There must exist a prime between $P_n$ and (including) $|(P_n\#)/2-2^y|$ as long as $|(P_n\#)/2-2^y| > P_n$.
The Primorial function is expressed as:
$p_n\#=\prod_{k=1}^{n}p_k$
I will show my solution through the following examples:
$P_7\#=2⋅3⋅5⋅7⋅11⋅13⋅19=570570$
$\frac{570570}{2}=285285$
From here on we know that:
$|(285285 - 3x⋅2)| , |(285285 - 5x⋅2)|, |(285285 - 7x⋅2)| , |(285285 - 11x⋅2)|, |(285285 - 13x⋅2)|, |(285285 - 19x⋅2)|$
Will all be composites with factors of $3,5,7,11,13,19$.
The only composite factors that we know for sure not to have factors of $3,5,7,11,13,19$ are $|285285-2^y|$.
Since I would like to achieve the smallest gap:
The closest bigger $2^y$ to $285285$ is $2^{17}=524288$.
The closest smaller $2^y$ to $285285$ is $2^{18}= 262144$.
$262144$ is closer, so I will use it:
$|285285-262144|= 23141$
If $23141$ is not a prime, then It has a prime factor that is bigger than $19$, and therefor there must be a prime between $19$ and $23141$ (including $23141$ itself).
Another example:
$P_2\#=2⋅3=6$
$\frac{6}{2}=3$
The closest bigger $2^y$ to $3$ is $2^2=4$.
$|3-4|= 1$ and $1<3$, so the next one is $2^3=8$ ( The condition is$|(P_n\#)/2-2^y|$ as long as $|(P_n\#)/2-2^y| > P_n$ ).
The closest smaller $2^y$ to $3$ does not exist.
I will use
$|3-8|= 5$
If $5$ is not a prime (pretending as if I don't know), then It has a prime factor that is bigger than $3$, and therefor there must be a prime between $3$ and $5$ (including $5$ itself).
At my current level that is the only way in which I could make sense of proving any gaps at all, and I would Just like to know if I am correct?
Thank you for your time.
In general, if $n=\prod p_i \pm \prod q_j$ where $p_i,q_j \in \mathbb P$ and $\forall i,j\ p_i\ne q_j$, then $n=\prod r_k$ where $r_k \ne p_i,q_j$ because $n$ will have a prime factorization and plainly none of $p_i,q_j$ can be a factor of $n$.
In your particular case, you saturate the primes up to $p_n$ by using the primorial, and then (1) remove the smallest prime $2$ from the product and (2) add or subtract a power of that prime. The numbers so obtained cannot be divisible by any of the $p_n$, but must have a prime factorization, so they must be composed of primes larger than $p_n$. Thus they will either be a prime themselves, or contain prime factors larger than $p_n$, no matter how close to $\frac{p_n\#}{2}$ they happen to be. Your reasoning is correct.
You could perform a similar algorithm on (for example) $\frac{p_n\#}{6} \pm 2^a3^b$ with similar results.