1.03/1.07 MAYBE 1.03/1.07 1.03/1.07 DP problem for innermost termination. 1.03/1.07 P = 1.03/1.07 init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1.03/1.07 f2#(I0, I1, I2) -> f2#(I1 + 1, I1 + 1, -1 * I2 - 1) [0 <= I1 - 1 /\ I1 + 1 <= I2 /\ -1 * I2 <= 1 /\ 0 <= I0 - 1 /\ I1 <= I2 - 1 /\ 1 <= I2 - 1 /\ 0 - I1 <= I2] 1.03/1.07 f2#(I3, I4, I5) -> f2#(I4 + 1, I4 + 1, -1 * I5 + 1) [0 <= I4 - 1 /\ 0 <= I3 - 1 /\ I5 <= 0 - I4 - 1 /\ 0 <= I5 - 1] 1.03/1.07 f2#(I6, I7, I8) -> f2#(I7 + 1, I7 + 1, -1 * I8 + 1) [0 <= I7 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 - I7 - 1 /\ I8 <= -1] 1.03/1.07 f2#(I9, I10, I11) -> f2#(I10 + 1, I10 + 1, 0) [0 <= I9 - 1 /\ I11 <= I10 + 1 - 1 /\ 0 <= I10 - 1 /\ I11 <= I10 /\ 0 - I10 <= I11 /\ 0 <= I11 - 1] 1.03/1.07 f2#(I12, I13, I14) -> f2#(I13 + 1, I13 + 1, 0) [0 <= I12 - 1 /\ I14 <= I13 + 1 - 1 /\ 0 <= I13 - 1 /\ I14 <= I13 /\ 0 - I13 <= I14 /\ I14 <= -1] 1.03/1.07 f1#(I15, I16, I17) -> f2#(1, 1, I16) [-1 <= I16 - 1 /\ 0 <= I15 - 1] 1.03/1.07 R = 1.03/1.07 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 1.03/1.07 f2(I0, I1, I2) -> f2(I1 + 1, I1 + 1, -1 * I2 - 1) [0 <= I1 - 1 /\ I1 + 1 <= I2 /\ -1 * I2 <= 1 /\ 0 <= I0 - 1 /\ I1 <= I2 - 1 /\ 1 <= I2 - 1 /\ 0 - I1 <= I2] 1.03/1.07 f2(I3, I4, I5) -> f2(I4 + 1, I4 + 1, -1 * I5 + 1) [0 <= I4 - 1 /\ 0 <= I3 - 1 /\ I5 <= 0 - I4 - 1 /\ 0 <= I5 - 1] 1.03/1.07 f2(I6, I7, I8) -> f2(I7 + 1, I7 + 1, -1 * I8 + 1) [0 <= I7 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 - I7 - 1 /\ I8 <= -1] 1.03/1.07 f2(I9, I10, I11) -> f2(I10 + 1, I10 + 1, 0) [0 <= I9 - 1 /\ I11 <= I10 + 1 - 1 /\ 0 <= I10 - 1 /\ I11 <= I10 /\ 0 - I10 <= I11 /\ 0 <= I11 - 1] 1.03/1.07 f2(I12, I13, I14) -> f2(I13 + 1, I13 + 1, 0) [0 <= I12 - 1 /\ I14 <= I13 + 1 - 1 /\ 0 <= I13 - 1 /\ I14 <= I13 /\ 0 - I13 <= I14 /\ I14 <= -1] 1.03/1.07 f1(I15, I16, I17) -> f2(1, 1, I16) [-1 <= I16 - 1 /\ 0 <= I15 - 1] 1.03/1.07 1.03/1.07 The dependency graph for this problem is: 1.03/1.07 0 -> 6 1.03/1.07 1 -> 3 1.03/1.07 2 -> 1.03/1.07 3 -> 1 1.03/1.07 4 -> 1.03/1.07 5 -> 1.03/1.07 6 -> 1, 4 1.03/1.07 Where: 1.03/1.07 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1.03/1.07 1) f2#(I0, I1, I2) -> f2#(I1 + 