2$ with $*$ any binary operator, you cannot say that if $\varphi$ is independent of $p$, then $\varphi_1$ and $\varphi_2$ are independent of $p$ as well. For example, take $\varphi = p \lor \neg p$. Then $\varphi$ is independent of $p$, but $p$ and $\neg p$ clearly are not!
This basically blocks the very induction you are trying to use, so I wouldn't use induction for this one at all. Instead:
To obtain $\varphi'$, replace all occurrences of $p$ in $\varphi$ with $\top$ (if you are allowed such a symbol as a statement (not a truth-value)), or $q \lor \neg q$ for some other variable $q$, for then we have for any $v$ that: $$[[\varphi']]^v=[[\varphi]]^{v[p/\top]} = [[\varphi]]^v$$
and clearly: $$VOC(\varphi') = VOC(\varphi) \setminus \{ p \}$$
