/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] eval_4#(I0, I1, I2, I3, B0) -> eval_2#(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] eval_3#(I4, I5, I6, I7, B1) -> eval_3#(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2] eval_1#(I32, I33, I34, I35, B8) -> eval_2#(I32, I33, I34, I35 - 1, B8) [2 > I34] eval_1#(I36, I37, I38, I39, B9) -> eval_2#(I36, I37, I38 - 1, I39, B9) [I38 >= 2] R = eval_4(i, j, l, r, n) -> eval_2(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] eval_4(I0, I1, I2, I3, B0) -> eval_2(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] eval_3(I4, I5, I6, I7, B1) -> eval_3(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] eval_3(I8, I9, I10, I11, B2) -> eval_4(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] eval_3(I12, I13, I14, I15, B3) -> eval_3(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3(I16, I17, I18, I19, B4) -> eval_3(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] eval_3(I20, I21, I22, I23, B5) -> eval_4(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] eval_3(I24, I25, I26, I27, B6) -> eval_4(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] eval_2(I28, I29, I30, I31, B7) -> eval_3(I30, 2 * I30, I30, I31, B7) [I31 >= 2] eval_1(I32, I33, I34, I35, B8) -> eval_2(I32, I33, I34, I35 - 1, B8) [2 > I34] eval_1(I36, I37, I38, I39, B9) -> eval_2(I36, I37, I38 - 1, I39, B9) [I38 >= 2] The dependency graph for this problem is: 0 -> 8 1 -> 8 2 -> 3 -> 0, 1 4 -> 2, 3, 4, 5, 6, 7 5 -> 2, 3, 4, 5, 6, 7 6 -> 0, 1 7 -> 0, 1 8 -> 2, 3, 4, 5, 6, 7 9 -> 8 10 -> 8 Where: 0) eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] 1) eval_4#(I0, I1, I2, I3, B0) -> eval_2#(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] 2) eval_3#(I4, I5, I6, I7, B1) -> eval_3#(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] 3) eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] 4) eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] 5) eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] 6) eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] 7) eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] 8) eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2] 9) eval_1#(I32, I33, I34, I35, B8) -> eval_2#(I32, I33, I34, I35 - 1, B8) [2 > I34] 10) eval_1#(I36, I37, I38, I39, B9) -> eval_2#(I36, I37, I38 - 1, I39, B9) [I38 >= 2] We have the following SCCs. { 0, 1, 3, 4, 5, 6, 7, 8 } DP problem for innermost termination. P = eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] eval_4#(I0, I1, I2, I3, B0) -> eval_2#(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2] R = eval_4(i, j, l, r, n) -> eval_2(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] eval_4(I0, I1, I2, I3, B0) -> eval_2(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] eval_3(I4, I5, I6, I7, B1) -> eval_3(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] eval_3(I8, I9, I10, I11, B2) -> eval_4(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] eval_3(I12, I13, I14, I15, B3) -> eval_3(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3(I16, I17, I18, I19, B4) -> eval_3(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] eval_3(I20, I21, I22, I23, B5) -> eval_4(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] eval_3(I24, I25, I26, I27, B6) -> eval_4(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] eval_2(I28, I29, I30, I31, B7) -> eval_3(I30, 2 * I30, I30, I31, B7) [I31 >= 2] eval_1(I32, I33, I34, I35, B8) -> eval_2(I32, I33, I34, I35 - 1, B8) [2 > I34] eval_1(I36, I37, I38, I39, B9) -> eval_2(I36, I37, I38 - 1, I39, B9) [I38 >= 2] We use the reverse value criterion with the projection function NU: NU[eval_3#(z1,z2,z3,z4,z5)] = z3 NU[eval_2#(z1,z2,z3,z4,z5)] = z3 NU[eval_4#(z1,z2,z3,z4,z5)] = z3 This gives the following inequalities: 2 > l && l >= 1 && r >= 2 ==> l >= l I2 >= 2 && I2 >= 1 && I3 >= 2 ==> I2 > I2 - 1 with I2 >= 0 I11 >= I9 && I9 > I11 - 1 ==> I10 >= I10 I15 >= I13 && I15 - 1 >= I13 && I13 >= 1 ==> I14 >= I14 I19 >= I17 && I19 - 1 >= I17 && I17 >= 1 ==> I18 >= I18 I23 >= I21 && I23 - 1 >= I21 ==> I22 >= I22 I27 >= I25 && I27 - 1 >= I25 ==> I26 >= I26 I31 >= 2 ==> I30 >= I30 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2] R = eval_4(i, j, l, r, n) -> eval_2(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] eval_4(I0, I1, I2, I3, B0) -> eval_2(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] eval_3(I4, I5, I6, I7, B1) -> eval_3(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] eval_3(I8, I9, I10, I11, B2) -> eval_4(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] eval_3(I12, I13, I14, I15, B3) -> eval_3(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3(I16, I17, I18, I19, B4) -> eval_3(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] eval_3(I20, I21, I22, I23, B5) -> eval_4(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] eval_3(I24, I25, I26, I27, B6) -> eval_4(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] eval_2(I28, I29, I30, I31, B7) -> eval_3(I30, 2 * I30, I30, I31, B7) [I31 >= 2] eval_1(I32, I33, I34, I35, B8) -> eval_2(I32, I33, I34, I35 - 1, B8) [2 > I34] eval_1(I36, I37, I38, I39, B9) -> eval_2(I36, I37, I38 - 1, I39, B9) [I38 >= 2] We use the reverse value criterion with the projection function NU: NU[eval_3#(z1,z2,z3,z4,z5)] = z4 NU[eval_2#(z1,z2,z3,z4,z5)] = z4 NU[eval_4#(z1,z2,z3,z4,z5)] = z4 This gives the following inequalities: 2 > l && l >= 1 && r >= 2 ==> r > r - 1 with r >= 0 I11 >= I9 && I9 > I11 - 1 ==> I11 >= I11 I15 >= I13 && I15 - 1 >= I13 && I13 >= 1 ==> I15 >= I15 I19 >= I17 && I19 - 1 >= I17 && I17 >= 1 ==> I19 >= I19 I23 >= I21 && I23 - 1 >= I21 ==> I23 >= I23 I27 >= I25 && I27 - 1 >= I25 ==> I27 >= I27 I31 >= 2 ==> I31 >= I31 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2] R = eval_4(i, j, l, r, n) -> eval_2(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] eval_4(I0, I1, I2, I3, B0) -> eval_2(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] eval_3(I4, I5, I6, I7, B1) -> eval_3(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] eval_3(I8, I9, I10, I11, B2) -> eval_4(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] eval_3(I12, I13, I14, I15, B3) -> eval_3(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3(I16, I17, I18, I19, B4) -> eval_3(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] eval_3(I20, I21, I22, I23, B5) -> eval_4(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] eval_3(I24, I25, I26, I27, B6) -> eval_4(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] eval_2(I28, I29, I30, I31, B7) -> eval_3(I30, 2 * I30, I30, I31, B7) [I31 >= 2] eval_1(I32, I33, I34, I35, B8) -> eval_2(I32, I33, I34, I35 - 1, B8) [2 > I34] eval_1(I36, I37, I38, I39, B9) -> eval_2(I36, I37, I38 - 1, I39, B9) [I38 >= 2] The dependency graph for this problem is: 3 -> 4 -> 3, 4, 5, 6, 7 5 -> 3, 4, 5, 6, 7 6 -> 7 -> 8 -> 3, 4, 5, 6, 7 Where: 3) eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] 4) eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] 5) eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] 6) eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] 7) eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] 8) eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2] We have the following SCCs. { 4, 5 } DP problem for innermost termination. P = eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] R = eval_4(i, j, l, r, n) -> eval_2(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2] eval_4(I0, I1, I2, I3, B0) -> eval_2(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2] eval_3(I4, I5, I6, I7, B1) -> eval_3(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1] eval_3(I8, I9, I10, I11, B2) -> eval_4(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1] eval_3(I12, I13, I14, I15, B3) -> eval_3(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1] eval_3(I16, I17, I18, I19, B4) -> eval_3(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1] eval_3(I20, I21, I22, I23, B5) -> eval_4(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21] eval_3(I24, I25, I26, I27, B6) -> eval_4(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25] eval_2(I28, I29, I30, I31, B7) -> eval_3(I30, 2 * I30, I30, I31, B7) [I31 >= 2] eval_1(I32, I33, I34, I35, B8) -> eval_2(I32, I33, I34, I35 - 1, B8) [2 > I34] eval_1(I36, I37, I38, I39, B9) -> eval_2(I36, I37, I38 - 1, I39, B9) [I38 >= 2] We use the reverse value criterion with the projection function NU: NU[eval_3#(z1,z2,z3,z4,z5)] = z4 + -1 * z2 This gives the following inequalities: I15 >= I13 && I15 - 1 >= I13 && I13 >= 1 ==> I15 + -1 * I13 > I15 + -1 * (2 * I13 + 2) with I15 + -1 * I13 >= 0 I19 >= I17 && I19 - 1 >= I17 && I17 >= 1 ==> I19 + -1 * I17 > I19 + -1 * (2 * I17) with I19 + -1 * I17 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.