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#(I0 + 1, I2) [0 <= I0 - 1] 0.00/0.36 f2#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 0] 0.00/0.36 f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1] 0.00/0.36 R = 0.00/0.36 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.36 f2(I0, I1) -> f2(I0 + 1, I2) [0 <= I0 - 1] 0.00/0.36 f2(I3, I4) -> f2(I3 - 1, I5) [I3 <= 0] 0.00/0.36 f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1] 0.00/0.36 0.00/0.36 The dependency graph for this problem is: 0.00/0.36 0 -> 3 0.00/0.36 1 -> 1 0.00/0.36 2 -> 2 0.00/0.36 3 -> 1, 2 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#(I0 + 1, I2) [0 <= I0 - 1] 0.00/0.36 2) f2#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 0] 0.00/0.36 3) f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1] 0.00/0.36 0.00/0.36 We have the following SCCs. 0.00/0.36 { 2 } 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#(I0 + 1, I2) [0 <= I0 - 1] 0.00/0.36 R = 0.00/0.36 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.36 f2(I0, I1) -> f2(I0 + 1, I2) [0 <= I0 - 1] 0.00/0.36 f2(I3, I4) -> f2(I3 - 1, I5) [I3 <= 0] 0.00/0.36 f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1] 0.00/0.36 0.00/3.34 EOF