/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 = init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] f2#(I13, I14, I15) -> f3#(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1#(I16, I17, I18) -> f2#(1, I19, I20) R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9] f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1(I16, I17, I18) -> f2(1, I19, I20) The dependency graph for this problem is: 0 -> 5 1 -> 1, 2, 3 2 -> 1, 2, 3 3 -> 4 4 -> 1, 2 5 -> 4 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] 2) f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] 3) f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] 4) f2#(I13, I14, I15) -> f3#(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] 5) f1#(I16, I17, I18) -> f2#(1, I19, I20) We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] f2#(I13, I14, I15) -> f3#(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9] f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1(I16, I17, I18) -> f2(1, I19, I20) We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = 99 + -1 * z1 NU[f3#(z1,z2,z3)] = 99 + -1 * (z1 + 1) This gives the following inequalities: I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1 ==> 99 + -1 * (I0 + 1) >= 99 + -1 * (I0 + 1) I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1 ==> 99 + -1 * (I3 + 1) >= 99 + -1 * (I3 + 1) I10 <= I9 ==> 99 + -1 * (I8 + 1) >= 99 + -1 * (I8 + 1) I13 <= 99 /\ 0 <= I13 - 1 ==> 99 + -1 * I13 > 99 + -1 * (I13 + 1) with 99 + -1 * I13 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9] f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1(I16, I17, I18) -> f2(1, I19, I20) The dependency graph for this problem is: 1 -> 1, 2, 3 2 -> 1, 2, 3 3 -> Where: 1) f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] 2) f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] 3) f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9] f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1(I16, I17, I18) -> f2(1, I19, I20) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3)] = 99 + -1 * z2 This gives the following inequalities: I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1 ==> 99 + -1 * I1 > 99 + -1 * (I1 + 1) with 99 + -1 * I1 >= 0 I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1 ==> 99 + -1 * I4 > 99 + -1 * (I4 + 1) with 99 + -1 * I4 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.