Correctness Analysis Clause Samples
Correctness Analysis. Theorem 1. GNi and GNi+1 can calculate the same symmetric key KTi+1, so that GNi+1 can get Mi.
Correctness Analysis. Given the group secret keys Ki and Kj generated by two vehicles Vi and Vj, we have: PIDi,1 = ri P, Di = dRSU ▇▇▇▇,▇ ▇▇▇▇,▇ = rj P, Dj = dRSU PIDj,1 Thus, ri−1 Di = dRSUP r−j 1 Dj = dRSUP = e( ∑ k=1 n = e( ∑ k=1 n = e( ∑ k=1 Di, ri−1 Di ) Di, dRSU P) Di, r−j 1 Dj ) Therefore, the two keys Ki and Kj are identical.