Given a weighted directed graph $$$G$$$ and $$$2$$$ different vertices $$$s$$$ and $$$t$$$ in it. How can we find all edges $$$e$$$ such that removing $$$e$$$ increases the distance between $$$s$$$ and $$$t$$$? Is there a better way then $$$O(E*E*log(V))$$$ time method where we test each edge by removing it? Can we also find all vertices $$$v$$$ defined analogously?

I had been solving High Score problem. After reading some blogs, I was able to conclude

Claim 1: Consider ALL negative cycles {C(i)} reachable from source S. In the Bellman Ford Algorithm, FOR EVERY cycle C(i), THERE EXISTS a vertex in C(i) whose distance is modified in the nth iteration of the Algorithm)

Claim 2: If the distance of a vertex is modified in the nth iteration, then it must be the part of a negative cycle

Can someone please verify if these claims are correct?

The Solution idea to High Score
Multiply all edge weights by -1, now we need to minimize our score
S: Source Vertex
D: Destination Vertex
C: A negative weight directed cycle

IF:
* In the nth iteration if there does not occur any relaxation for any edge, then there is no negative cycle reachable from S hence the answer dist(D) from Bellman Ford is the Correct answer
ELSE:
* There must exist a negative weight cycle reachable from S. But this does not imply that the Minimum Score to reach D is -INFINITY
==> It is possible that if we enter the cycle C, it is not possible to reach D so it we can't keep reducing our score by cycling through the cycle.
* So we need to check if there exits a negative weight cycle C such that it is possible to reach C from S and then reach D from C.
If such a C exists, the minimum score to reach D from S is -INFINITY
Else bellman ford Gave us th ecorrect answer after the nth ( or (n-1)th) iteration
All test cases passed
Could solmeone Acknowledge if Claim 1 and Claim 2 are correct? I believe they need to be for the correctness of the above mentioned solution.