1.59/1.62 MAYBE 1.59/1.62 1.59/1.62 DP problem for innermost termination. 1.59/1.62 P = 1.59/1.62 f4#(x1, x2, x3) -> f3#(x1, x2, x3) 1.59/1.62 f3#(I0, I1, I2) -> f1#(I0, I1, I2) 1.59/1.62 f2#(I3, I4, I5) -> f1#(I3, I4, I5) 1.59/1.62 f1#(I6, I7, I8) -> f2#(I6 + I7, I7 - I8, 1 + I8) [1 <= I6] 1.59/1.62 R = 1.59/1.62 f4(x1, x2, x3) -> f3(x1, x2, x3) 1.59/1.62 f3(I0, I1, I2) -> f1(I0, I1, I2) 1.59/1.62 f2(I3, I4, I5) -> f1(I3, I4, I5) 1.59/1.62 f1(I6, I7, I8) -> f2(I6 + I7, I7 - I8, 1 + I8) [1 <= I6] 1.59/1.62 1.59/1.62 The dependency graph for this problem is: 1.59/1.62 0 -> 1 1.59/1.62 1 -> 3 1.59/1.62 2 -> 3 1.59/1.62 3 -> 2 1.59/1.62 Where: 1.59/1.62 0) f4#(x1, x2, x3) -> f3#(x1, x2, x3) 1.59/1.62 1) f3#(I0, I1, I2) -> f1#(I0, I1, I2) 1.59/1.62 2) f2#(I3, I4, I5) -> f1#(I3, I4, I5) 1.59/1.62 3) f1#(I6, I7, I8) -> f2#(I6 + I7, I7 - I8, 1 + I8) [1 <= I6] 1.59/1.62 1.59/1.62 We have the following SCCs. 1.59/1.62 { 2, 3 } 1.59/1.62 1.59/1.62 DP problem for innermost termination. 1.59/1.62 P = 1.59/1.62 f2#(I3, I4, I5) -> f1#(I3, I4, I5) 1.59/1.62 f1#(I6, I7, I8) -> f2#(I6 + I7, I7 - I8, 1 + I8) [1 <= I6] 1.59/1.62 R = 1.59/1.62 f4(x1, x2, x3) -> f3(x1, x2, x3) 1.59/1.62 f3(I0, I1, I2) -> f1(I0, I1, I2) 1.59/1.62 f2(I3, I4, I5) -> f1(I3, I4, I5) 1.59/1.62 f1(I6, I7, I8) -> f2(I6 + I7, I7 - I8, 1 + I8) [1 <= I6] 1.59/1.62 1.59/4.59 EOF