0.00/0.43 MAYBE 0.00/0.43 0.00/0.43 DP problem for innermost termination. 0.00/0.43 P = 0.00/0.43 init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.43 f2#(I0, I1) -> f2#(I0 - 1, I2) [5 <= I0 - 1] 0.00/0.43 f2#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 4 /\ 0 <= I3 - 1] 0.00/0.43 f2#(I6, I7) -> f2#(5, I8) [5 = I6] 0.00/0.43 f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.00/0.43 R = 0.00/0.43 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.43 f2(I0, I1) -> f2(I0 - 1, I2) [5 <= I0 - 1] 0.00/0.43 f2(I3, I4) -> f2(I3 - 1, I5) [I3 <= 4 /\ 0 <= I3 - 1] 0.00/0.43 f2(I6, I7) -> f2(5, I8) [5 = I6] 0.00/0.43 f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.00/0.43 0.00/0.43 The dependency graph for this problem is: 0.00/0.43 0 -> 4 0.00/0.43 1 -> 1, 3 0.00/0.43 2 -> 2 0.00/0.43 3 -> 3 0.00/0.43 4 -> 1, 2, 3 0.00/0.43 Where: 0.00/0.43 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.00/0.43 1) f2#(I0, I1) -> f2#(I0 - 1, I2) [5 <= I0 - 1] 0.00/0.43 2) f2#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 4 /\ 0 <= I3 - 1] 0.00/0.43 3) f2#(I6, I7) -> f2#(5, I8) [5 = I6] 0.00/0.43 4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.00/0.43 0.00/0.43 We have the following SCCs. 0.00/0.43 { 2 } 0.00/0.43 { 1 } 0.00/0.43 { 3 } 0.00/0.43 0.00/0.43 DP problem for innermost termination. 0.00/0.43 P = 0.00/0.43 f2#(I6, I7) -> f2#(5, I8) [5 = I6] 0.00/0.43 R = 0.00/0.43 init(x1, x2) -> f1(rnd1, rnd2) 0.00/0.43 f2(I0, I1) -> f2(I0 - 1, I2) [5 <= I0 - 1] 0.00/0.43 f2(I3, I4) -> f2(I3 - 1, I5) [I3 <= 4 /\ 0 <= I3 - 1] 0.00/0.43 f2(I6, I7) -> f2(5, I8) [5 = I6] 0.00/0.43 f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.00/0.43 0.00/3.41 EOF