/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 = f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) f9#(I0, I1, I2, I3, I4) -> f1#(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3] f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [1 <= I8] f2#(I10, I11, I12, I13, I14) -> f3#(0, I11, I15, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ I15 = I15 /\ rnd5 = I15] f4#(I16, I17, I18, I19, I20) -> f1#(I16, 0, I21, I22, I20) [I20 <= 0 /\ I23 = 1 /\ I21 = I21 /\ I22 = I21] f4#(I24, I25, I26, I27, I28) -> f3#(I24, I25, I26, I27, I28) [1 <= I28] f8#(I29, I30, I31, I32, I33) -> f7#(I29, I30, I31, I32, I33) f7#(I34, I35, I36, I37, I38) -> f8#(I34, I35, I36, I37, I38) f3#(I44, I45, I46, I47, I48) -> f4#(I44, I45, I46, I47, I48) f1#(I49, I50, I51, I52, I53) -> f2#(I49, I50, I51, I52, I53) R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f1(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3] f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [1 <= I8] f2(I10, I11, I12, I13, I14) -> f3(0, I11, I15, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ I15 = I15 /\ rnd5 = I15] f4(I16, I17, I18, I19, I20) -> f1(I16, 0, I21, I22, I20) [I20 <= 0 /\ I23 = 1 /\ I21 = I21 /\ I22 = I21] f4(I24, I25, I26, I27, I28) -> f3(I24, I25, I26, I27, I28) [1 <= I28] f8(I29, I30, I31, I32, I33) -> f7(I29, I30, I31, I32, I33) f7(I34, I35, I36, I37, I38) -> f8(I34, I35, I36, I37, I38) f5(I39, I40, I41, I42, I43) -> f6(I39, I40, I41, I42, I43) f3(I44, I45, I46, I47, I48) -> f4(I44, I45, I46, I47, I48) f1(I49, I50, I51, I52, I53) -> f2(I49, I50, I51, I52, I53) The dependency graph for this problem is: 0 -> 1 1 -> 9 2 -> 7 3 -> 8 4 -> 9 5 -> 8 6 -> 7 7 -> 6 8 -> 4, 5 9 -> 2, 3 Where: 0) f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 1) f9#(I0, I1, I2, I3, I4) -> f1#(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3] 2) f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [1 <= I8] 3) f2#(I10, I11, I12, I13, I14) -> f3#(0, I11, I15, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ I15 = I15 /\ rnd5 = I15] 4) f4#(I16, I17, I18, I19, I20) -> f1#(I16, 0, I21, I22, I20) [I20 <= 0 /\ I23 = 1 /\ I21 = I21 /\ I22 = I21] 5) f4#(I24, I25, I26, I27, I28) -> f3#(I24, I25, I26, I27, I28) [1 <= I28] 6) f8#(I29, I30, I31, I32, I33) -> f7#(I29, I30, I31, I32, I33) 7) f7#(I34, I35, I36, I37, I38) -> f8#(I34, I35, I36, I37, I38) 8) f3#(I44, I45, I46, I47, I48) -> f4#(I44, I45, I46, I47, I48) 9) f1#(I49, I50, I51, I52, I53) -> f2#(I49, I50, I51, I52, I53) We have the following SCCs. { 3, 4, 5, 8, 9 } { 6, 7 } DP problem for innermost termination. P = f8#(I29, I30, I31, I32, I33) -> f7#(I29, I30, I31, I32, I33) f7#(I34, I35, I36, I37, I38) -> f8#(I34, I35, I36, I37, I38) R = f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f1(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3] f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [1 <= I8] f2(I10, I11, I12, I13, I14) -> f3(0, I11, I15, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ I15 = I15 /\ rnd5 = I15] f4(I16, I17, I18, I19, I20) -> f1(I16, 0, I21, I22, I20) [I20 <= 0 /\ I23 = 1 /\ I21 = I21 /\ I22 = I21] f4(I24, I25, I26, I27, I28) -> f3(I24, I25, I26, I27, I28) [1 <= I28] f8(I29, I30, I31, I32, I33) -> f7(I29, I30, I31, I32, I33) f7(I34, I35, I36, I37, I38) -> f8(I34, I35, I36, I37, I38) f5(I39, I40, I41, I42, I43) -> f6(I39, I40, I41, I42, I43) f3(I44, I45, I46, I47, I48) -> f4(I44, I45, I46, I47, I48) f1(I49, I50, I51, I52, I53) -> f2(I49, I50, I51, I52, I53)