0.00/0.14 MAYBE 0.00/0.14 0.00/0.14 DP problem for innermost termination. 0.00/0.14 P = 0.00/0.14 f4#(x1) -> f3#(x1) 0.00/0.14 f3#(I0) -> f1#(1) [y1 = 1 /\ y2 = 0] 0.00/0.14 f2#(I1) -> f1#(I1) 0.00/0.14 f1#(I2) -> f2#(I2) 0.00/0.14 R = 0.00/0.14 f4(x1) -> f3(x1) 0.00/0.14 f3(I0) -> f1(1) [y1 = 1 /\ y2 = 0] 0.00/0.14 f2(I1) -> f1(I1) 0.00/0.14 f1(I2) -> f2(I2) 0.00/0.14 0.00/0.14 The dependency graph for this problem is: 0.00/0.14 0 -> 1 0.00/0.14 1 -> 3 0.00/0.14 2 -> 3 0.00/0.14 3 -> 2 0.00/0.14 Where: 0.00/0.14 0) f4#(x1) -> f3#(x1) 0.00/0.14 1) f3#(I0) -> f1#(1) [y1 = 1 /\ y2 = 0] 0.00/0.14 2) f2#(I1) -> f1#(I1) 0.00/0.14 3) f1#(I2) -> f2#(I2) 0.00/0.14 0.00/0.14 We have the following SCCs. 0.00/0.14 { 2, 3 } 0.00/0.14 0.00/0.14 DP problem for innermost termination. 0.00/0.14 P = 0.00/0.14 f2#(I1) -> f1#(I1) 0.00/0.14 f1#(I2) -> f2#(I2) 0.00/0.14 R = 0.00/0.14 f4(x1) -> f3(x1) 0.00/0.14 f3(I0) -> f1(1) [y1 = 1 /\ y2 = 0] 0.00/0.14 f2(I1) -> f1(I1) 0.00/0.14 f1(I2) -> f2(I2) 0.00/0.14 0.00/3.12 EOF