/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 = init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) f6#(I0, I1, I2, I3) -> f6#(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0] f5#(I8, I9, I10, I11) -> f5#(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1] f5#(I15, I16, I17, I18) -> f6#(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1] f4#(I23, I24, I25, I26) -> f4#(I23 - 1, I23, I27, I28) [0 <= I24 - 1] f4#(I29, I30, I31, I32) -> f5#(2, I33, I34, I35) [I30 <= 0] f3#(I36, I37, I38, I39) -> f4#(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0] f3#(I42, I43, I44, I45) -> f3#(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42] f2#(I47, I48, I49, I50) -> f3#(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0] f2#(I52, I53, I54, I55) -> f2#(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52] f1#(I57, I58, I59, I60) -> f2#(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f6(I0, I1, I2, I3) -> f6(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0] f5(I8, I9, I10, I11) -> f5(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1] f5(I15, I16, I17, I18) -> f6(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1] f4(I23, I24, I25, I26) -> f4(I23 - 1, I23, I27, I28) [0 <= I24 - 1] f4(I29, I30, I31, I32) -> f5(2, I33, I34, I35) [I30 <= 0] f3(I36, I37, I38, I39) -> f4(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0] f3(I42, I43, I44, I45) -> f3(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42] f2(I47, I48, I49, I50) -> f3(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0] f2(I52, I53, I54, I55) -> f2(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52] f1(I57, I58, I59, I60) -> f2(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57] The dependency graph for this problem is: 0 -> 10 1 -> 1 2 -> 2, 3 3 -> 1 4 -> 4, 5 5 -> 2, 3 6 -> 4, 5 7 -> 6, 7 8 -> 6, 7 9 -> 8, 9 10 -> 8, 9 Where: 0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 1) f6#(I0, I1, I2, I3) -> f6#(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0] 2) f5#(I8, I9, I10, I11) -> f5#(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1] 3) f5#(I15, I16, I17, I18) -> f6#(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1] 4) f4#(I23, I24, I25, I26) -> f4#(I23 - 1, I23, I27, I28) [0 <= I24 - 1] 5) f4#(I29, I30, I31, I32) -> f5#(2, I33, I34, I35) [I30 <= 0] 6) f3#(I36, I37, I38, I39) -> f4#(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0] 7) f3#(I42, I43, I44, I45) -> f3#(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42] 8) f2#(I47, I48, I49, I50) -> f3#(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0] 9) f2#(I52, I53, I54, I55) -> f2#(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52] 10) f1#(I57, I58, I59, I60) -> f2#(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57] We have the following SCCs. { 9 } { 7 } { 4 } { 2 } { 1 } DP problem for innermost termination. P = f6#(I0, I1, I2, I3) -> f6#(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f6(I0, I1, I2, I3) -> f6(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0] f5(I8, I9, I10, I11) -> f5(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1] f5(I15, I16, I17, I18) -> f6(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1] f4(I23, I24, I25, I26) -> f4(I23 - 1, I23, I27, I28) [0 <= I24 - 1] f4(I29, I30, I31, I32) -> f5(2, I33, I34, I35) [I30 <= 0] f3(I36, I37, I38, I39) -> f4(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0] f3(I42, I43, I44, I45) -> f3(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42] f2(I47, I48, I49, I50) -> f3(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0] f2(I52, I53, I54, I55) -> f2(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52] f1(I57, I58, I59, I60) -> f2(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57] We use the basic value criterion with the projection function NU: NU[f6#(z1,z2,z3,z4)] = z1 This gives the following inequalities: -1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0 ==> I0 >! I4 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f5#(I8, I9, I10, I11) -> f5#(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f6(I0, I1, I2, I3) -> f6(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0] f5(I8, I9, I10, I11) -> f5(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1] f5(I15, I16, I17, I18) -> f6(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1] f4(I23, I24, I25, I26) -> f4(I23 - 1, I23, I27, I28) [0 <= I24 - 1] f4(I29, I30, I31, I32) -> f5(2, I33, I34, I35) [I30 <= 0] f3(I36, I37, I38, I39) -> f4(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0] f3(I42, I43, I44, I45) -> f3(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42] f2(I47, I48, I49, I50) -> f3(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0] f2(I52, I53, I54, I55) -> f2(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52] f1(I57, I58, I59, I60) -> f2(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57] We use the basic value criterion with the projection function NU: NU[f5#(z1,z2,z3,z4)] = z1 This gives the following inequalities: I8 <= 2 /\ -1 <= I8 - 1 ==> I8 >! I8 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f4#(I23, I24, I25, I26) -> f4#(I23 - 1, I23, I27, I28) [0 <= I24 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f6(I0, I1, I2, I3) -> f6(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0] f5(I8, I9, I10, I11) -> f5(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1] f5(I15, I16, I17, I18) -> f6(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1] f4(I23, I24, I25, I26) -> f4(I23 - 1, I23, I27, I28) [0 <= I24 - 1] f4(I29, I30, I31, I32) -> f5(2, I33, I34, I35) [I30 <= 0] f3(I36, I37, I38, I39) -> f4(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0] f3(I42, I43, I44, I45) -> f3(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42] f2(I47, I48, I49, I50) -> f3(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0] f2(I52, I53, I54, I55) -> f2(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52] f1(I57, I58, I59, I60) -> f2(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57]