/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 = f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) f6#(I0, I1, I2, I3) -> f1#(I0, 2, I2, I3) f2#(I4, I5, I6, I7) -> f3#(I4, I5, I5, I7) [I5 <= I4] f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f1(I0, 2, I2, I3) f2(I4, I5, I6, I7) -> f3(I4, I5, I5, I7) [I5 <= I4] f2(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9] f4(I12, I13, I14, I15) -> f1(I12, 1 + I13, I14, I15) f4(I16, I17, I18, I19) -> f3(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27) The dependency graph for this problem is: 0 -> 1 1 -> 6 2 -> 5 3 -> 6 4 -> 5 5 -> 3, 4 6 -> 2 Where: 0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 1) f6#(I0, I1, I2, I3) -> f1#(I0, 2, I2, I3) 2) f2#(I4, I5, I6, I7) -> f3#(I4, I5, I5, I7) [I5 <= I4] 3) f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) 4) f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] 5) f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) 6) f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) We have the following SCCs. { 2, 3, 4, 5, 6 } DP problem for innermost termination. P = f2#(I4, I5, I6, I7) -> f3#(I4, I5, I5, I7) [I5 <= I4] f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f1(I0, 2, I2, I3) f2(I4, I5, I6, I7) -> f3(I4, I5, I5, I7) [I5 <= I4] f2(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9] f4(I12, I13, I14, I15) -> f1(I12, 1 + I13, I14, I15) f4(I16, I17, I18, I19) -> f3(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27) We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3)] = x0 - x1 NU[f4#(x0,x1,x2,x3)] = x0 - x1 - 1 NU[f3#(x0,x1,x2,x3)] = x0 - x1 - 1 NU[f2#(x0,x1,x2,x3)] = x0 - x1 This gives the following inequalities: I5 <= I4 ==> I4 - I5 > I4 - I5 - 1 with I4 - I5 >= 0 ==> I12 - I13 - 1 >= I12 - (1 + I13) rnd4 = rnd4 ==> I16 - I17 - 1 >= I16 - I17 - 1 ==> I20 - I21 - 1 >= I20 - I21 - 1 ==> I24 - I25 >= I24 - I25 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f1(I0, 2, I2, I3) f2(I4, I5, I6, I7) -> f3(I4, I5, I5, I7) [I5 <= I4] f2(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9] f4(I12, I13, I14, I15) -> f1(I12, 1 + I13, I14, I15) f4(I16, I17, I18, I19) -> f3(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27) The dependency graph for this problem is: 3 -> 6 4 -> 5 5 -> 3, 4 6 -> Where: 3) f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) 4) f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] 5) f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) 6) f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) We have the following SCCs. { 4, 5 } DP problem for innermost termination. P = f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f1(I0, 2, I2, I3) f2(I4, I5, I6, I7) -> f3(I4, I5, I5, I7) [I5 <= I4] f2(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9] f4(I12, I13, I14, I15) -> f1(I12, 1 + I13, I14, I15) f4(I16, I17, I18, I19) -> f3(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4] f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27)