0.61/0.66 MAYBE 0.61/0.66 0.61/0.66 DP problem for innermost termination. 0.61/0.66 P = 0.61/0.66 init#(x1, x2) -> f1#(rnd1, rnd2) 0.61/0.66 f2#(I0, I1) -> f2#(-1 * I0 - 2, I2) [I0 <= 0 /\ 3 <= -1 * I0 /\ I0 <= -2 - 1 /\ I0 <= -1 /\ I0 <= -1 - 1] 0.61/0.66 f2#(I3, I4) -> f2#(-1 * I3 + 2, I5) [-1 * I3 <= -1 - 1 /\ 2 <= I3 - 1] 0.61/0.66 f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] 0.61/0.66 f2#(I9, I10) -> f2#(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] 0.61/0.66 f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 0.61/0.66 R = 0.61/0.66 init(x1, x2) -> f1(rnd1, rnd2) 0.61/0.66 f2(I0, I1) -> f2(-1 * I0 - 2, I2) [I0 <= 0 /\ 3 <= -1 * I0 /\ I0 <= -2 - 1 /\ I0 <= -1 /\ I0 <= -1 - 1] 0.61/0.66 f2(I3, I4) -> f2(-1 * I3 + 2, I5) [-1 * I3 <= -1 - 1 /\ 2 <= I3 - 1] 0.61/0.66 f2(I6, I7) -> f2(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] 0.61/0.66 f2(I9, I10) -> f2(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] 0.61/0.66 f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 0.61/0.66 0.61/0.66 The dependency graph for this problem is: 0.61/0.66 0 -> 5 0.61/0.66 1 -> 2, 3 0.61/0.66 2 -> 1, 4 0.61/0.66 3 -> 4 0.61/0.66 4 -> 3 0.61/0.66 5 -> 2, 3 0.61/0.66 Where: 0.61/0.66 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.61/0.66 1) f2#(I0, I1) -> f2#(-1 * I0 - 2, I2) [I0 <= 0 /\ 3 <= -1 * I0 /\ I0 <= -2 - 1 /\ I0 <= -1 /\ I0 <= -1 - 1] 0.61/0.66 2) f2#(I3, I4) -> f2#(-1 * I3 + 2, I5) [-1 * I3 <= -1 - 1 /\ 2 <= I3 - 1] 0.61/0.66 3) f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] 0.61/0.66 4) f2#(I9, I10) -> f2#(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] 0.61/0.66 5) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 0.61/0.66 0.61/0.66 We have the following SCCs. 0.61/0.66 { 1, 2 } 0.61/0.66 { 3, 4 } 0.61/0.66 0.61/0.66 DP problem for innermost termination. 0.61/0.66 P = 0.61/0.66 f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] 0.61/0.66 f2#(I9, I10) -> f2#(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] 0.61/0.66 R = 0.61/0.66 init(x1, x2) -> f1(rnd1, rnd2) 0.61/0.66 f2(I0, I1) -> f2(-1 * I0 - 2, I2) [I0 <= 0 /\ 3 <= -1 * I0 /\ I0 <= -2 - 1 /\ I0 <= -1 /\ I0 <= -1 - 1] 0.61/0.66 f2(I3, I4) -> f2(-1 * I3 + 2, I5) [-1 * I3 <= -1 - 1 /\ 2 <= I3 - 1] 0.61/0.66 f2(I6, I7) -> f2(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] 0.61/0.66 f2(I9, I10) -> f2(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] 0.61/0.66 f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 0.61/0.66 0.61/3.64 EOF