/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f7#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) f5#(I6, I7, I8, I9, I10, I11) -> f2#(I6, I7, I8, I8, 2, 2) f2#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1, I14, I15, I16, I17) [2 <= I17] f2#(I18, I19, I20, I21, I22, I23) -> f3#(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f3#(I30, I31, I32, I33, I34, I35) -> f4#(1, I31, I32, I33, I34, I35) f3#(I36, I37, I38, I39, I40, I41) -> f4#(0, I37, I38, I39, I40, I41) [I37 <= 0] f1#(I42, I43, I44, I45, I46, I47) -> f2#(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] f1#(I48, I49, I50, I51, I52, I53) -> f2#(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] R = f7(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f6(I0, I1, I2, I3, I4, I5) [I0 <= 0] f5(I6, I7, I8, I9, I10, I11) -> f2(I6, I7, I8, I8, 2, 2) f2(I12, I13, I14, I15, I16, I17) -> f3(I12, 1, I14, I15, I16, I17) [2 <= I17] f2(I18, I19, I20, I21, I22, I23) -> f3(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3(I24, I25, I26, I27, I28, I29) -> f1(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f3(I30, I31, I32, I33, I34, I35) -> f4(1, I31, I32, I33, I34, I35) f3(I36, I37, I38, I39, I40, I41) -> f4(0, I37, I38, I39, I40, I41) [I37 <= 0] f1(I42, I43, I44, I45, I46, I47) -> f2(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] f1(I48, I49, I50, I51, I52, I53) -> f2(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 4, 5 3 -> 4, 5, 6 4 -> 8 5 -> 6 -> 7 -> 2, 3 8 -> 2, 3 Where: 0) f7#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) 1) f5#(I6, I7, I8, I9, I10, I11) -> f2#(I6, I7, I8, I8, 2, 2) 2) f2#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1, I14, I15, I16, I17) [2 <= I17] 3) f2#(I18, I19, I20, I21, I22, I23) -> f3#(I18, 0, I20, I21, I22, I23) [I23 <= 1] 4) f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] 5) f3#(I30, I31, I32, I33, I34, I35) -> f4#(1, I31, I32, I33, I34, I35) 6) f3#(I36, I37, I38, I39, I40, I41) -> f4#(0, I37, I38, I39, I40, I41) [I37 <= 0] 7) f1#(I42, I43, I44, I45, I46, I47) -> f2#(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] 8) f1#(I48, I49, I50, I51, I52, I53) -> f2#(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] We have the following SCCs. { 2, 3, 4, 8 } DP problem for innermost termination. P = f2#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1, I14, I15, I16, I17) [2 <= I17] f2#(I18, I19, I20, I21, I22, I23) -> f3#(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f1#(I48, I49, I50, I51, I52, I53) -> f2#(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] R = f7(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f6(I0, I1, I2, I3, I4, I5) [I0 <= 0] f5(I6, I7, I8, I9, I10, I11) -> f2(I6, I7, I8, I8, 2, 2) f2(I12, I13, I14, I15, I16, I17) -> f3(I12, 1, I14, I15, I16, I17) [2 <= I17] f2(I18, I19, I20, I21, I22, I23) -> f3(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3(I24, I25, I26, I27, I28, I29) -> f1(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f3(I30, I31, I32, I33, I34, I35) -> f4(1, I31, I32, I33, I34, I35) f3(I36, I37, I38, I39, I40, I41) -> f4(0, I37, I38, I39, I40, I41) [I37 <= 0] f1(I42, I43, I44, I45, I46, I47) -> f2(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] f1(I48, I49, I50, I51, I52, I53) -> f2(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3,x4,x5)] = -x5 NU[f3#(x0,x1,x2,x3,x4,x5)] = -x5 NU[f2#(x0,x1,x2,x3,x4,x5)] = -x5 + 1 This gives the following inequalities: 2 <= I17 ==> -I17 + 1 >= -I17 I23 <= 1 ==> -I23 + 1 > -I23 with -I23 + 1 >= 0 1 <= rnd4 /\ rnd4 = rnd4 ==> -I29 >= -I29 1 <= I51 ==> -I53 >= -(1 + I53) + 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1, I14, I15, I16, I17) [2 <= I17] f3#(I24, I25, I26, I27, I28, I29) -> f1#(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f1#(I48, I49, I50, I51, I52, I53) -> f2#(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51] R = f7(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f6(I0, I1, I2, I3, I4, I5) [I0 <= 0] f5(I6, I7, I8, I9, I10, I11) -> f2(I6, I7, I8, I8, 2, 2) f2(I12, I13, I14, I15, I16, I17) -> f3(I12, 1, I14, I15, I16, I17) [2 <= I17] f2(I18, I19, I20, I21, I22, I23) -> f3(I18, 0, I20, I21, I22, I23) [I23 <= 1] f3(I24, I25, I26, I27, I28, I29) -> f1(I24, I25, I26, rnd4, I28, I29) [1 <= rnd4 /\ rnd4 = rnd4] f3(I30, I31, I32, I33, I34, I35) -> f4(1, I31, I32, I33, I34, I35) f3(I36, I37, I38, I39, I40, I41) -> f4(0, I37, I38, I39, I40, I41) [I37 <= 0] f1(I42, I43, I44, I45, I46, I47) -> f2(I42, I43, I44, I45, I46, -1 + I47) [I45 <= 0] f1(I48, I49, I50, I51, I52, I53) -> f2(I48, I49, I50, I51, I52, 1 + I53) [1 <= I51]