Proof of correctness: Dijkstra's Algorithm

Notations:

D(S,u) = the minimum distance computed by Dijkstra's algorithm between nodes S and u
d(S,u) = the actual minimum distance between nodes S and u

Proposition 13.23 (Goodrich)

In Dijkstra's Algorithm, when (at the moment that) a vertex u is included into the ReachedSet, we have that:

Proof: (by contradiction)

Suppose the statement is false
Then there are some vertices where when (at the moment that) a vertex u is included into the ReachedSet:

Let x = the first of these vertices that was included into the ReachSet

In other words:

x = the first vertex such that when (at the moment that) vertex x is included into the ReachedSet:
This implies that all previous vertices z that were included into the ReachedSet, that:

Let's look at the moment when this node x is included into the ReachedSet

Denote:

The P be the (real) shortest path from S to x
Let z = the first vertex that is not in the "ReachSet" and is on the shortest path
Let y = the predecessor vertex of z on the shortest path P

Graphical depiction of the situation:

We can conclude the following:

D(S,y) = d(s,y) ....... (1)
(The min. distance S ⇒ y computed by the algorithm = actual min. distance S ⇒ y because y in included before x)
D(S,z) = D(S,y) + linkcost(y,z)
= d(s,y) + linkcost(y,z) ....... (2)
D(S,x) ≤ D(S,z) ....... (3)
(Because the next vertex included by the algorithm is vertex x)

Improtant fact:

A sub path of a shortest path is itself a shortest path
I.e.:
shortest path "S → x" = shortest path "S → z" ∪ shortest path "z → x"
Therefore:

We can now conclude that:

D(S,x) ≤ D(S,z) ............. (3) = d(s,y) + linkcost(y,z) (Equation (2)) ≤ d(S,y) + linkcost(y,z) + d(z,x) = d(S,x)

Ths result is a contradiction to the assumption:
Therefore, the assumption that "the statement false" is false
The statement must be true