/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) f6#(I0, I1, I2, I3, I4) -> f1#(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2#(I5, I6, I7, I8, I9) -> f1#(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10] f2#(I10, I11, I12, I13, I14) -> f5#(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10] f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) f3#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10] f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10] f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10] f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10] f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26] f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) The dependency graph for this problem is: 0 -> 1 1 -> 6 2 -> 6 3 -> 4 4 -> 5 5 -> 4 6 -> 2, 3 Where: 0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 1) f6#(I0, I1, I2, I3, I4) -> f1#(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3] 2) f2#(I5, I6, I7, I8, I9) -> f1#(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10] 3) f2#(I10, I11, I12, I13, I14) -> f5#(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10] 4) f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) 5) f3#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10] 6) f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) We have the following SCCs. { 2, 6 } { 4, 5 } DP problem for innermost termination. P = f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) f3#(I20, I21, I22, I23, I24) -> f5#(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10] R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10] f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10] f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10] f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26] f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2) NU[f5#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2) This gives the following inequalities: ==> 10 + -1 * (1 + I16) >= 10 + -1 * (1 + I16) 1 + I21 <= 10 ==> 10 + -1 * (1 + I21) > 10 + -1 * (1 + (1 + I21)) with 10 + -1 * (1 + I21) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10] f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10] f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10] f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26] f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) The dependency graph for this problem is: 4 -> Where: 4) f5#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, I19) We have the following SCCs. DP problem for innermost termination. P = f2#(I5, I6, I7, I8, I9) -> f1#(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10] f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10] f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10] f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10] f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26] f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z1) NU[f2#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z1) This gives the following inequalities: 1 + I5 <= 10 ==> 10 + -1 * (1 + I5) > 10 + -1 * (1 + (1 + I5)) with 10 + -1 * (1 + I5) >= 0 ==> 10 + -1 * (1 + I30) >= 10 + -1 * (1 + I30) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f1(0, I1, rnd3, rnd4, I4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2(I5, I6, I7, I8, I9) -> f1(1 + I5, I6, I7, I8, I9) [1 + I5 <= 10] f2(I10, I11, I12, I13, I14) -> f5(I10, 0, I12, I13, rnd5) [rnd5 = rnd5 /\ 10 <= I10] f5(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, I19) f3(I20, I21, I22, I23, I24) -> f5(I20, 1 + I21, I22, I23, I24) [1 + I21 <= 10] f3(I25, I26, I27, I28, I29) -> f4(I25, I26, I27, I28, I29) [10 <= I26] f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, I33, I34) The dependency graph for this problem is: 6 -> Where: 6) f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, I33, I34) We have the following SCCs.