1.19/1.55 MAYBE 1.19/1.55 1.19/1.55 DP problem for innermost termination. 1.19/1.55 P = 1.19/1.55 f4#(x1, x2) -> f3#(x1, x2) 1.19/1.55 f3#(I0, I1) -> f1#(I0, I1) 1.19/1.55 f2#(I2, I3) -> f1#(I2, I3) 1.19/1.55 f1#(I4, I5) -> f2#(I4, I5) [I5 <= I4 /\ I4 <= I5] 1.19/1.55 R = 1.19/1.55 f4(x1, x2) -> f3(x1, x2) 1.19/1.55 f3(I0, I1) -> f1(I0, I1) 1.19/1.55 f2(I2, I3) -> f1(I2, I3) 1.19/1.55 f1(I4, I5) -> f2(I4, I5) [I5 <= I4 /\ I4 <= I5] 1.19/1.55 1.19/1.55 The dependency graph for this problem is: 1.19/1.55 0 -> 1 1.19/1.55 1 -> 3 1.19/1.55 2 -> 3 1.19/1.55 3 -> 2 1.19/1.55 Where: 1.19/1.55 0) f4#(x1, x2) -> f3#(x1, x2) 1.19/1.55 1) f3#(I0, I1) -> f1#(I0, I1) 1.19/1.55 2) f2#(I2, I3) -> f1#(I2, I3) 1.19/1.55 3) f1#(I4, I5) -> f2#(I4, I5) [I5 <= I4 /\ I4 <= I5] 1.19/1.55 1.19/1.55 We have the following SCCs. 1.19/1.55 { 2, 3 } 1.19/1.55 1.19/1.55 DP problem for innermost termination. 1.19/1.55 P = 1.19/1.55 f2#(I2, I3) -> f1#(I2, I3) 1.19/1.55 f1#(I4, I5) -> f2#(I4, I5) [I5 <= I4 /\ I4 <= I5] 1.19/1.55 R = 1.19/1.55 f4(x1, x2) -> f3(x1, x2) 1.19/1.55 f3(I0, I1) -> f1(I0, I1) 1.19/1.55 f2(I2, I3) -> f1(I2, I3) 1.19/1.55 f1(I4, I5) -> f2(I4, I5) [I5 <= I4 /\ I4 <= I5] 1.19/1.55 1.19/4.53 EOF