--- i1 --- --- i2 --- ... --- ir-1 --- --- ir --- ... -- ik --- ---- j1 --- --- j2 --- ... --- jr-1 --- --- jr --- ... ---jk --- ...Suppose that you know that the first r-1 of the it finish, respectively, no later than the first r-1 of the jt. Suppose there is an ir and a jr. Since ir-1 finishes no later than jr-1, and jr is compatible with jr-1; jr is compatible with ir-1 too. Therefore, jr is among the candidates considered when the EFT algorithm chooses ir. This shows that ir must have a finish time no later than jr. The foregoing considerations amount to a proof by induction that each it has a finish time no later than the corresponding jt for t = 1, 2, ..., k.