0.69/0.76 YES 0.69/0.76 0.69/0.76 DP problem for innermost termination. 0.69/0.76 P = 0.69/0.76 f5#(x1, x2, x3) -> f4#(x1, x2, x3) 0.69/0.76 f4#(I0, I1, I2) -> f1#(I0, I1, I2) 0.69/0.76 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 0.69/0.76 f1#(I6, I7, I8) -> f3#(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7] 0.69/0.76 R = 0.69/0.76 f5(x1, x2, x3) -> f4(x1, x2, x3) 0.69/0.76 f4(I0, I1, I2) -> f1(I0, I1, I2) 0.69/0.76 f3(I3, I4, I5) -> f1(I3, I4, I5) 0.69/0.76 f1(I6, I7, I8) -> f3(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7] 0.69/0.76 f1(I9, I10, I11) -> f2(rnd1, I10, I11) [rnd1 = rnd1 /\ I10 + I11 <= 0 /\ 0 <= -1 + I11 /\ 0 <= -1 + I10] 0.69/0.76 f1(I12, I13, I14) -> f2(I15, I13, I14) [I15 = I15 /\ I14 <= 0 /\ 0 <= -1 + I13] 0.69/0.76 f1(I16, I17, I18) -> f2(I19, I17, I18) [I19 = I19 /\ I17 <= 0] 0.69/0.76 0.69/0.76 The dependency graph for this problem is: 0.69/0.76 0 -> 1 0.69/0.76 1 -> 3 0.69/0.76 2 -> 3 0.69/0.76 3 -> 2 0.69/0.76 Where: 0.69/0.76 0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 0.69/0.76 1) f4#(I0, I1, I2) -> f1#(I0, I1, I2) 0.69/0.76 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 0.69/0.76 3) f1#(I6, I7, I8) -> f3#(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7] 0.69/0.76 0.69/0.76 We have the following SCCs. 0.69/0.76 { 2, 3 } 0.69/0.76 0.69/0.76 DP problem for innermost termination. 0.69/0.76 P = 0.69/0.76 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 0.69/0.76 f1#(I6, I7, I8) -> f3#(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7] 0.69/0.76 R = 0.69/0.76 f5(x1, x2, x3) -> f4(x1, x2, x3) 0.69/0.76 f4(I0, I1, I2) -> f1(I0, I1, I2) 0.69/0.76 f3(I3, I4, I5) -> f1(I3, I4, I5) 0.69/0.76 f1(I6, I7, I8) -> f3(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7] 0.69/0.76 f1(I9, I10, I11) -> f2(rnd1, I10, I11) [rnd1 = rnd1 /\ I10 + I11 <= 0 /\ 0 <= -1 + I11 /\ 0 <= -1 + I10] 0.69/0.76 f1(I12, I13, I14) -> f2(I15, I13, I14) [I15 = I15 /\ I14 <= 0 /\ 0 <= -1 + I13] 0.69/0.76 f1(I16, I17, I18) -> f2(I19, I17, I18) [I19 = I19 /\ I17 <= 0] 0.69/0.76 0.69/0.76 We use the basic value criterion with the projection function NU: 0.69/0.76 NU[f1#(z1,z2,z3)] = z3 0.69/0.76 NU[f3#(z1,z2,z3)] = z3 0.69/0.76 0.69/0.76 This gives the following inequalities: 0.69/0.76 ==> I5 (>! \union =) I5 0.69/0.76 0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7 ==> I8 >! 1 + -2 + I8 0.69/0.76 0.69/0.76 We remove all the strictly oriented dependency pairs. 0.69/0.76 0.69/0.76 DP problem for innermost termination. 0.69/0.76 P = 0.69/0.76 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 0.69/0.76 R = 0.69/0.76 f5(x1, x2, x3) -> f4(x1, x2, x3) 0.69/0.76 f4(I0, I1, I2) -> f1(I0, I1, I2) 0.69/0.76 f3(I3, I4, I5) -> f1(I3, I4, I5) 0.69/0.76 f1(I6, I7, I8) -> f3(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7] 0.69/0.76 f1(I9, I10, I11) -> f2(rnd1, I10, I11) [rnd1 = rnd1 /\ I10 + I11 <= 0 /\ 0 <= -1 + I11 /\ 0 <= -1 + I10] 0.69/0.76 f1(I12, I13, I14) -> f2(I15, I13, I14) [I15 = I15 /\ I14 <= 0 /\ 0 <= -1 + I13] 0.69/0.76 f1(I16, I17, I18) -> f2(I19, I17, I18) [I19 = I19 /\ I17 <= 0] 0.69/0.76 0.69/0.76 The dependency graph for this problem is: 0.69/0.76 2 -> 0.69/0.76 Where: 0.69/0.76 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 0.69/0.76 0.69/0.76 We have the following SCCs. 0.69/0.76 0.69/3.74 EOF