/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) f2#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) f3#(I21, I22, I23, I24, I25, I26, I27) -> f5#(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] R = f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) f2(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I13, I11, I12, I13) [I7 <= I8] f5(I14, I15, I16, I17, I18, I19, I20) -> f3(I14, I15, I16, I17, I18, I19, I20) f3(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] f3(I28, I29, I30, I31, I32, I33, I34) -> f4(I28, I29, I30, I34, I32, I33, I34) [I28 <= I29] f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] The dependency graph for this problem is: 0 -> 4 1 -> 3 2 -> 3 3 -> 2 4 -> 1 Where: 0) f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 1) f2#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] 2) f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) 3) f3#(I21, I22, I23, I24, I25, I26, I27) -> f5#(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] 4) f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) f3#(I21, I22, I23, I24, I25, I26, I27) -> f5#(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] R = f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) f2(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I13, I11, I12, I13) [I7 <= I8] f5(I14, I15, I16, I17, I18, I19, I20) -> f3(I14, I15, I16, I17, I18, I19, I20) f3(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] f3(I28, I29, I30, I31, I32, I33, I34) -> f4(I28, I29, I30, I34, I32, I33, I34) [I28 <= I29] f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = z1 + -1 * (1 + z2) NU[f5#(z1,z2,z3,z4,z5,z6,z7)] = z1 + -1 * (1 + z2) This gives the following inequalities: ==> I14 + -1 * (1 + I15) >= I14 + -1 * (1 + I15) 1 + I22 <= I21 ==> I21 + -1 * (1 + I22) > I21 + -1 * (1 + (1 + I22)) with I21 + -1 * (1 + I22) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) R = f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) f2(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I13, I11, I12, I13) [I7 <= I8] f5(I14, I15, I16, I17, I18, I19, I20) -> f3(I14, I15, I16, I17, I18, I19, I20) f3(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] f3(I28, I29, I30, I31, I32, I33, I34) -> f4(I28, I29, I30, I34, I32, I33, I34) [I28 <= I29] f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] The dependency graph for this problem is: 2 -> Where: 2) f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) We have the following SCCs.