/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 = f8#(x1, x2, x3) -> f1#(x1, x2, x3) f7#(I0, I1, I2) -> f2#(I0, I1, I2) f6#(I3, I4, I5) -> f7#(I3, I4, -1 + I5) f5#(I6, I7, I8) -> f6#(I6, I7, I8) [1 <= I7] f5#(I9, I10, I11) -> f6#(I9, I10, I11) [1 + I10 <= 0] f2#(I12, I13, I14) -> f5#(I12, rnd2, I14) [rnd2 = rnd2 /\ -1 * I14 <= 0] f4#(I15, I16, I17) -> f2#(I15, I16, I17) f2#(I18, I19, I20) -> f4#(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] f1#(I25, I26, I27) -> f2#(I25, I26, I27) R = f8(x1, x2, x3) -> f1(x1, x2, x3) f7(I0, I1, I2) -> f2(I0, I1, I2) f6(I3, I4, I5) -> f7(I3, I4, -1 + I5) f5(I6, I7, I8) -> f6(I6, I7, I8) [1 <= I7] f5(I9, I10, I11) -> f6(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f5(I12, rnd2, I14) [rnd2 = rnd2 /\ -1 * I14 <= 0] f4(I15, I16, I17) -> f2(I15, I16, I17) f2(I18, I19, I20) -> f4(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] f2(I22, I23, I24) -> f3(rnd1, I23, I24) [rnd1 = rnd1 /\ 0 <= -1 - I24] f1(I25, I26, I27) -> f2(I25, I26, I27) The dependency graph for this problem is: 0 -> 8 1 -> 5, 7 2 -> 1 3 -> 2 4 -> 2 5 -> 3, 4 6 -> 5, 7 7 -> 6 8 -> 5, 7 Where: 0) f8#(x1, x2, x3) -> f1#(x1, x2, x3) 1) f7#(I0, I1, I2) -> f2#(I0, I1, I2) 2) f6#(I3, I4, I5) -> f7#(I3, I4, -1 + I5) 3) f5#(I6, I7, I8) -> f6#(I6, I7, I8) [1 <= I7] 4) f5#(I9, I10, I11) -> f6#(I9, I10, I11) [1 + I10 <= 0] 5) f2#(I12, I13, I14) -> f5#(I12, rnd2, I14) [rnd2 = rnd2 /\ -1 * I14 <= 0] 6) f4#(I15, I16, I17) -> f2#(I15, I16, I17) 7) f2#(I18, I19, I20) -> f4#(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] 8) f1#(I25, I26, I27) -> f2#(I25, I26, I27) We have the following SCCs. { 1, 2, 3, 4, 5, 6, 7 } DP problem for innermost termination. P = f7#(I0, I1, I2) -> f2#(I0, I1, I2) f6#(I3, I4, I5) -> f7#(I3, I4, -1 + I5) f5#(I6, I7, I8) -> f6#(I6, I7, I8) [1 <= I7] f5#(I9, I10, I11) -> f6#(I9, I10, I11) [1 + I10 <= 0] f2#(I12, I13, I14) -> f5#(I12, rnd2, I14) [rnd2 = rnd2 /\ -1 * I14 <= 0] f4#(I15, I16, I17) -> f2#(I15, I16, I17) f2#(I18, I19, I20) -> f4#(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] R = f8(x1, x2, x3) -> f1(x1, x2, x3) f7(I0, I1, I2) -> f2(I0, I1, I2) f6(I3, I4, I5) -> f7(I3, I4, -1 + I5) f5(I6, I7, I8) -> f6(I6, I7, I8) [1 <= I7] f5(I9, I10, I11) -> f6(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f5(I12, rnd2, I14) [rnd2 = rnd2 /\ -1 * I14 <= 0] f4(I15, I16, I17) -> f2(I15, I16, I17) f2(I18, I19, I20) -> f4(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] f2(I22, I23, I24) -> f3(rnd1, I23, I24) [rnd1 = rnd1 /\ 0 <= -1 - I24] f1(I25, I26, I27) -> f2(I25, I26, I27) We use the extended value criterion with the projection function NU: NU[f4#(x0,x1,x2)] = x2 NU[f5#(x0,x1,x2)] = x2 - 1 NU[f6#(x0,x1,x2)] = x2 - 1 NU[f2#(x0,x1,x2)] = x2 NU[f7#(x0,x1,x2)] = x2 This gives the following inequalities: ==> I2 >= I2 ==> I5 - 1 >= (-1 + I5) 1 <= I7 ==> I8 - 1 >= I8 - 1 1 + I10 <= 0 ==> I11 - 1 >= I11 - 1 rnd2 = rnd2 /\ -1 * I14 <= 0 ==> I14 > I14 - 1 with I14 >= 0 ==> I17 >= I17 0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0 ==> I20 >= I20 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f7#(I0, I1, I2) -> f2#(I0, I1, I2) f6#(I3, I4, I5) -> f7#(I3, I4, -1 + I5) f5#(I6, I7, I8) -> f6#(I6, I7, I8) [1 <= I7] f5#(I9, I10, I11) -> f6#(I9, I10, I11) [1 + I10 <= 0] f4#(I15, I16, I17) -> f2#(I15, I16, I17) f2#(I18, I19, I20) -> f4#(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] R = f8(x1, x2, x3) -> f1(x1, x2, x3) f7(I0, I1, I2) -> f2(I0, I1, I2) f6(I3, I4, I5) -> f7(I3, I4, -1 + I5) f5(I6, I7, I8) -> f6(I6, I7, I8) [1 <= I7] f5(I9, I10, I11) -> f6(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f5(I12, rnd2, I14) [rnd2 = rnd2 /\ -1 * I14 <= 0] f4(I15, I16, I17) -> f2(I15, I16, I17) f2(I18, I19, I20) -> f4(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] f2(I22, I23, I24) -> f3(rnd1, I23, I24) [rnd1 = rnd1 /\ 0 <= -1 - I24] f1(I25, I26, I27) -> f2(I25, I26, I27) The dependency graph for this problem is: 1 -> 7 2 -> 1 3 -> 2 4 -> 2 6 -> 7 7 -> 6 Where: 1) f7#(I0, I1, I2) -> f2#(I0, I1, I2) 2) f6#(I3, I4, I5) -> f7#(I3, I4, -1 + I5) 3) f5#(I6, I7, I8) -> f6#(I6, I7, I8) [1 <= I7] 4) f5#(I9, I10, I11) -> f6#(I9, I10, I11) [1 + I10 <= 0] 6) f4#(I15, I16, I17) -> f2#(I15, I16, I17) 7) f2#(I18, I19, I20) -> f4#(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] We have the following SCCs. { 6, 7 } DP problem for innermost termination. P = f4#(I15, I16, I17) -> f2#(I15, I16, I17) f2#(I18, I19, I20) -> f4#(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] R = f8(x1, x2, x3) -> f1(x1, x2, x3) f7(I0, I1, I2) -> f2(I0, I1, I2) f6(I3, I4, I5) -> f7(I3, I4, -1 + I5) f5(I6, I7, I8) -> f6(I6, I7, I8) [1 <= I7] f5(I9, I10, I11) -> f6(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f5(I12, rnd2, I14) [rnd2 = rnd2 /\ -1 * I14 <= 0] f4(I15, I16, I17) -> f2(I15, I16, I17) f2(I18, I19, I20) -> f4(I18, I21, I20) [0 <= I21 /\ I21 <= 0 /\ I21 = I21 /\ -1 * I20 <= 0] f2(I22, I23, I24) -> f3(rnd1, I23, I24) [rnd1 = rnd1 /\ 0 <= -1 - I24] f1(I25, I26, I27) -> f2(I25, I26, I27)