0.71/0.81 YES 0.71/0.81 0.71/0.81 DP problem for innermost termination. 0.71/0.81 P = 0.71/0.81 f7#(x1, x2, x3) -> f6#(x1, x2, x3) 0.71/0.81 f6#(I0, I1, I2) -> f4#(0, I1, I2) 0.71/0.81 f4#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 <= 0] 0.71/0.81 f3#(I9, I10, I11) -> f1#(I9, I10, I11) 0.71/0.81 f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14] 0.71/0.81 R = 0.71/0.81 f7(x1, x2, x3) -> f6(x1, x2, x3) 0.71/0.81 f6(I0, I1, I2) -> f4(0, I1, I2) 0.71/0.81 f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4] 0.71/0.81 f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0] 0.71/0.81 f3(I9, I10, I11) -> f1(I9, I10, I11) 0.71/0.81 f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14] 0.71/0.81 f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0] 0.71/0.81 0.71/0.81 The dependency graph for this problem is: 0.71/0.81 0 -> 1 0.71/0.81 1 -> 2 0.71/0.81 2 -> 3 0.71/0.81 3 -> 4 0.71/0.81 4 -> 3 0.71/0.81 Where: 0.71/0.81 0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 0.71/0.81 1) f6#(I0, I1, I2) -> f4#(0, I1, I2) 0.71/0.81 2) f4#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 <= 0] 0.71/0.81 3) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 0.71/0.81 4) f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14] 0.71/0.81 0.71/0.81 We have the following SCCs. 0.71/0.81 { 3, 4 } 0.71/0.81 0.71/0.81 DP problem for innermost termination. 0.71/0.81 P = 0.71/0.81 f3#(I9, I10, I11) -> f1#(I9, I10, I11) 0.71/0.81 f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14] 0.71/0.81 R = 0.71/0.81 f7(x1, x2, x3) -> f6(x1, x2, x3) 0.71/0.81 f6(I0, I1, I2) -> f4(0, I1, I2) 0.71/0.81 f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4] 0.71/0.81 f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0] 0.71/0.81 f3(I9, I10, I11) -> f1(I9, I10, I11) 0.71/0.81 f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14] 0.71/0.81 f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0] 0.71/0.81 0.71/0.81 We use the basic value criterion with the projection function NU: 0.71/0.81 NU[f1#(z1,z2,z3)] = z3 0.71/0.81 NU[f3#(z1,z2,z3)] = z3 0.71/0.81 0.71/0.81 This gives the following inequalities: 0.71/0.81 ==> I11 (>! \union =) I11 0.71/0.81 1 <= I14 ==> I14 >! -1 + I14 0.71/0.81 0.71/0.81 We remove all the strictly oriented dependency pairs. 0.71/0.81 0.71/0.81 DP problem for innermost termination. 0.71/0.81 P = 0.71/0.81 f3#(I9, I10, I11) -> f1#(I9, I10, I11) 0.71/0.81 R = 0.71/0.81 f7(x1, x2, x3) -> f6(x1, x2, x3) 0.71/0.81 f6(I0, I1, I2) -> f4(0, I1, I2) 0.71/0.81 f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4] 0.71/0.81 f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0] 0.71/0.81 f3(I9, I10, I11) -> f1(I9, I10, I11) 0.71/0.81 f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14] 0.71/0.81 f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0] 0.71/0.81 0.71/0.81 The dependency graph for this problem is: 0.71/0.81 3 -> 0.71/0.81 Where: 0.71/0.81 3) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 0.71/0.81 0.71/0.81 We have the following SCCs. 0.71/0.81 0.71/3.79 EOF