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

*$\ (p\ q)\,\ \Leftarrow:\Rightarrow$ $$ 2\cdot p+1\ \le\ x\ \le\ 2\cdot q-1.$$*

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

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.