1.39/1.41 MAYBE 1.39/1.41 1.39/1.41 DP problem for innermost termination. 1.39/1.41 P = 1.39/1.41 b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) 1.39/1.41 b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 1.39/1.41 Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 1.39/1.41 b10#(I6, I7, I8) -> b14#(I6, I7, I8) 1.39/1.41 R = 1.39/1.41 b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) 1.39/1.41 b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) 1.39/1.41 Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 1.39/1.41 b10(I6, I7, I8) -> b14(I6, I7, I8) 1.39/1.41 1.39/1.41 This problem is converted using chaining, where edges between chained DPs are removed. 1.39/1.41 1.39/1.41 DP problem for innermost termination. 1.39/1.41 P = 1.39/1.41 b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) 1.39/1.41 b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 1.39/1.41 Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 1.39/1.41 b10#(I6, I7, I8) -> b14#(I6, I7, I8) 1.39/1.41 b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && 1 < sv14_14] 1.39/1.41 R = 1.39/1.41 b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) 1.39/1.41 b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) 1.39/1.41 Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 1.39/1.41 b10(I6, I7, I8) -> b14(I6, I7, I8) 1.39/1.41 1.39/1.41 The dependency graph for this problem is: 1.39/1.41 0 -> 1.39/1.41 1 -> 3 1.39/1.41 2 -> 1 1.39/1.41 3 -> 4, 0 1.39/1.41 4 -> 1 1.39/1.41 Where: 1.39/1.41 0) b14#(sv14_14, sv23_37, sv24_38) -> Cond_b14#(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) 1.39/1.41 1) b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 1.39/1.41 2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 1.39/1.41 3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 1.39/1.41 4) b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && 1 < sv14_14] 1.39/1.41 1.39/1.41 We have the following SCCs. 1.39/1.41 { 1, 3, 4 } 1.39/1.41 1.39/1.41 DP problem for innermost termination. 1.39/1.41 P = 1.39/1.41 b15#(I0, I1, I2) -> b10#(I0, I1 - I0, I2 + 1) 1.39/1.41 b10#(I6, I7, I8) -> b14#(I6, I7, I8) 1.39/1.41 b14#(sv14_14, sv23_37, sv24_38) -> b15#(sv14_14, sv23_37, sv24_38) [sv23_37 >= sv14_14 && 1 < sv14_14] 1.39/1.41 R = 1.39/1.41 b14(sv14_14, sv23_37, sv24_38) -> Cond_b14(sv23_37 >= sv14_14 && 1 < sv14_14, sv14_14, sv23_37, sv24_38) 1.39/1.41 b15(I0, I1, I2) -> b10(I0, I1 - I0, I2 + 1) 1.39/1.41 Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 1.39/1.41 b10(I6, I7, I8) -> b14(I6, I7, I8) 1.39/1.41 1.39/4.39 EOF