 Solution

Start with $P_1=I\vee E$ and repeatedly apply part (b), keeping a fresh output matrix. A simple implementation is

P := I OR E
repeat:
    Q := P OR boolean_product(P, E)
    if Q = P: return P
    P := Q

At the start of each iteration, $P=P_k$ for the current length bound $k$. The update produces $P_{k+1}$, proving the loop invariant by mathematical induction. If $Q=P$, every edge extension of a known graph path is already known. Repeating the same update then makes no further change, so every longer graph path is represented as well.

Every reachable pair of distinct vertices has a simple path of at most $n-1$ edges: a repeated vertex allows the intervening part to be deleted. Thus the process stabilizes after finitely many updates and returnsThis is the reflexive transitive closure. An alternative stopping rule is to stop once the length bound is at least $n-1$, avoiding a final equality-detection iteration. For $n=0$ or $1$, $P_1$ already suffices.