/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f6#(x1, x2, x3) -> f5#(x1, x2, x3) f5#(I0, I1, I2) -> f1#(I0, I1, I2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 3, 5 2 -> 3, 5 3 -> 2 4 -> 3, 5 5 -> 4 Where: 0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 1) f5#(I0, I1, I2) -> f1#(I0, I1, I2) 2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 3) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] 4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 5) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f4#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0 NU[f1#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0 NU[f4#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0 This gives the following inequalities: ==> -1 - I4 + I5 + -1 * 0 >= -1 - I4 + I5 + -1 * 0 0 <= -1 - I7 + I8 ==> -1 - I7 + I8 + -1 * 0 > -1 - (1 + I7) + I8 + -1 * 0 with -1 - I7 + I8 + -1 * 0 >= 0 ==> -1 - I10 + I11 + -1 * 0 >= -1 - I10 + I11 + -1 * 0 0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0 ==> -1 - I13 + I14 + -1 * 0 >= -1 - (1 + I13) + I14 + -1 * 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I3, I4, I5) -> f1#(I3, I4, I5) f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0] The dependency graph for this problem is: 2 -> 5 4 -> 5 5 -> 4 Where: 2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 5) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] We have the following SCCs. { 4, 5 } DP problem for innermost termination. P = f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0]