/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 = f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16] f1#(I20, I21, I22, I23, I24) -> f4#(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16] f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 5, 7 2 -> 5, 7 3 -> 2 4 -> 3 5 -> 4 6 -> 5, 7 7 -> 6 Where: 0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 1) f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) 4) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16] 5) f1#(I20, I21, I22, I23, I24) -> f4#(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] 6) f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 7) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] We have the following SCCs. { 2, 3, 4, 5, 6, 7 } DP problem for innermost termination. P = f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16] f1#(I20, I21, I22, I23, I24) -> f4#(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16] f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1 NU[f4#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2 NU[f5#(x0,x1,x2,x3,x4)] = -x3 + x4 - 2 NU[f1#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1 NU[f6#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1 This gives the following inequalities: ==> -I8 + I9 - 1 >= -I8 + I9 - 1 ==> -I13 + I14 - 2 >= -(1 + I13) + I14 - 1 I16 = I16 ==> -I18 + I19 - 2 >= -I18 + I19 - 2 rnd3 = rnd3 /\ 0 <= -1 - I23 + I24 ==> -I23 + I24 - 1 > -I23 + I24 - 2 with -I23 + I24 - 1 >= 0 ==> -I28 + I29 - 1 >= -I28 + I29 - 1 0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34 ==> -I33 + I34 - 1 >= -I33 + (-1 + I34) - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16] f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16] f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0] The dependency graph for this problem is: 2 -> 7 3 -> 2 4 -> 3 6 -> 7 7 -> 6 Where: 2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) 4) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16] 6) f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 7) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] We have the following SCCs. { 6, 7 } DP problem for innermost termination. P = f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16] f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = -1 - z4 + z5 + -1 * 0 NU[f3#(z1,z2,z3,z4,z5)] = -1 - z4 + z5 + -1 * 0 This gives the following inequalities: ==> -1 - I28 + I29 + -1 * 0 >= -1 - I28 + I29 + -1 * 0 0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34 ==> -1 - I33 + I34 + -1 * 0 > -1 - I33 + (-1 + I34) + -1 * 0 with -1 - I33 + I34 + -1 * 0 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16] f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, -1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0] The dependency graph for this problem is: 6 -> Where: 6) f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) We have the following SCCs.