7.90/7.83 YES 7.90/7.83 7.90/7.83 DP problem for innermost termination. 7.90/7.83 P = 7.90/7.83 f5#(x1, x2, x3) -> f1#(x1, x2, x3) 7.90/7.83 f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 f4#(I3, I4, I5) -> f3#(I3, I4, I5) 7.90/7.83 f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8] 7.90/7.83 f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 f1#(I12, I13, I14) -> f2#(I12, I13, I14) 7.90/7.83 R = 7.90/7.83 f5(x1, x2, x3) -> f1(x1, x2, x3) 7.90/7.83 f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 f4(I3, I4, I5) -> f3(I3, I4, I5) 7.90/7.83 f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] 7.90/7.83 f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 f1(I12, I13, I14) -> f2(I12, I13, I14) 7.90/7.83 7.90/7.83 The dependency graph for this problem is: 7.90/7.83 0 -> 5 7.90/7.83 1 -> 3, 4 7.90/7.83 2 -> 3, 4 7.90/7.83 3 -> 2 7.90/7.83 4 -> 1 7.90/7.83 5 -> 1 7.90/7.83 Where: 7.90/7.83 0) f5#(x1, x2, x3) -> f1#(x1, x2, x3) 7.90/7.83 1) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 2) f4#(I3, I4, I5) -> f3#(I3, I4, I5) 7.90/7.83 3) f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8] 7.90/7.83 4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 5) f1#(I12, I13, I14) -> f2#(I12, I13, I14) 7.90/7.83 7.90/7.83 We have the following SCCs. 7.90/7.83 { 1, 2, 3, 4 } 7.90/7.83 7.90/7.83 DP problem for innermost termination. 7.90/7.83 P = 7.90/7.83 f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 f4#(I3, I4, I5) -> f3#(I3, I4, I5) 7.90/7.83 f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8] 7.90/7.83 f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 R = 7.90/7.83 f5(x1, x2, x3) -> f1(x1, x2, x3) 7.90/7.83 f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 f4(I3, I4, I5) -> f3(I3, I4, I5) 7.90/7.83 f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] 7.90/7.83 f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 f1(I12, I13, I14) -> f2(I12, I13, I14) 7.90/7.83 7.90/7.83 We use the reverse value criterion with the projection function NU: 7.90/7.83 NU[f4#(z1,z2,z3)] = z3 + -1 * (1 + z2) 7.90/7.83 NU[f3#(z1,z2,z3)] = z3 + -1 * (1 + z2) 7.90/7.83 NU[f2#(z1,z2,z3)] = z3 + -1 * (1 + z2) 7.90/7.83 7.90/7.83 This gives the following inequalities: 7.90/7.83 1 + I0 <= I1 ==> I2 + -1 * (1 + I1) >= I2 + -1 * (1 + I1) 7.90/7.83 ==> I5 + -1 * (1 + I4) >= I5 + -1 * (1 + I4) 7.90/7.83 1 + I7 <= I8 ==> I8 + -1 * (1 + I7) > I8 + -1 * (1 + (1 + I7)) with I8 + -1 * (1 + I7) >= 0 7.90/7.83 I11 <= I10 ==> I11 + -1 * (1 + I10) >= I11 + -1 * (1 + I10) 7.90/7.83 7.90/7.83 We remove all the strictly oriented dependency pairs. 7.90/7.83 7.90/7.83 DP problem for innermost termination. 7.90/7.83 P = 7.90/7.83 f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 f4#(I3, I4, I5) -> f3#(I3, I4, I5) 7.90/7.83 f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 R = 7.90/7.83 f5(x1, x2, x3) -> f1(x1, x2, x3) 7.90/7.83 f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 f4(I3, I4, I5) -> f3(I3, I4, I5) 7.90/7.83 f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] 7.90/7.83 f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 f1(I12, I13, I14) -> f2(I12, I13, I14) 7.90/7.83 7.90/7.83 The dependency graph for this problem is: 7.90/7.83 1 -> 4 7.90/7.83 2 -> 4 7.90/7.83 4 -> 1 7.90/7.83 Where: 7.90/7.83 1) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 2) f4#(I3, I4, I5) -> f3#(I3, I4, I5) 7.90/7.83 4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 7.90/7.83 We have the following SCCs. 7.90/7.83 { 1, 4 } 7.90/7.83 7.90/7.83 DP problem for innermost termination. 7.90/7.83 P = 7.90/7.83 f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 R = 7.90/7.83 f5(x1, x2, x3) -> f1(x1, x2, x3) 7.90/7.83 f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 f4(I3, I4, I5) -> f3(I3, I4, I5) 7.90/7.83 f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] 7.90/7.83 f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 f1(I12, I13, I14) -> f2(I12, I13, I14) 7.90/7.83 7.90/7.83 We use the reverse value criterion with the projection function NU: 7.90/7.83 NU[f3#(z1,z2,z3)] = z2 + -1 * (1 + (1 + z1)) 7.90/7.83 NU[f2#(z1,z2,z3)] = z2 + -1 * (1 + z1) 7.90/7.83 7.90/7.83 This gives the following inequalities: 7.90/7.83 1 + I0 <= I1 ==> I1 + -1 * (1 + I0) > I1 + -1 * (1 + (1 + I0)) with I1 + -1 * (1 + I0) >= 0 7.90/7.83 I11 <= I10 ==> I10 + -1 * (1 + (1 + I9)) >= I10 + -1 * (1 + (1 + I9)) 7.90/7.83 7.90/7.83 We remove all the strictly oriented dependency pairs. 7.90/7.83 7.90/7.83 DP problem for innermost termination. 7.90/7.83 P = 7.90/7.83 f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 R = 7.90/7.83 f5(x1, x2, x3) -> f1(x1, x2, x3) 7.90/7.83 f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] 7.90/7.83 f4(I3, I4, I5) -> f3(I3, I4, I5) 7.90/7.83 f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] 7.90/7.83 f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 f1(I12, I13, I14) -> f2(I12, I13, I14) 7.90/7.83 7.90/7.83 The dependency graph for this problem is: 7.90/7.83 4 -> 7.90/7.83 Where: 7.90/7.83 4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 7.90/7.83 7.90/7.83 We have the following SCCs. 7.90/7.83 7.90/10.80 EOF