0.00/0.46 MAYBE 0.00/0.46 0.00/0.46 DP problem for innermost termination. 0.00/0.46 P = 0.00/0.46 init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.46 f2#(I0, I1) -> f2#(35, I2) [I0 <= 35 /\ 30 <= I0 - 1] 0.00/0.46 f2#(I3, I4) -> f2#(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 35 /\ I3 <= 30] 0.00/0.46 f2#(I6, I7) -> f2#(0, I8) [35 <= I6 - 1] 0.00/0.46 f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.00/0.46 R = 0.00/0.46 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.46 f2(I0, I1) -> f2(35, I2) [I0 <= 35 /\ 30 <= I0 - 1] 0.00/0.46 f2(I3, I4) -> f2(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 35 /\ I3 <= 30] 0.00/0.46 f2(I6, I7) -> f2(0, I8) [35 <= I6 - 1] 0.00/0.46 f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.00/0.46 0.00/0.46 The dependency graph for this problem is: 0.00/0.46 0 -> 4 0.00/0.46 1 -> 1 0.00/0.46 2 -> 2 0.00/0.46 3 -> 0.00/0.46 4 -> 1, 2, 3 0.00/0.46 Where: 0.00/0.46 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.46 1) f2#(I0, I1) -> f2#(35, I2) [I0 <= 35 /\ 30 <= I0 - 1] 0.00/0.46 2) f2#(I3, I4) -> f2#(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 35 /\ I3 <= 30] 0.00/0.46 3) f2#(I6, I7) -> f2#(0, I8) [35 <= I6 - 1] 0.00/0.46 4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.00/0.46 0.00/0.46 We have the following SCCs. 0.00/0.46 { 2 } 0.00/0.46 { 1 } 0.00/0.46 0.00/0.46 DP problem for innermost termination. 0.00/0.46 P = 0.00/0.46 f2#(I0, I1) -> f2#(35, I2) [I0 <= 35 /\ 30 <= I0 - 1] 0.00/0.46 R = 0.00/0.46 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.46 f2(I0, I1) -> f2(35, I2) [I0 <= 35 /\ 30 <= I0 - 1] 0.00/0.46 f2(I3, I4) -> f2(I3 - 1, I5) [0 <= I3 - 1 /\ I3 <= 35 /\ I3 <= 30] 0.00/0.46 f2(I6, I7) -> f2(0, I8) [35 <= I6 - 1] 0.00/0.46 f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.00/0.46 0.00/3.44 EOF