Rank of formulas if unique readability fails. Recursively define  $V_{n}$ for each natural number $n$ in the following manner:
$V_{0}=$ The set of all propositional atoms
$V_{n+1}=V_{n} \cup\{x:x=p \land q \space \text{or}\space p\lor q\space \text{or}\space p\to q\space \text{or}\space \neg p \space \text{for some }p,q\in V_{n}\}$
[I'm delibrately avoiding brackets in order for unique readibility to fail]
Let $V=\bigcup V_{n}$ denote the union of all the $V_{n}$; given $y\in V$ define the rank of $y$ to be the minimum $n$ such that $y\in V_{n}$.  Is it true that the rank of $p\land q$ is always greater or equal to the rank of $p$ and that of $q$ ?,if brackets are added then unique readability holds so this is obviously true(and one does not need the “equals to condition” ), but I wonder if this holds.
 A: Yes, this is true. It suffices to show that for any binary connective $\square$, if $p\square q\in V_n$, then $p\in V_n$ and $q\in V_n$. Let's prove this by induction on $n$.
The base case $n = 0$ is vacuous: $p\square q$ is not a propositional atom, so it is not in $V_0$.
For the inductive step, assume $p\square q\in V_{n+1}$. By definition of $V_{n+1}$, there are three cases.
Case 1: $p\square q\in V_n$. By induction, $p\in V_n$ and $q\in V_n$. Since $V_n\subseteq V_{n+1}$, $p\in V_{n+1}$ and $q\in V_{n+1}$.
Case 2: $p\square q = \lnot r$, with $r\in V_n$. Then $p\square q$ begins with $\lnot$, so we can write it as $\lnot p'\square q$ with $p = \lnot p'$ and $r = p'\square q$. By induction, $p'\in V_n$ and $q\in V_n\subseteq V_{n+1}$. Then also $p = \lnot p'\in V_{n+1}$.
Case 3: $p\square q = p'\lozenge q'$, where $\lozenge$ is a binary connective and $p',q'\in V_n$. If $p = p'$ and $q = q'$, then we are done since $V_n\subseteq V_{n+1}$. Otherwise, there is a string $r$ such that either $p\square q = p'\lozenge r \square q$ or $p\square q = p\square r \lozenge q'$. Let's deal with the first case (the second is similar). In this case we have $p = p'\lozenge r$ and $q' = r\square q$. Since $q'\in V_n$, by induction $r\in V_n$ and $q\in V_n\subseteq V_{n+1}$. And since $p'\in V_n$ and $r\in V_n$, $p\in V_{n+1}$. This completes the proof.
