/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 = f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) f6#(I0, I1, I2, I3, I4) -> f3#(0, I1, I2, I3, I4) f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) f4#(I10, I11, I12, I13, I14) -> f2#(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200] f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) f1#(I25, I26, I27, I28, I29) -> f3#(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100] f1#(I30, I31, I32, I33, I34) -> f2#(I30, 100, I32, I33, I34) [100 <= I30] R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200] f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16] f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100] f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30] The dependency graph for this problem is: 0 -> 1 1 -> 4 2 -> 3 3 -> 2 4 -> 5, 6 5 -> 4 6 -> 2 Where: 0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 1) f6#(I0, I1, I2, I3, I4) -> f3#(0, I1, I2, I3, I4) 2) f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) 3) f4#(I10, I11, I12, I13, I14) -> f2#(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200] 4) f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 5) f1#(I25, I26, I27, I28, I29) -> f3#(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100] 6) f1#(I30, I31, I32, I33, I34) -> f2#(I30, 100, I32, I33, I34) [100 <= I30] We have the following SCCs. { 4, 5 } { 2, 3 } DP problem for innermost termination. P = f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) f4#(I10, I11, I12, I13, I14) -> f2#(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200] R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200] f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16] f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100] f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30] We use the reverse value criterion with the projection function NU: NU[f4#(z1,z2,z3,z4,z5)] = 200 + -1 * (1 + z2) NU[f2#(z1,z2,z3,z4,z5)] = 200 + -1 * (1 + z2) This gives the following inequalities: ==> 200 + -1 * (1 + I6) >= 200 + -1 * (1 + I6) 1 + I11 <= 200 ==> 200 + -1 * (1 + I11) > 200 + -1 * (1 + (1 + I11)) with 200 + -1 * (1 + I11) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200] f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16] f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100] f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30] The dependency graph for this problem is: 2 -> Where: 2) f2#(I5, I6, I7, I8, I9) -> f4#(I5, I6, I7, I8, I9) We have the following SCCs. DP problem for innermost termination. P = f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) f1#(I25, I26, I27, I28, I29) -> f3#(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100] R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200] f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16] f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100] f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = 100 + -1 * (1 + z1) NU[f3#(z1,z2,z3,z4,z5)] = 100 + -1 * (1 + z1) This gives the following inequalities: ==> 100 + -1 * (1 + I20) >= 100 + -1 * (1 + I20) 1 + I25 <= 100 ==> 100 + -1 * (1 + I25) > 100 + -1 * (1 + (1 + I25)) with 100 + -1 * (1 + I25) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f3(0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f4(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f2(I10, 1 + I11, I12, I13, I11) [1 + I11 <= 200] f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [200 <= I16] f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) f1(I25, I26, I27, I28, I29) -> f3(1 + I25, I26, I25, I25, I29) [1 + I25 <= 100] f1(I30, I31, I32, I33, I34) -> f2(I30, 100, I32, I33, I34) [100 <= I30] The dependency graph for this problem is: 4 -> Where: 4) f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) We have the following SCCs.