/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 = init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) f5#(I0, I1, I2) -> f5#(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0] f5#(I4, I5, I6) -> f5#(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1] f3#(I8, I9, I10) -> f5#(I9, I10, I11) [0 <= I8 - 1] f4#(I12, I13, I14) -> f3#(I12, I14, I13) f2#(I15, I16, I17) -> f3#(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1] f2#(I19, I20, I21) -> f3#(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1] f2#(I23, I24, I25) -> f3#(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1] f2#(I26, I27, I28) -> f3#(1, 0, I26) [1 = I27] f2#(I29, I30, I31) -> f2#(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1] f1#(I33, I34, I35) -> f2#(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f5(I0, I1, I2) -> f5(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0] f5(I4, I5, I6) -> f5(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1] f3(I8, I9, I10) -> f5(I9, I10, I11) [0 <= I8 - 1] f4(I12, I13, I14) -> f3(I12, I14, I13) f2(I15, I16, I17) -> f3(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1] f2(I19, I20, I21) -> f3(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1] f2(I23, I24, I25) -> f3(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1] f2(I26, I27, I28) -> f3(1, 0, I26) [1 = I27] f2(I29, I30, I31) -> f2(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1] f1(I33, I34, I35) -> f2(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1] The dependency graph for this problem is: 0 -> 10 1 -> 1 2 -> 1, 2 3 -> 1, 2 4 -> 3 5 -> 3 6 -> 3 7 -> 3 8 -> 3 9 -> 5, 6, 7, 8, 9 10 -> 5, 6, 7, 8, 9 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f5#(I0, I1, I2) -> f5#(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0] 2) f5#(I4, I5, I6) -> f5#(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1] 3) f3#(I8, I9, I10) -> f5#(I9, I10, I11) [0 <= I8 - 1] 4) f4#(I12, I13, I14) -> f3#(I12, I14, I13) 5) f2#(I15, I16, I17) -> f3#(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1] 6) f2#(I19, I20, I21) -> f3#(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1] 7) f2#(I23, I24, I25) -> f3#(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1] 8) f2#(I26, I27, I28) -> f3#(1, 0, I26) [1 = I27] 9) f2#(I29, I30, I31) -> f2#(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1] 10) f1#(I33, I34, I35) -> f2#(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1] We have the following SCCs. { 9 } { 2 } { 1 } DP problem for innermost termination. P = f5#(I0, I1, I2) -> f5#(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f5(I0, I1, I2) -> f5(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0] f5(I4, I5, I6) -> f5(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1] f3(I8, I9, I10) -> f5(I9, I10, I11) [0 <= I8 - 1] f4(I12, I13, I14) -> f3(I12, I14, I13) f2(I15, I16, I17) -> f3(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1] f2(I19, I20, I21) -> f3(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1] f2(I23, I24, I25) -> f3(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1] f2(I26, I27, I28) -> f3(1, 0, I26) [1 = I27] f2(I29, I30, I31) -> f2(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1] f1(I33, I34, I35) -> f2(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1] We use the basic value criterion with the projection function NU: NU[f5#(z1,z2,z3)] = z1 This gives the following inequalities: 0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0 ==> I0 >! I0 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f5#(I4, I5, I6) -> f5#(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f5(I0, I1, I2) -> f5(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0] f5(I4, I5, I6) -> f5(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1] f3(I8, I9, I10) -> f5(I9, I10, I11) [0 <= I8 - 1] f4(I12, I13, I14) -> f3(I12, I14, I13) f2(I15, I16, I17) -> f3(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1] f2(I19, I20, I21) -> f3(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1] f2(I23, I24, I25) -> f3(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1] f2(I26, I27, I28) -> f3(1, 0, I26) [1 = I27] f2(I29, I30, I31) -> f2(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1] f1(I33, I34, I35) -> f2(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1] We use the basic value criterion with the projection function NU: NU[f5#(z1,z2,z3)] = z2 This gives the following inequalities: I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1 ==> I5 >! I5 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I29, I30, I31) -> f2#(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f5(I0, I1, I2) -> f5(I0 - 1, I1, I3) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ I1 <= 0] f5(I4, I5, I6) -> f5(I4, I5 - 1, I7) [I5 - 1 <= I5 - 1 /\ 0 <= I5 - 1] f3(I8, I9, I10) -> f5(I9, I10, I11) [0 <= I8 - 1] f4(I12, I13, I14) -> f3(I12, I14, I13) f2(I15, I16, I17) -> f3(I16, I18, 0) [0 = I15 /\ I16 - 1 <= I16 - 1 /\ 0 <= I16 - 1] f2(I19, I20, I21) -> f3(I20, I22, I19) [I20 - 1 <= I20 - 1 /\ 0 <= I20 - 1] f2(I23, I24, I25) -> f3(I24, 0, 0) [0 = I23 /\ I24 - 1 <= I24 - 1 /\ 0 <= I24 - 1] f2(I26, I27, I28) -> f3(1, 0, I26) [1 = I27] f2(I29, I30, I31) -> f2(I29, I30 - 1, I32) [I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1] f1(I33, I34, I35) -> f2(I36, I37, I38) [0 <= I33 - 1 /\ -1 <= I36 - 1 /\ 1 <= I34 - 1 /\ -1 <= I37 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = z2 This gives the following inequalities: I30 - 1 <= I30 - 1 /\ 0 <= I30 - 1 ==> I30 >! I30 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.