2.97/2.98 MAYBE 2.97/2.98 2.97/2.98 DP problem for innermost termination. 2.97/2.98 P = 2.97/2.98 f5#(x1, x2) -> f4#(x1, x2) 2.97/2.98 f4#(I0, I1) -> f1#(1, rnd2) [rnd2 = rnd2] 2.97/2.98 f3#(I2, I3) -> f1#(I2, I3) 2.97/2.98 f1#(I4, I5) -> f3#(-1 + I4, -10 + I5) [101 <= I5 /\ 1 <= I4] 2.97/2.98 f2#(I6, I7) -> f1#(I6, I7) 2.97/2.98 f1#(I8, I9) -> f2#(1 + I8, 11 + I9) [I9 <= 100 /\ 1 <= I8] 2.97/2.98 R = 2.97/2.98 f5(x1, x2) -> f4(x1, x2) 2.97/2.98 f4(I0, I1) -> f1(1, rnd2) [rnd2 = rnd2] 2.97/2.98 f3(I2, I3) -> f1(I2, I3) 2.97/2.98 f1(I4, I5) -> f3(-1 + I4, -10 + I5) [101 <= I5 /\ 1 <= I4] 2.97/2.98 f2(I6, I7) -> f1(I6, I7) 2.97/2.98 f1(I8, I9) -> f2(1 + I8, 11 + I9) [I9 <= 100 /\ 1 <= I8] 2.97/2.98 2.97/2.98 The dependency graph for this problem is: 2.97/2.98 0 -> 1 2.97/2.98 1 -> 3, 5 2.97/2.98 2 -> 3, 5 2.97/2.98 3 -> 2 2.97/2.98 4 -> 3, 5 2.97/2.98 5 -> 4 2.97/2.98 Where: 2.97/2.98 0) f5#(x1, x2) -> f4#(x1, x2) 2.97/2.98 1) f4#(I0, I1) -> f1#(1, rnd2) [rnd2 = rnd2] 2.97/2.98 2) f3#(I2, I3) -> f1#(I2, I3) 2.97/2.98 3) f1#(I4, I5) -> f3#(-1 + I4, -10 + I5) [101 <= I5 /\ 1 <= I4] 2.97/2.98 4) f2#(I6, I7) -> f1#(I6, I7) 2.97/2.98 5) f1#(I8, I9) -> f2#(1 + I8, 11 + I9) [I9 <= 100 /\ 1 <= I8] 2.97/2.98 2.97/2.98 We have the following SCCs. 2.97/2.98 { 2, 3, 4, 5 } 2.97/2.98 2.97/2.98 DP problem for innermost termination. 2.97/2.98 P = 2.97/2.98 f3#(I2, I3) -> f1#(I2, I3) 2.97/2.98 f1#(I4, I5) -> f3#(-1 + I4, -10 + I5) [101 <= I5 /\ 1 <= I4] 2.97/2.98 f2#(I6, I7) -> f1#(I6, I7) 2.97/2.98 f1#(I8, I9) -> f2#(1 + I8, 11 + I9) [I9 <= 100 /\ 1 <= I8] 2.97/2.98 R = 2.97/2.98 f5(x1, x2) -> f4(x1, x2) 2.97/2.98 f4(I0, I1) -> f1(1, rnd2) [rnd2 = rnd2] 2.97/2.98 f3(I2, I3) -> f1(I2, I3) 2.97/2.98 f1(I4, I5) -> f3(-1 + I4, -10 + I5) [101 <= I5 /\ 1 <= I4] 2.97/2.98 f2(I6, I7) -> f1(I6, I7) 2.97/2.98 f1(I8, I9) -> f2(1 + I8, 11 + I9) [I9 <= 100 /\ 1 <= I8] 2.97/2.98 2.97/5.96 EOF