0.00/0.43 MAYBE 0.00/0.43 0.00/0.43 DP problem for innermost termination. 0.00/0.43 P = 0.00/0.43 init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.43 f3#(I0, I1) -> f3#(I0 - 1, I0 - 1) [I0 = I1] 0.00/0.43 f2#(I2, I3) -> f3#(0, 1) [-1 <= y1 - 1 /\ 0 <= I3 - 1 /\ y1 - 2 * y2 = 0 /\ 0 <= I2 - 1 /\ y1 - 2 * y2 <= 1 /\ 0 <= y1 - 2 * y2] 0.00/0.43 f1#(I4, I5) -> f2#(I4, I5) [-1 <= I6 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 * I7 = 0 /\ 0 <= I4 - 1] 0.00/0.43 f2#(I8, I9) -> f3#(1, 0) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ I10 - 2 * I11 = 1 /\ 0 <= I8 - 1 /\ I10 - 2 * I11 <= 1 /\ 0 <= I10 - 2 * I11] 0.00/0.43 f1#(I12, I13) -> f2#(I12, I13) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ I14 - 2 * I15 = 1 /\ 0 <= I12 - 1] 0.00/0.43 R = 0.00/0.43 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.43 f3(I0, I1) -> f3(I0 - 1, I0 - 1) [I0 = I1] 0.00/0.43 f2(I2, I3) -> f3(0, 1) [-1 <= y1 - 1 /\ 0 <= I3 - 1 /\ y1 - 2 * y2 = 0 /\ 0 <= I2 - 1 /\ y1 - 2 * y2 <= 1 /\ 0 <= y1 - 2 * y2] 0.00/0.43 f1(I4, I5) -> f2(I4, I5) [-1 <= I6 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 * I7 = 0 /\ 0 <= I4 - 1] 0.00/0.43 f2(I8, I9) -> f3(1, 0) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ I10 - 2 * I11 = 1 /\ 0 <= I8 - 1 /\ I10 - 2 * I11 <= 1 /\ 0 <= I10 - 2 * I11] 0.00/0.43 f1(I12, I13) -> f2(I12, I13) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ I14 - 2 * I15 = 1 /\ 0 <= I12 - 1] 0.00/0.43 0.00/0.43 The dependency graph for this problem is: 0.00/0.43 0 -> 3, 5 0.00/0.43 1 -> 1 0.00/0.43 2 -> 0.00/0.43 3 -> 2, 4 0.00/0.43 4 -> 0.00/0.43 5 -> 2, 4 0.00/0.43 Where: 0.00/0.43 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.43 1) f3#(I0, I1) -> f3#(I0 - 1, I0 - 1) [I0 = I1] 0.00/0.43 2) f2#(I2, I3) -> f3#(0, 1) [-1 <= y1 - 1 /\ 0 <= I3 - 1 /\ y1 - 2 * y2 = 0 /\ 0 <= I2 - 1 /\ y1 - 2 * y2 <= 1 /\ 0 <= y1 - 2 * y2] 0.00/0.43 3) f1#(I4, I5) -> f2#(I4, I5) [-1 <= I6 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 * I7 = 0 /\ 0 <= I4 - 1] 0.00/0.43 4) f2#(I8, I9) -> f3#(1, 0) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ I10 - 2 * I11 = 1 /\ 0 <= I8 - 1 /\ I10 - 2 * I11 <= 1 /\ 0 <= I10 - 2 * I11] 0.00/0.43 5) f1#(I12, I13) -> f2#(I12, I13) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ I14 - 2 * I15 = 1 /\ 0 <= I12 - 1] 0.00/0.43 0.00/0.43 We have the following SCCs. 0.00/0.43 { 1 } 0.00/0.43 0.00/0.43 DP problem for innermost termination. 0.00/0.43 P = 0.00/0.43 f3#(I0, I1) -> f3#(I0 - 1, I0 - 1) [I0 = I1] 0.00/0.43 R = 0.00/0.43 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.43 f3(I0, I1) -> f3(I0 - 1, I0 - 1) [I0 = I1] 0.00/0.43 f2(I2, I3) -> f3(0, 1) [-1 <= y1 - 1 /\ 0 <= I3 - 1 /\ y1 - 2 * y2 = 0 /\ 0 <= I2 - 1 /\ y1 - 2 * y2 <= 1 /\ 0 <= y1 - 2 * y2] 0.00/0.43 f1(I4, I5) -> f2(I4, I5) [-1 <= I6 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 * I7 = 0 /\ 0 <= I4 - 1] 0.00/0.43 f2(I8, I9) -> f3(1, 0) [-1 <= I10 - 1 /\ 0 <= I9 - 1 /\ I10 - 2 * I11 = 1 /\ 0 <= I8 - 1 /\ I10 - 2 * I11 <= 1 /\ 0 <= I10 - 2 * I11] 0.00/0.43 f1(I12, I13) -> f2(I12, I13) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ I14 - 2 * I15 = 1 /\ 0 <= I12 - 1] 0.00/0.43 0.00/3.41 EOF