Answer should be $E,$ i.e., all statements $A,B,C,D$ are correct.
Explanation:
$\omega\left ( P \right ) =\sum_{e} \omega_e$
$\omega'\left ( P \right ) =\sum_{e} \omega_e'$
$\qquad =\sum_{e} (\omega (e) + \phi(v) - \phi(u))$
$\qquad =\sum \omega(e) + \sum ( \phi(v) - \phi(u) )$
Given that $P$ is a path from $s$ to $v.$ Let $P$ be comprised of $S \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \cdots \rightarrow V_n \rightarrow V $
$\require{cancel} \omega'\left ( P \right ) =\sum_{e} \omega(e) + \phi(v) -\cancel {\phi(v_n)}$
$\qquad +\cancel {\phi(v_n)} - \cancel{\phi(v_{n-1})}$
$\qquad + $
$\qquad \vdots$
$\qquad +\cancel {\phi(v_1)} - \phi(v_s) $
$\omega'(P)=\sum \omega(e) + \phi(v) - \phi(s)$
Since, $s$ is the source (in the context of the shortest path) $\phi (s) = 0.$
$ \therefore \omega'(P) = \sum \omega(e) + \phi(v) = \omega(P) + \phi(v)$
If $P$ is the shortest path in $G,$ $\omega'(P) = \sum \omega(e) + \phi(v)$
$\qquad = \phi(v) + \phi(v)$
$\qquad = 2. \phi(v)$
$\qquad = 2. \sum_{e} \omega_e$
$\qquad =2. \omega(P)$
(C is correct)
$\omega'(P) = \omega(P) + \phi(v)$
$\qquad \le \omega(P) + \omega(P)$
$\qquad = 2* \omega(P)$
or $\omega'(P) \le 2\omega(P)$
(D is correct)
If $P$ is the shortest path in $G,$ (Say from $s$ to $v)$
- $\omega(P) = \phi(v)$
- $\omega'(P) = \omega(P) + \phi(v)$
Say $P'$ is the shortest path in $G'$ (not equal to $P)$
$\omega'(P') \le \omega'(P)$
$\Rightarrow \omega(P') + \phi(v) \le \omega(P) + \phi(v)$
$\Rightarrow \omega(P') \le \omega(P) \Rightarrow$ Contradiction as we assumed $P$ to be shortest path in $G$
$\therefore$ $P$ is the shortest path in $G'$ as well.
(A is correct)
Similarly say $P$ is the shortest path in $G'$ $(s$ to $v)$
$\omega'(P) = \omega(P) + \phi(v)$
We claim $P$ is the shortest path in $G$ as well. By contradiction let us assume $P'$ is the shortest path in $G.$
$\omega(P') \le \omega(P)$
$\omega(P') + \phi(v) \le \omega(P) + \phi(v)$
$\omega'(P') \le \omega'(P) \Rightarrow$ Contradiction
Here, what we assumed was wrong and $P$ is the shortest path in $G$
(B is correct)
$\therefore$ E is the answer