0.49/0.53 MAYBE 0.49/0.53 0.49/0.53 DP problem for innermost termination. 0.49/0.53 P = 0.49/0.53 init#(x1, x2) -> f1#(rnd1, rnd2) 0.49/0.53 f2#(I0, I1) -> f2#(I0 + 1, I1 - 1) [I0 <= I1 - 1 /\ 51 <= I1 - 1] 0.49/0.53 f2#(I2, I3) -> f2#(I2 - 1, I3 + 1) [I2 <= I3 - 1 /\ I3 <= 51] 0.49/0.53 f1#(I4, I5) -> f2#(0, 100) 0.49/0.53 R = 0.49/0.53 init(x1, x2) -> f1(rnd1, rnd2) 0.49/0.53 f2(I0, I1) -> f2(I0 + 1, I1 - 1) [I0 <= I1 - 1 /\ 51 <= I1 - 1] 0.49/0.53 f2(I2, I3) -> f2(I2 - 1, I3 + 1) [I2 <= I3 - 1 /\ I3 <= 51] 0.49/0.53 f1(I4, I5) -> f2(0, 100) 0.49/0.53 0.49/0.53 The dependency graph for this problem is: 0.49/0.53 0 -> 3 0.49/0.53 1 -> 1, 2 0.49/0.53 2 -> 1, 2 0.49/0.53 3 -> 1 0.49/0.53 Where: 0.49/0.53 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.49/0.53 1) f2#(I0, I1) -> f2#(I0 + 1, I1 - 1) [I0 <= I1 - 1 /\ 51 <= I1 - 1] 0.49/0.53 2) f2#(I2, I3) -> f2#(I2 - 1, I3 + 1) [I2 <= I3 - 1 /\ I3 <= 51] 0.49/0.53 3) f1#(I4, I5) -> f2#(0, 100) 0.49/0.53 0.49/0.53 We have the following SCCs. 0.49/0.53 { 1, 2 } 0.49/0.53 0.49/0.53 DP problem for innermost termination. 0.49/0.53 P = 0.49/0.53 f2#(I0, I1) -> f2#(I0 + 1, I1 - 1) [I0 <= I1 - 1 /\ 51 <= I1 - 1] 0.49/0.53 f2#(I2, I3) -> f2#(I2 - 1, I3 + 1) [I2 <= I3 - 1 /\ I3 <= 51] 0.49/0.53 R = 0.49/0.53 init(x1, x2) -> f1(rnd1, rnd2) 0.49/0.53 f2(I0, I1) -> f2(I0 + 1, I1 - 1) [I0 <= I1 - 1 /\ 51 <= I1 - 1] 0.49/0.53 f2(I2, I3) -> f2(I2 - 1, I3 + 1) [I2 <= I3 - 1 /\ I3 <= 51] 0.49/0.53 f1(I4, I5) -> f2(0, 100) 0.49/0.53 0.49/0.53 EOF