0.00/0.01 YES 0.00/0.01 0.00/0.01 DP problem for innermost termination. 0.00/0.01 P = 0.00/0.01 f4#(x1, x2) -> f3#(x1, x2) 0.00/0.01 f3#(I0, I1) -> f1#(I0, I1) 0.00/0.01 f2#(I2, I3) -> f1#(I2, I3) 0.00/0.01 f1#(I4, I5) -> f2#(I4, I5) [1 + I5 <= I4 /\ 1 + I4 <= I5] 0.00/0.01 R = 0.00/0.01 f4(x1, x2) -> f3(x1, x2) 0.00/0.01 f3(I0, I1) -> f1(I0, I1) 0.00/0.01 f2(I2, I3) -> f1(I2, I3) 0.00/0.01 f1(I4, I5) -> f2(I4, I5) [1 + I5 <= I4 /\ 1 + I4 <= I5] 0.00/0.01 0.00/0.01 The dependency graph for this problem is: 0.00/0.01 0 -> 1 0.00/0.01 1 -> 0.00/0.01 2 -> 0.00/0.01 3 -> 0.00/0.01 Where: 0.00/0.01 0) f4#(x1, x2) -> f3#(x1, x2) 0.00/0.01 1) f3#(I0, I1) -> f1#(I0, I1) 0.00/0.01 2) f2#(I2, I3) -> f1#(I2, I3) 0.00/0.01 3) f1#(I4, I5) -> f2#(I4, I5) [1 + I5 <= I4 /\ 1 + I4 <= I5] 0.00/0.01 0.00/0.01 We have the following SCCs. 0.00/0.01 0.00/0.01 EOF