0.00/0.38 MAYBE 0.00/0.38 0.00/0.38 DP problem for innermost termination. 0.00/0.38 P = 0.00/0.38 init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.38 f2#(I0, I1) -> f2#(I1 - 1, I1 + 1) [I0 = I1 - 2 /\ I1 - 2 <= I1 - 1] 0.00/0.38 f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1 /\ I2 <= I3 - 2 - 1] 0.00/0.38 f1#(I4, I5) -> f2#(0, 100) 0.00/0.38 R = 0.00/0.38 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.38 f2(I0, I1) -> f2(I1 - 1, I1 + 1) [I0 = I1 - 2 /\ I1 - 2 <= I1 - 1] 0.00/0.38 f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1 /\ I2 <= I3 - 2 - 1] 0.00/0.38 f1(I4, I5) -> f2(0, 100) 0.00/0.38 0.00/0.38 The dependency graph for this problem is: 0.00/0.38 0 -> 3 0.00/0.38 1 -> 1 0.00/0.38 2 -> 1, 2 0.00/0.38 3 -> 2 0.00/0.38 Where: 0.00/0.38 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.38 1) f2#(I0, I1) -> f2#(I1 - 1, I1 + 1) [I0 = I1 - 2 /\ I1 - 2 <= I1 - 1] 0.00/0.38 2) f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1 /\ I2 <= I3 - 2 - 1] 0.00/0.38 3) f1#(I4, I5) -> f2#(0, 100) 0.00/0.38 0.00/0.38 We have the following SCCs. 0.00/0.38 { 2 } 0.00/0.38 { 1 } 0.00/0.38 0.00/0.38 DP problem for innermost termination. 0.00/0.38 P = 0.00/0.38 f2#(I0, I1) -> f2#(I1 - 1, I1 + 1) [I0 = I1 - 2 /\ I1 - 2 <= I1 - 1] 0.00/0.38 R = 0.00/0.38 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.38 f2(I0, I1) -> f2(I1 - 1, I1 + 1) [I0 = I1 - 2 /\ I1 - 2 <= I1 - 1] 0.00/0.38 f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1 /\ I2 <= I3 - 2 - 1] 0.00/0.38 f1(I4, I5) -> f2(0, 100) 0.00/0.38 0.00/3.36 EOF