0.49/0.62 MAYBE 0.49/0.62 0.49/0.62 DP problem for innermost termination. 0.49/0.62 P = 0.49/0.62 init#(x1, x2) -> f1#(rnd1, rnd2) 0.49/0.62 f2#(I0, I1) -> f2#(1, I1) [0 = I0 /\ 0 <= I1 - 1] 0.49/0.62 f2#(I2, I3) -> f2#(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1] 0.49/0.62 f2#(I4, I5) -> f2#(I4 + 1, I5) [I4 <= I5 - 1 /\ 0 <= I4 - 1] 0.49/0.62 f2#(I6, I7) -> f2#(0, 0) [0 = I7 /\ 0 = I6] 0.49/0.62 f1#(I8, I9) -> f2#(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1] 0.49/0.62 R = 0.49/0.62 init(x1, x2) -> f1(rnd1, rnd2) 0.49/0.62 f2(I0, I1) -> f2(1, I1) [0 = I0 /\ 0 <= I1 - 1] 0.49/0.62 f2(I2, I3) -> f2(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1] 0.49/0.62 f2(I4, I5) -> f2(I4 + 1, I5) [I4 <= I5 - 1 /\ 0 <= I4 - 1] 0.49/0.62 f2(I6, I7) -> f2(0, 0) [0 = I7 /\ 0 = I6] 0.49/0.62 f1(I8, I9) -> f2(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1] 0.49/0.62 0.49/0.62 The dependency graph for this problem is: 0.49/0.62 0 -> 5 0.49/0.62 1 -> 2, 3 0.49/0.62 2 -> 1, 4 0.49/0.62 3 -> 2, 3 0.49/0.62 4 -> 4 0.49/0.62 5 -> 1, 2, 3 0.49/0.62 Where: 0.49/0.62 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.49/0.62 1) f2#(I0, I1) -> f2#(1, I1) [0 = I0 /\ 0 <= I1 - 1] 0.49/0.62 2) f2#(I2, I3) -> f2#(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1] 0.49/0.62 3) f2#(I4, I5) -> f2#(I4 + 1, I5) [I4 <= I5 - 1 /\ 0 <= I4 - 1] 0.49/0.62 4) f2#(I6, I7) -> f2#(0, 0) [0 = I7 /\ 0 = I6] 0.49/0.62 5) f1#(I8, I9) -> f2#(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1] 0.49/0.62 0.49/0.62 We have the following SCCs. 0.49/0.62 { 1, 2, 3 } 0.49/0.62 { 4 } 0.49/0.62 0.49/0.62 DP problem for innermost termination. 0.49/0.62 P = 0.49/0.62 f2#(I6, I7) -> f2#(0, 0) [0 = I7 /\ 0 = I6] 0.49/0.62 R = 0.49/0.62 init(x1, x2) -> f1(rnd1, rnd2) 0.49/0.62 f2(I0, I1) -> f2(1, I1) [0 = I0 /\ 0 <= I1 - 1] 0.49/0.62 f2(I2, I3) -> f2(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1] 0.49/0.62 f2(I4, I5) -> f2(I4 + 1, I5) [I4 <= I5 - 1 /\ 0 <= I4 - 1] 0.49/0.62 f2(I6, I7) -> f2(0, 0) [0 = I7 /\ 0 = I6] 0.49/0.62 f1(I8, I9) -> f2(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1] 0.49/0.62 0.49/3.60 EOF