/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 = f13#(x1, x2, x3) -> f12#(x1, x2, x3) f12#(I0, I1, I2) -> f1#(1, I1, I2) f2#(I3, I4, I5) -> f3#(I3, 1, I5) [I3 <= 5] f2#(I6, I7, I8) -> f5#(1, I7, I8) [6 <= I6] f4#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [I10 <= 5] f4#(I12, I13, I14) -> f1#(1 + I12, I13, I14) [6 <= I13] f10#(I15, I16, I17) -> f9#(I15, I16, I17) f6#(I18, I19, I20) -> f7#(I18, 1, I20) [I18 <= 5] f8#(I24, I25, I26) -> f10#(I24, I25, 1) [I25 <= 5] f8#(I27, I28, I29) -> f5#(1 + I27, I28, I29) [6 <= I28] f9#(I30, I31, I32) -> f10#(I30, I31, 1 + I32) [I32 <= 5] f9#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) [6 <= I35] f7#(I36, I37, I38) -> f8#(I36, I37, I38) f5#(I39, I40, I41) -> f6#(I39, I40, I41) f3#(I42, I43, I44) -> f4#(I42, I43, I44) f1#(I45, I46, I47) -> f2#(I45, I46, I47) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) The dependency graph for this problem is: 0 -> 1 1 -> 15 2 -> 14 3 -> 13 4 -> 14 5 -> 15 6 -> 10, 11 7 -> 12 8 -> 6 9 -> 13 10 -> 6 11 -> 12 12 -> 8, 9 13 -> 7 14 -> 4, 5 15 -> 2, 3 Where: 0) f13#(x1, x2, x3) -> f12#(x1, x2, x3) 1) f12#(I0, I1, I2) -> f1#(1, I1, I2) 2) f2#(I3, I4, I5) -> f3#(I3, 1, I5) [I3 <= 5] 3) f2#(I6, I7, I8) -> f5#(1, I7, I8) [6 <= I6] 4) f4#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [I10 <= 5] 5) f4#(I12, I13, I14) -> f1#(1 + I12, I13, I14) [6 <= I13] 6) f10#(I15, I16, I17) -> f9#(I15, I16, I17) 7) f6#(I18, I19, I20) -> f7#(I18, 1, I20) [I18 <= 5] 8) f8#(I24, I25, I26) -> f10#(I24, I25, 1) [I25 <= 5] 9) f8#(I27, I28, I29) -> f5#(1 + I27, I28, I29) [6 <= I28] 10) f9#(I30, I31, I32) -> f10#(I30, I31, 1 + I32) [I32 <= 5] 11) f9#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) [6 <= I35] 12) f7#(I36, I37, I38) -> f8#(I36, I37, I38) 13) f5#(I39, I40, I41) -> f6#(I39, I40, I41) 14) f3#(I42, I43, I44) -> f4#(I42, I43, I44) 15) f1#(I45, I46, I47) -> f2#(I45, I46, I47) We have the following SCCs. { 2, 4, 5, 14, 15 } { 6, 7, 8, 9, 10, 11, 12, 13 } DP problem for innermost termination. P = f10#(I15, I16, I17) -> f9#(I15, I16, I17) f6#(I18, I19, I20) -> f7#(I18, 1, I20) [I18 <= 5] f8#(I24, I25, I26) -> f10#(I24, I25, 1) [I25 <= 5] f8#(I27, I28, I29) -> f5#(1 + I27, I28, I29) [6 <= I28] f9#(I30, I31, I32) -> f10#(I30, I31, 1 + I32) [I32 <= 5] f9#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) [6 <= I35] f7#(I36, I37, I38) -> f8#(I36, I37, I38) f5#(I39, I40, I41) -> f6#(I39, I40, I41) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) We use the extended value criterion with the projection function NU: NU[f5#(x0,x1,x2)] = -x0 + 5 NU[f8#(x0,x1,x2)] = -x0 + 4 NU[f7#(x0,x1,x2)] = -x0 + 4 NU[f6#(x0,x1,x2)] = -x0 + 5 NU[f9#(x0,x1,x2)] = -x0 + 4 NU[f10#(x0,x1,x2)] = -x0 + 4 This gives the following inequalities: ==> -I15 + 4 >= -I15 + 4 I18 <= 5 ==> -I18 + 5 > -I18 + 4 with -I18 + 5 >= 0 I25 <= 5 ==> -I24 + 4 >= -I24 + 4 6 <= I28 ==> -I27 + 4 >= -(1 + I27) + 5 I32 <= 5 ==> -I30 + 4 >= -I30 + 4 6 <= I35 ==> -I33 + 4 >= -I33 + 4 ==> -I36 + 4 >= -I36 + 4 ==> -I39 + 5 >= -I39 + 5 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f10#(I15, I16, I17) -> f9#(I15, I16, I17) f8#(I24, I25, I26) -> f10#(I24, I25, 1) [I25 <= 5] f8#(I27, I28, I29) -> f5#(1 + I27, I28, I29) [6 <= I28] f9#(I30, I31, I32) -> f10#(I30, I31, 1 + I32) [I32 <= 5] f9#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) [6 <= I35] f7#(I36, I37, I38) -> f8#(I36, I37, I38) f5#(I39, I40, I41) -> f6#(I39, I40, I41) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) The dependency graph for this problem is: 6 -> 10, 11 8 -> 6 9 -> 13 10 -> 6 11 -> 12 12 -> 8, 9 13 -> Where: 6) f10#(I15, I16, I17) -> f9#(I15, I16, I17) 8) f8#(I24, I25, I26) -> f10#(I24, I25, 1) [I25 <= 5] 9) f8#(I27, I28, I29) -> f5#(1 + I27, I28, I29) [6 <= I28] 10) f9#(I30, I31, I32) -> f10#(I30, I31, 1 + I32) [I32 <= 5] 11) f9#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) [6 <= I35] 12) f7#(I36, I37, I38) -> f8#(I36, I37, I38) 13) f5#(I39, I40, I41) -> f6#(I39, I40, I41) We have the following SCCs. { 6, 8, 10, 11, 12 } DP problem for innermost termination. P = f10#(I15, I16, I17) -> f9#(I15, I16, I17) f8#(I24, I25, I26) -> f10#(I24, I25, 1) [I25 <= 5] f9#(I30, I31, I32) -> f10#(I30, I31, 1 + I32) [I32 <= 5] f9#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) [6 <= I35] f7#(I36, I37, I38) -> f8#(I36, I37, I38) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) We use the extended value criterion with the projection function NU: NU[f7#(x0,x1,x2)] = -x1 + 5 NU[f8#(x0,x1,x2)] = -x1 + 5 NU[f9#(x0,x1,x2)] = -x1 + 4 NU[f10#(x0,x1,x2)] = -x1 + 4 This gives the following inequalities: ==> -I16 + 4 >= -I16 + 4 I25 <= 5 ==> -I25 + 5 > -I25 + 4 with -I25 + 5 >= 0 I32 <= 5 ==> -I31 + 4 >= -I31 + 4 6 <= I35 ==> -I34 + 4 >= -(1 + I34) + 5 ==> -I37 + 5 >= -I37 + 5 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f10#(I15, I16, I17) -> f9#(I15, I16, I17) f9#(I30, I31, I32) -> f10#(I30, I31, 1 + I32) [I32 <= 5] f9#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) [6 <= I35] f7#(I36, I37, I38) -> f8#(I36, I37, I38) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) The dependency graph for this problem is: 6 -> 10, 11 10 -> 6 11 -> 12 12 -> Where: 6) f10#(I15, I16, I17) -> f9#(I15, I16, I17) 10) f9#(I30, I31, I32) -> f10#(I30, I31, 1 + I32) [I32 <= 5] 11) f9#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) [6 <= I35] 12) f7#(I36, I37, I38) -> f8#(I36, I37, I38) We have the following SCCs. { 6, 10 } DP problem for innermost termination. P = f10#(I15, I16, I17) -> f9#(I15, I16, I17) f9#(I30, I31, I32) -> f10#(I30, I31, 1 + I32) [I32 <= 5] R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) We use the reverse value criterion with the projection function NU: NU[f9#(z1,z2,z3)] = 5 + -1 * z3 NU[f10#(z1,z2,z3)] = 5 + -1 * z3 This gives the following inequalities: ==> 5 + -1 * I17 >= 5 + -1 * I17 I32 <= 5 ==> 5 + -1 * I32 > 