0.00/0.36 MAYBE 0.00/0.36 0.00/0.36 DP problem for innermost termination. 0.00/0.36 P = 0.00/0.36 init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.36 f2#(I0, I1) -> f2#(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1] 0.00/0.36 f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1] 0.00/0.36 R = 0.00/0.36 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.36 f2(I0, I1) -> f2(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1] 0.00/0.36 f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1] 0.00/0.36 0.00/0.36 The dependency graph for this problem is: 0.00/0.36 0 -> 2 0.00/0.36 1 -> 1 0.00/0.36 2 -> 1 0.00/0.36 Where: 0.00/0.36 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.36 1) f2#(I0, I1) -> f2#(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1] 0.00/0.36 2) f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1] 0.00/0.36 0.00/0.36 We have the following SCCs. 0.00/0.36 { 1 } 0.00/0.36 0.00/0.36 DP problem for innermost termination. 0.00/0.36 P = 0.00/0.36 f2#(I0, I1) -> f2#(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1] 0.00/0.36 R = 0.00/0.36 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.36 f2(I0, I1) -> f2(I1, I0) [0 <= I0 - 1 /\ 0 <= I1 - 1] 0.00/0.36 f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1] 0.00/0.36 0.00/3.34 EOF