7.15/7.07 YES 7.15/7.07 7.15/7.07 DP problem for innermost termination. 7.15/7.07 P = 7.15/7.07 f6#(x1, x2, x3) -> f5#(x1, x2, x3) 7.15/7.07 f5#(I0, I1, I2) -> f1#(I0, I1, I2) 7.15/7.07 f4#(I3, I4, I5) -> f1#(I3, I4, I5) 7.15/7.07 f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 f3#(I9, I10, I11) -> f1#(I9, I10, I11) 7.15/7.07 f1#(I12, I13, I14) -> f3#(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] 7.15/7.07 R = 7.15/7.07 f6(x1, x2, x3) -> f5(x1, x2, x3) 7.15/7.07 f5(I0, I1, I2) -> f1(I0, I1, I2) 7.15/7.07 f4(I3, I4, I5) -> f1(I3, I4, I5) 7.15/7.07 f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 f3(I9, I10, I11) -> f1(I9, I10, I11) 7.15/7.07 f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] 7.15/7.07 f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0] 7.15/7.07 7.15/7.07 The dependency graph for this problem is: 7.15/7.07 0 -> 1 7.15/7.07 1 -> 3, 5 7.15/7.07 2 -> 3, 5 7.15/7.07 3 -> 2 7.15/7.07 4 -> 3, 5 7.15/7.07 5 -> 4 7.15/7.07 Where: 7.15/7.07 0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 7.15/7.07 1) f5#(I0, I1, I2) -> f1#(I0, I1, I2) 7.15/7.07 2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 7.15/7.07 3) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 7.15/7.07 5) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] 7.15/7.07 7.15/7.07 We have the following SCCs. 7.15/7.07 { 2, 3, 4, 5 } 7.15/7.07 7.15/7.07 DP problem for innermost termination. 7.15/7.07 P = 7.15/7.07 f4#(I3, I4, I5) -> f1#(I3, I4, I5) 7.15/7.07 f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 f3#(I9, I10, I11) -> f1#(I9, I10, I11) 7.15/7.07 f1#(I12, I13, I14) -> f3#(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] 7.15/7.07 R = 7.15/7.07 f6(x1, x2, x3) -> f5(x1, x2, x3) 7.15/7.07 f5(I0, I1, I2) -> f1(I0, I1, I2) 7.15/7.07 f4(I3, I4, I5) -> f1(I3, I4, I5) 7.15/7.07 f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 f3(I9, I10, I11) -> f1(I9, I10, I11) 7.15/7.07 f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] 7.15/7.07 f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0] 7.15/7.07 7.15/7.07 We use the reverse value criterion with the projection function NU: 7.15/7.07 NU[f3#(z1,z2,z3)] = 2 - z2 + -1 * 0 7.15/7.07 NU[f1#(z1,z2,z3)] = 2 - z2 + -1 * 0 7.15/7.07 NU[f4#(z1,z2,z3)] = 2 - z2 + -1 * 0 7.15/7.07 7.15/7.07 This gives the following inequalities: 7.15/7.07 ==> 2 - I4 + -1 * 0 >= 2 - I4 + -1 * 0 7.15/7.07 0 <= 1 - I8 ==> 2 - I7 + -1 * 0 >= 2 - (1 + I7) + -1 * 0 7.15/7.07 ==> 2 - I10 + -1 * 0 >= 2 - I10 + -1 * 0 7.15/7.07 0 <= 2 - I13 /\ 2 - I14 <= 0 ==> 2 - I13 + -1 * 0 > 2 - (1 + I13) + -1 * 0 with 2 - I13 + -1 * 0 >= 0 7.15/7.07 7.15/7.07 We remove all the strictly oriented dependency pairs. 7.15/7.07 7.15/7.07 DP problem for innermost termination. 7.15/7.07 P = 7.15/7.07 f4#(I3, I4, I5) -> f1#(I3, I4, I5) 7.15/7.07 f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 f3#(I9, I10, I11) -> f1#(I9, I10, I11) 7.15/7.07 R = 7.15/7.07 f6(x1, x2, x3) -> f5(x1, x2, x3) 7.15/7.07 f5(I0, I1, I2) -> f1(I0, I1, I2) 7.15/7.07 f4(I3, I4, I5) -> f1(I3, I4, I5) 7.15/7.07 f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 f3(I9, I10, I11) -> f1(I9, I10, I11) 7.15/7.07 f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] 7.15/7.07 f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0] 7.15/7.07 7.15/7.07 The dependency graph for this problem is: 7.15/7.07 2 -> 3 7.15/7.07 3 -> 2 7.15/7.07 4 -> 3 7.15/7.07 Where: 7.15/7.07 2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 7.15/7.07 3) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 7.15/7.07 7.15/7.07 We have the following SCCs. 7.15/7.07 { 2, 3 } 7.15/7.07 7.15/7.07 DP problem for innermost termination. 7.15/7.07 P = 7.15/7.07 f4#(I3, I4, I5) -> f1#(I3, I4, I5) 7.15/7.07 f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 R = 7.15/7.07 f6(x1, x2, x3) -> f5(x1, x2, x3) 7.15/7.07 f5(I0, I1, I2) -> f1(I0, I1, I2) 7.15/7.07 f4(I3, I4, I5) -> f1(I3, I4, I5) 7.15/7.07 f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 f3(I9, I10, I11) -> f1(I9, I10, I11) 7.15/7.07 f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] 7.15/7.07 f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0] 7.15/7.07 7.15/7.07 We use the reverse value criterion with the projection function NU: 7.15/7.07 NU[f1#(z1,z2,z3)] = 1 - z3 + -1 * 0 7.15/7.07 NU[f4#(z1,z2,z3)] = 1 - z3 + -1 * 0 7.15/7.07 7.15/7.07 This gives the following inequalities: 7.15/7.07 ==> 1 - I5 + -1 * 0 >= 1 - I5 + -1 * 0 7.15/7.07 0 <= 1 - I8 ==> 1 - I8 + -1 * 0 > 1 - (1 + I8) + -1 * 0 with 1 - I8 + -1 * 0 >= 0 7.15/7.07 7.15/7.07 We remove all the strictly oriented dependency pairs. 7.15/7.07 7.15/7.07 DP problem for innermost termination. 7.15/7.07 P = 7.15/7.07 f4#(I3, I4, I5) -> f1#(I3, I4, I5) 7.15/7.07 R = 7.15/7.07 f6(x1, x2, x3) -> f5(x1, x2, x3) 7.15/7.07 f5(I0, I1, I2) -> f1(I0, I1, I2) 7.15/7.07 f4(I3, I4, I5) -> f1(I3, I4, I5) 7.15/7.07 f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 7.15/7.07 f3(I9, I10, I11) -> f1(I9, I10, I11) 7.15/7.07 f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] 7.15/7.07 f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0] 7.15/7.07 7.15/7.07 The dependency graph for this problem is: 7.15/7.07 2 -> 7.15/7.07 Where: 7.15/7.07 2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 7.15/7.07 7.15/7.07 We have the following SCCs. 7.15/7.07 7.15/10.05 EOF