**NOTATION**:

$u_0\ \ldots\ u_n\,\ $ and $\ v_0\ \ldots\ v_n\,\ $ are arbitrary finite strictly increasing sequences of non-negative integers, of the same sequence's length $\ n+1$.

$p(0)=2, p(1)=3, \ldots\ $ is the strictly increasing sequence of **all**
(consecutive) primes.

$ p(u_0)\ \ldots\ p(u_n)\,\ $ and $\,\ p(v_0)\ \ldots\ p(v_n)\ $ are
arbitrary finite strictly increasing sequences of primes (not necessarily
consecutive), where the length of the two sequences is the same, namely
$\ n+1.$

**The goal:** We will see that under certain additional assumptions, the
above two finite sequences are identical; in particular when one of them is assured to exist as an extension of a different shorter sequence.

Wojowu has posted his answer in the Q&A style. He deserves a more rounded presentation. Let me be the first one to do it. It'll be but a presentation. The results are exclusively due to **Wojowu (Wojciech Wawrów)**.

**Theorem 1** Assuming our NOTATION, if there exists a prime $\ q\ $
such that for every integer $\ x\ $ there exists at least one integer (index)
$\ a(x)\ $ such that
$$ p(u_{a(x)})\ \equiv x\ \mod q $$
and if there exists an integer $\ s\in\mathbb Z\ $ such that

$$\forall_{t=0}^n\quad p(v_t) = p(u_{s+t}) $$
and
$$ \forall_{t=1}^n\quad p(v_t)-p(v_{t-1})\ =\ p(u_t)-p(u_{t-1}) $$

then $\ s=0$.

**Remark** The last assumption can be written equivalently as

$$ \forall_{t=1}^n\quad p(v_t)-p(u_t)\ =\ p(v_{t-1})-p(u_{t-1}) $$

**Proof** (*of Theorem 1*) Let the assumptions of the theorem hold. By the above *Remark*,
for every integer $\ x\ $ there exists at least one integer (index)
$\ b(x)\ $ such that
$$ p(v_{b(x)})\ \equiv x\ \mod q $$

The terms in each prime sequence that are $\ 0 \mod q\ $
are simply equal to $\ q.\ $ If integer $\ s\ $ were different from $\ 0\ $
then the same prime $\ q\ $ would appear twice -- $\ s\ $ places apart -- in the increasing sequence of all primes; that would be a contradiction.
**End of Proof**

**More NOTATION**: Given integers $\ 0\le m\le n,\ $ prime sequence
$ p(u_0)\ \ldots\ p(u_n)\,\ $ is called to be the minimal $n$-extension
of a strictly increasing sequence $ p(u_0)\ \ldots\ p(u_m)\,\
\Leftarrow:\Rightarrow$
$$ \forall_{k=m+1}^n\quad u_k=u_m+k-m, $$

**Theorem 2** For every strictly increasing finite prime sequence $ p(u_0)\ \ldots\ p(u_m)\,\ $ there exists a minimal extension $ p(u_0)\ \ldots\ p(u_n)\,\ $ which is lonely, meaning that it is the only prime sequence $ p(v_0)\ \ldots\ p(v_n)\,\ $ such that there exists an integer $\ s\in\mathbb Z\ $ such that

$$\forall_{t=0}^n\quad p(v_t) = p(u_{s+t}) $$
and
$$ \forall_{t=1}^n\quad p(v_t)-p(v_{t-1})\ =\ p(u_t)-p(u_{t-1}). $$

**Proof** For $n$ large enough all residue classes \mod p(u_0) appear
in $ p(u_0)\ \ldots\ p(u_n)\,\ $ (by Dirichlet Theorem). This, by Theorem 1,
makes sequence $ p(u_0)\ \ldots\ p(u_n)\,\ $ lonely. **End of Proof**

answerbelow (it was my first choice but I don't mean to write here purely for my own sake).. $\endgroup$ – Wlod AA Sep 24 at 4:21