5 + -1 * (1 + I32) with 5 + -1 * I32 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f10#(I15, I16, I17) -> f9#(I15, I16, I17) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) The dependency graph for this problem is: 6 -> Where: 6) f10#(I15, I16, I17) -> f9#(I15, I16, I17) We have the following SCCs. DP problem for innermost termination. P = f2#(I3, I4, I5) -> f3#(I3, 1, I5) [I3 <= 5] f4#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [I10 <= 5] f4#(I12, I13, I14) -> f1#(1 + I12, I13, I14) [6 <= I13] f3#(I42, I43, I44) -> f4#(I42, I43, I44) f1#(I45, I46, I47) -> f2#(I45, I46, I47) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2)] = -x0 + 5 NU[f4#(x0,x1,x2)] = -x0 + 4 NU[f3#(x0,x1,x2)] = -x0 + 4 NU[f2#(x0,x1,x2)] = -x0 + 5 This gives the following inequalities: I3 <= 5 ==> -I3 + 5 > -I3 + 4 with -I3 + 5 >= 0 I10 <= 5 ==> -I9 + 4 >= -I9 + 4 6 <= I13 ==> -I12 + 4 >= -(1 + I12) + 5 ==> -I42 + 4 >= -I42 + 4 ==> -I45 + 5 >= -I45 + 5 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [I10 <= 5] f4#(I12, I13, I14) -> f1#(1 + I12, I13, I14) [6 <= I13] f3#(I42, I43, I44) -> f4#(I42, I43, I44) f1#(I45, I46, I47) -> f2#(I45, I46, I47) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) The dependency graph for this problem is: 4 -> 14 5 -> 15 14 -> 4, 5 15 -> Where: 4) f4#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [I10 <= 5] 5) f4#(I12, I13, I14) -> f1#(1 + I12, I13, I14) [6 <= I13] 14) f3#(I42, I43, I44) -> f4#(I42, I43, I44) 15) f1#(I45, I46, I47) -> f2#(I45, I46, I47) We have the following SCCs. { 4, 14 } DP problem for innermost termination. P = f4#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [I10 <= 5] f3#(I42, I43, I44) -> f4#(I42, I43, I44) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3)] = 5 + -1 * z2 NU[f4#(z1,z2,z3)] = 5 + -1 * z2 This gives the following inequalities: I10 <= 5 ==> 5 + -1 * I10 > 5 + -1 * (1 + I10) with 5 + -1 * I10 >= 0 ==> 5 + -1 * I43 >= 5 + -1 * I43 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I42, I43, I44) -> f4#(I42, I43, I44) R = f13(x1, x2, x3) -> f12(x1, x2, x3) f12(I0, I1, I2) -> f1(1, I1, I2) f2(I3, I4, I5) -> f3(I3, 1, I5) [I3 <= 5] f2(I6, I7, I8) -> f5(1, I7, I8) [6 <= I6] f4(I9, I10, I11) -> f3(I9, 1 + I10, I11) [I10 <= 5] f4(I12, I13, I14) -> f1(1 + I12, I13, I14) [6 <= I13] f10(I15, I16, I17) -> f9(I15, I16, I17) f6(I18, I19, I20) -> f7(I18, 1, I20) [I18 <= 5] f6(I21, I22, I23) -> f11(I21, I22, I23) [6 <= I21] f8(I24, I25, I26) -> f10(I24, I25, 1) [I25 <= 5] f8(I27, I28, I29) -> f5(1 + I27, I28, I29) [6 <= I28] f9(I30, I31, I32) -> f10(I30, I31, 1 + I32) [I32 <= 5] f9(I33, I34, I35) -> f7(I33, 1 + I34, I35) [6 <= I35] f7(I36, I37, I38) -> f8(I36, I37, I38) f5(I39, I40, I41) -> f6(I39, I40, I41) f3(I42, I43, I44) -> f4(I42, I43, I44) f1(I45, I46, I47) -> f2(I45, I46, I47) The dependency graph for this problem is: 14 -> Where: 14) f3#(I42, I43, I44) -> f4#(I42, I43, I44) We have the following SCCs.