/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 = f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) f9#(I0, I1, I2, I3) -> f2#(I0, 0, I2, I3) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] R = f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0] f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8] f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37] f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46] The dependency graph for this problem is: 0 -> 1 1 -> 8 2 -> 5 3 -> 5 4 -> 6 5 -> 6 6 -> 7 7 -> 10, 11 8 -> 9 9 -> 7 10 -> 2, 3, 4 11 -> 8 Where: 0) f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 1) f9#(I0, I1, I2, I3) -> f2#(I0, 0, I2, I3) 2) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] 3) f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] 4) f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] 5) f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 6) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 7) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 8) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 9) f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35] 10) f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] 11) f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 } DP problem for innermost termination. P = f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] R = f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0] f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8] f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37] f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46] We use the extended value criterion with the projection function NU: NU[f4#(x0,x1,x2,x3)] = -x1 + x3 + 1 NU[f2#(x0,x1,x2,x3)] = -x1 + x3 + 1 NU[f1#(x0,x1,x2,x3)] = -x1 + x3 NU[f6#(x0,x1,x2,x3)] = -x1 + x3 NU[f7#(x0,x1,x2,x3)] = -x1 + x3 NU[f8#(x0,x1,x2,x3)] = -x1 + x3 NU[f3#(x0,x1,x2,x3)] = -x1 + x3 This gives the following inequalities: 1 + I4 <= 0 ==> -I5 + I7 >= -I5 + I7 1 <= I8 ==> -I9 + I11 >= -I9 + I11 0 <= I12 /\ I12 <= 0 ==> -I13 + I15 >= -I13 + I15 ==> -I17 + I19 >= -I17 + (-1 + I19) ==> -I21 + I23 >= -I21 + I23 ==> -I25 + I27 >= -I25 + I27 ==> -I29 + I31 + 1 >= -I29 + I31 + 1 1 + I33 <= I35 ==> -I33 + I35 + 1 > -I33 + I35 with -I33 + I35 + 1 >= 0 rnd1 = rnd1 /\ 1 + I42 <= I43 ==> -I41 + I43 >= -I41 + I43 I47 <= I46 ==> -I45 + I47 >= -(1 + I45) + I47 + 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] R = f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0] f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8] f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37] f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46] The dependency graph for this problem is: 2 -> 5 3 -> 5 4 -> 6 5 -> 6 6 -> 7 7 -> 10, 11 8 -> 10 -> 2, 3, 4 11 -> 8 Where: 2) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] 3) f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] 4) f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] 5) f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 6) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 7) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 8) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 10) f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] 11) f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 10 } DP problem for innermost termination. P = f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] R = f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0] f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8] f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37] f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46] We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3)] = -x2 + x3 - 1 NU[f6#(x0,x1,x2,x3)] = -x2 + x3 - 1 NU[f7#(x0,x1,x2,x3)] = -x2 + x3 - 2 NU[f8#(x0,x1,x2,x3)] = -x2 + x3 - 2 NU[f3#(x0,x1,x2,x3)] = -x2 + x3 - 2 This gives the following inequalities: 1 + I4 <= 0 ==> -I6 + I7 - 2 >= -I6 + I7 - 2 1 <= I8 ==> -I10 + I11 - 2 >= -I10 + I11 - 2 0 <= I12 /\ I12 <= 0 ==> -I14 + I15 - 2 >= -I14 + I15 - 2 ==> -I18 + I19 - 2 >= -(-1 + I18) + (-1 + I19) - 2 ==> -I22 + I23 - 2 >= -(1 + I22) + I23 - 1 ==> -I26 + I27 - 1 >= -I26 + I27 - 1 rnd1 = rnd1 /\ 1 + I42 <= I43 ==> -I42 + I43 - 1 > -I42 + I43 - 2 with -I42 + I43 - 1 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) R = f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0] f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8] f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35] f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37] f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46] The dependency graph for this problem is: 2 -> 5 3 -> 5 4 -> 6 5 -> 6 6 -> 7 7 -> Where: 2) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] 3) f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] 4) f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] 5) f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 6) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 7) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) We have the following SCCs.