Primes above the distant prime neighbors

Question

Let $\ \mathbb P\ $ be the set of all natural primes. Pair $\ (p\ q)\ $ are prime neighbors $\ \Leftarrow:\Rightarrow$ $$ \{x\in\mathbb Z: p\le x\le q\}\cap\mathbb P\,\ =\,\ \{p\,\ q\} $$ Prime $\ x\in\mathbb P\ $ is said to be above neighbors $\ (p\ q)\,\ \Leftarrow:\Rightarrow$ $$ 2\cdot p+1\ \le\ x\ \le\ 2\cdot q-1.$$

EXAMPLE: There are no primes above prime neighbors $\ (101\ 103),\ $ i.e. $$ \{x\in\mathbb P:\ 2\cdot101+1\ \le\ x\ \le\ 2\cdot103-1\}\ \cap\ \mathbb P \quad =\quad \emptyset $$

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Prime neighbors $\ (p\ q)\ $ are said to be distant $\ \Leftarrow:\Rightarrow\ $ for every prime neighbors $\ (x\ y)\ $ such that $\ x<p\ $ we have $\ y-x<q-p$.

QUESTION (a conjecture):   There is a prime above every pair of distant prime neighbors.

Answer 1

This is probably true, and will almost certainly be hard to prove. The prime neighbours satisfying your hypothesis grow more quickly than one might expect, and the $p$ are listed in A002386. The corresponding gaps $q-p$ also grow more quickly than one might expect, and are listed in A005250. So an automatic search is easy to do, and shows that the number of primes "above" is also growing. In particular, your question has a positive answer for primes up to $10^{10}$.