1.76/1.82 YES 1.76/1.82 1.76/1.82 DP problem for innermost termination. 1.76/1.82 P = 1.76/1.82 f5#(x1, x2, x3) -> f4#(x1, x2, x3) 1.76/1.82 f4#(I0, I1, I2) -> f3#(I0, 1, I2) 1.76/1.82 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 R = 1.76/1.82 f5(x1, x2, x3) -> f4(x1, x2, x3) 1.76/1.82 f4(I0, I1, I2) -> f3(I0, 1, I2) 1.76/1.82 f3(I3, I4, I5) -> f1(I3, I4, I5) 1.76/1.82 f1(I6, I7, I8) -> f3(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10] 1.76/1.82 1.76/1.82 The dependency graph for this problem is: 1.76/1.82 0 -> 1 1.76/1.82 1 -> 2 1.76/1.82 2 -> 3 1.76/1.82 3 -> 2 1.76/1.82 Where: 1.76/1.82 0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 1.76/1.82 1) f4#(I0, I1, I2) -> f3#(I0, 1, I2) 1.76/1.82 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 3) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 1.76/1.82 We have the following SCCs. 1.76/1.82 { 2, 3 } 1.76/1.82 1.76/1.82 DP problem for innermost termination. 1.76/1.82 P = 1.76/1.82 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 R = 1.76/1.82 f5(x1, x2, x3) -> f4(x1, x2, x3) 1.76/1.82 f4(I0, I1, I2) -> f3(I0, 1, I2) 1.76/1.82 f3(I3, I4, I5) -> f1(I3, I4, I5) 1.76/1.82 f1(I6, I7, I8) -> f3(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10] 1.76/1.82 1.76/1.82 We use the reverse value criterion with the projection function NU: 1.76/1.82 NU[f1#(z1,z2,z3)] = z1 + -1 * z2 1.76/1.82 NU[f3#(z1,z2,z3)] = z1 + -1 * z2 1.76/1.82 1.76/1.82 This gives the following inequalities: 1.76/1.82 ==> I3 + -1 * I4 >= I3 + -1 * I4 1.76/1.82 I7 <= I6 ==> I6 + -1 * I7 > I6 + -1 * (1 + I7) with I6 + -1 * I7 >= 0 1.76/1.82 1.76/1.82 We remove all the strictly oriented dependency pairs. 1.76/1.82 1.76/1.82 DP problem for innermost termination. 1.76/1.82 P = 1.76/1.82 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 R = 1.76/1.82 f5(x1, x2, x3) -> f4(x1, x2, x3) 1.76/1.82 f4(I0, I1, I2) -> f3(I0, 1, I2) 1.76/1.82 f3(I3, I4, I5) -> f1(I3, I4, I5) 1.76/1.82 f1(I6, I7, I8) -> f3(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10] 1.76/1.82 1.76/1.82 The dependency graph for this problem is: 1.76/1.82 2 -> 1.76/1.82 Where: 1.76/1.82 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 1.76/1.82 We have the following SCCs. 1.76/1.82 1.76/4.80 EOF