1, I1 + 1, -1 * I2 - 1) [0 <= I1 - 1 /\ I1 + 1 <= I2 /\ -1 * I2 <= 1 /\ 0 <= I0 - 1 /\ I1 <= I2 - 1 /\ 1 <= I2 - 1 /\ 0 - I1 <= I2] 1.03/1.07 2) f2#(I3, I4, I5) -> f2#(I4 + 1, I4 + 1, -1 * I5 + 1) [0 <= I4 - 1 /\ 0 <= I3 - 1 /\ I5 <= 0 - I4 - 1 /\ 0 <= I5 - 1] 1.03/1.07 3) f2#(I6, I7, I8) -> f2#(I7 + 1, I7 + 1, -1 * I8 + 1) [0 <= I7 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 - I7 - 1 /\ I8 <= -1] 1.03/1.07 4) f2#(I9, I10, I11) -> f2#(I10 + 1, I10 + 1, 0) [0 <= I9 - 1 /\ I11 <= I10 + 1 - 1 /\ 0 <= I10 - 1 /\ I11 <= I10 /\ 0 - I10 <= I11 /\ 0 <= I11 - 1] 1.03/1.07 5) f2#(I12, I13, I14) -> f2#(I13 + 1, I13 + 1, 0) [0 <= I12 - 1 /\ I14 <= I13 + 1 - 1 /\ 0 <= I13 - 1 /\ I14 <= I13 /\ 0 - I13 <= I14 /\ I14 <= -1] 1.03/1.07 6) f1#(I15, I16, I17) -> f2#(1, 1, I16) [-1 <= I16 - 1 /\ 0 <= I15 - 1] 1.03/1.07 1.03/1.07 We have the following SCCs. 1.03/1.07 { 1, 3 } 1.03/1.07 1.03/1.07 DP problem for innermost termination. 1.03/1.07 P = 1.03/1.07 f2#(I0, I1, I2) -> f2#(I1 + 1, I1 + 1, -1 * I2 - 1) [0 <= I1 - 1 /\ I1 + 1 <= I2 /\ -1 * I2 <= 1 /\ 0 <= I0 - 1 /\ I1 <= I2 - 1 /\ 1 <= I2 - 1 /\ 0 - I1 <= I2] 1.03/1.07 f2#(I6, I7, I8) -> f2#(I7 + 1, I7 + 1, -1 * I8 + 1) [0 <= I7 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 - I7 - 1 /\ I8 <= -1] 1.03/1.07 R = 1.03/1.07 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 1.03/1.07 f2(I0, I1, I2) -> f2(I1 + 1, I1 + 1, -1 * I2 - 1) [0 <= I1 - 1 /\ I1 + 1 <= I2 /\ -1 * I2 <= 1 /\ 0 <= I0 - 1 /\ I1 <= I2 - 1 /\ 1 <= I2 - 1 /\ 0 - I1 <= I2] 1.03/1.07 f2(I3, I4, I5) -> f2(I4 + 1, I4 + 1, -1 * I5 + 1) [0 <= I4 - 1 /\ 0 <= I3 - 1 /\ I5 <= 0 - I4 - 1 /\ 0 <= I5 - 1] 1.03/1.07 f2(I6, I7, I8) -> f2(I7 + 1, I7 + 1, -1 * I8 + 1) [0 <= I7 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 - I7 - 1 /\ I8 <= -1] 1.03/1.07 f2(I9, I10, I11) -> f2(I10 + 1, I10 + 1, 0) [0 <= I9 - 1 /\ I11 <= I10 + 1 - 1 /\ 0 <= I10 - 1 /\ I11 <= I10 /\ 0 - I10 <= I11 /\ 0 <= I11 - 1] 1.03/1.07 f2(I12, I13, I14) -> f2(I13 + 1, I13 + 1, 0) [0 <= I12 - 1 /\ I14 <= I13 + 1 - 1 /\ 0 <= I13 - 1 /\ I14 <= I13 /\ 0 - I13 <= I14 /\ I14 <= -1] 1.03/1.07 f1(I15, I16, I17) -> f2(1, 1, I16) [-1 <= I16 - 1 /\ 0 <= I15 - 1] 1.03/1.07 1.03/4.05 EOF