/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9] f2#(I12, I13, I14, I15) -> f5#(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] f1#(I29, I30, I31, I32) -> f2#(I29, I30, I31, I32) R = f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9] f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28] f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) The dependency graph for this problem is: 0 -> 7 1 -> 4, 6 2 -> 1 3 -> 2 4 -> 3 5 -> 4, 6 6 -> 5 7 -> 4, 6 Where: 0) f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 2) f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9] 4) f2#(I12, I13, I14, I15) -> f5#(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] 5) f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 6) f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] 7) f1#(I29, I30, I31, I32) -> f2#(I29, I30, I31, I32) We have the following SCCs. { 1, 2, 3, 4, 5, 6 } DP problem for innermost termination. P = f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9] f2#(I12, I13, I14, I15) -> f5#(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] R = f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9] f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28] f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) We use the extended value criterion with the projection function NU: NU[f4#(x0,x1,x2,x3)] = x3 NU[f5#(x0,x1,x2,x3)] = x3 - 1 NU[f6#(x0,x1,x2,x3)] = x3 - 1 NU[f2#(x0,x1,x2,x3)] = x3 NU[f7#(x0,x1,x2,x3)] = x3 This gives the following inequalities: ==> I3 >= I3 ==> I7 - 1 >= (-1 + I7) I9 = I9 ==> I11 - 1 >= I11 - 1 rnd3 = rnd3 /\ -1 * I15 <= 0 ==> I15 > I15 - 1 with I15 >= 0 ==> I19 >= I19 0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0 ==> I23 >= I23 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9] f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] R = f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9] f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28] f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) The dependency graph for this problem is: 1 -> 6 2 -> 1 3 -> 2 5 -> 6 6 -> 5 Where: 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 2) f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9] 5) f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 6) f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] We have the following SCCs. { 5, 6 } DP problem for innermost termination. P = f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] R = f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9] f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28] f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32)