3.59/3.61 MAYBE 3.59/3.61 3.59/3.61 DP problem for innermost termination. 3.59/3.61 P = 3.59/3.61 f5#(x1, x2) -> f4#(x1, x2) 3.59/3.61 f4#(I0, I1) -> f1#(I0, I1) 3.59/3.61 f3#(I2, I3) -> f1#(I2, I3) 3.59/3.61 f1#(I4, I5) -> f3#(0, -1) [I5 <= 1 /\ 1 <= I4] 3.59/3.61 f2#(I6, I7) -> f1#(I6, I7) 3.59/3.61 f1#(I8, I9) -> f2#(1 + I8, -2 + I9) [2 <= I9 /\ 0 <= I8] 3.59/3.61 R = 3.59/3.61 f5(x1, x2) -> f4(x1, x2) 3.59/3.61 f4(I0, I1) -> f1(I0, I1) 3.59/3.61 f3(I2, I3) -> f1(I2, I3) 3.59/3.61 f1(I4, I5) -> f3(0, -1) [I5 <= 1 /\ 1 <= I4] 3.59/3.61 f2(I6, I7) -> f1(I6, I7) 3.59/3.61 f1(I8, I9) -> f2(1 + I8, -2 + I9) [2 <= I9 /\ 0 <= I8] 3.59/3.61 3.59/3.61 The dependency graph for this problem is: 3.59/3.61 0 -> 1 3.59/3.61 1 -> 3, 5 3.59/3.61 2 -> 3, 5 3.59/3.61 3 -> 2 3.59/3.61 4 -> 3, 5 3.59/3.61 5 -> 4 3.59/3.61 Where: 3.59/3.61 0) f5#(x1, x2) -> f4#(x1, x2) 3.59/3.61 1) f4#(I0, I1) -> f1#(I0, I1) 3.59/3.61 2) f3#(I2, I3) -> f1#(I2, I3) 3.59/3.61 3) f1#(I4, I5) -> f3#(0, -1) [I5 <= 1 /\ 1 <= I4] 3.59/3.61 4) f2#(I6, I7) -> f1#(I6, I7) 3.59/3.61 5) f1#(I8, I9) -> f2#(1 + I8, -2 + I9) [2 <= I9 /\ 0 <= I8] 3.59/3.61 3.59/3.61 We have the following SCCs. 3.59/3.61 { 2, 3, 4, 5 } 3.59/3.61 3.59/3.61 DP problem for innermost termination. 3.59/3.61 P = 3.59/3.61 f3#(I2, I3) -> f1#(I2, I3) 3.59/3.61 f1#(I4, I5) -> f3#(0, -1) [I5 <= 1 /\ 1 <= I4] 3.59/3.62 f2#(I6, I7) -> f1#(I6, I7) 3.59/3.62 f1#(I8, I9) -> f2#(1 + I8, -2 + I9) [2 <= I9 /\ 0 <= I8] 3.59/3.62 R = 3.59/3.62 f5(x1, x2) -> f4(x1, x2) 3.59/3.62 f4(I0, I1) -> f1(I0, I1) 3.59/3.62 f3(I2, I3) -> f1(I2, I3) 3.59/3.62 f1(I4, I5) -> f3(0, -1) [I5 <= 1 /\ 1 <= I4] 3.59/3.62 f2(I6, I7) -> f1(I6, I7) 3.59/3.62 f1(I8, I9) -> f2(1 + I8, -2 + I9) [2 <= I9 /\ 0 <= I8] 3.59/3.62 3.59/6.59 EOF