25.02/24.67 YES 25.02/24.67 25.02/24.67 DP problem for innermost termination. 25.02/24.67 P = 25.02/24.67 f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 25.02/24.67 f10#(I0, I1, I2, I3, I4) -> f1#(4, 0, I2, I3, I4) 25.02/24.67 f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f2#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 f8#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) 25.02/24.67 f9#(I30, I31, I32, I33, I34) -> f6#(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f5#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I47, I48, I49) 25.02/24.67 f5#(I50, I51, I52, I53, I54) -> f3#(I50, I51, I52, I53, I54) 25.02/24.67 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 R = 25.02/24.67 f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 25.02/24.67 f10(I0, I1, I2, I3, I4) -> f1(4, 0, I2, I3, I4) 25.02/24.67 f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 f8(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8(I20, I21, I22, I23, I24) -> f1(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) 25.02/24.67 f9(I30, I31, I32, I33, I34) -> f6(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7(I35, I36, I37, I38, I39) -> f8(I35, I36, I37, I38, I39) 25.02/24.67 f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f5(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) 25.02/24.67 f5(I50, I51, I52, I53, I54) -> f3(I50, I51, I52, I53, I54) 25.02/24.67 f3(I55, I56, I57, I58, I59) -> f4(I55, I56, I57, I58, I59) 25.02/24.67 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 The dependency graph for this problem is: 25.02/24.67 0 -> 1 25.02/24.67 1 -> 12 25.02/24.67 2 -> 8 25.02/24.67 3 -> 10, 11 25.02/24.67 4 -> 6, 7 25.02/24.67 5 -> 12 25.02/24.67 6 -> 9 25.02/24.67 7 -> 9 25.02/24.67 8 -> 4, 5 25.02/24.67 9 -> 8 25.02/24.67 10 -> 25.02/24.67 11 -> 25.02/24.67 12 -> 2, 3 25.02/24.67 Where: 25.02/24.67 0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 25.02/24.67 1) f10#(I0, I1, I2, I3, I4) -> f1#(4, 0, I2, I3, I4) 25.02/24.67 2) f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 3) f2#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 4) f8#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 5) f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 6) f9#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) 25.02/24.67 7) f9#(I30, I31, I32, I33, I34) -> f6#(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 8) f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 9) f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) 25.02/24.67 10) f5#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I47, I48, I49) 25.02/24.67 11) f5#(I50, I51, I52, I53, I54) -> f3#(I50, I51, I52, I53, I54) 25.02/24.67 12) f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 We have the following SCCs. 25.02/24.67 { 2, 4, 5, 6, 7, 8, 9, 12 } 25.02/24.67 25.02/24.67 DP problem for innermost termination. 25.02/24.67 P = 25.02/24.67 f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f8#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) 25.02/24.67 f9#(I30, I31, I32, I33, I34) -> f6#(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 R = 25.02/24.67 f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 25.02/24.67 f10(I0, I1, I2, I3, I4) -> f1(4, 0, I2, I3, I4) 25.02/24.67 f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 f8(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8(I20, I21, I22, I23, I24) -> f1(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) 25.02/24.67 f9(I30, I31, I32, I33, I34) -> f6(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7(I35, I36, I37, I38, I39) -> f8(I35, I36, I37, I38, I39) 25.02/24.67 f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f5(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) 25.02/24.67 f5(I50, I51, I52, I53, I54) -> f3(I50, I51, I52, I53, I54) 25.02/24.67 f3(I55, I56, I57, I58, I59) -> f4(I55, I56, I57, I58, I59) 25.02/24.67 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 We use the extended value criterion with the projection function NU: 25.02/24.67 NU[f6#(x0,x1,x2,x3,x4)] = x0 - x1 - 2 25.02/24.67 NU[f1#(x0,x1,x2,x3,x4)] = x0 - x1 - 1 25.02/24.67 NU[f9#(x0,x1,x2,x3,x4)] = x0 - x1 - 2 25.02/24.67 NU[f8#(x0,x1,x2,x3,x4)] = x0 - x1 - 1 25.02/24.67 NU[f7#(x0,x1,x2,x3,x4)] = x0 - x1 - 1 25.02/24.67 NU[f2#(x0,x1,x2,x3,x4)] = x0 - x1 - 1 25.02/24.67 25.02/24.67 This gives the following inequalities: 25.02/24.67 0 <= I5 ==> I5 - I6 - 1 >= I5 - I6 - 1 25.02/24.67 1 + I16 <= I15 ==> I15 - I16 - 1 > I15 - I16 - 2 with I15 - I16 - 1 >= 0 25.02/24.67 I20 <= I21 ==> I20 - I21 - 1 >= (-1 + I20) - I21 - 1 25.02/24.67 ==> I25 - I26 - 2 >= I25 - I26 - 2 25.02/24.67 rnd3 = rnd3 ==> I30 - I31 - 2 >= I30 - I31 - 2 25.02/24.67 ==> I35 - I36 - 1 >= I35 - I36 - 1 25.02/24.67 ==> I40 - I41 - 2 >= I40 - (1 + I41) - 1 25.02/24.67 ==> I60 - I61 - 1 >= I60 - I61 - 1 25.02/24.67 25.02/24.67 We remove all the strictly oriented dependency pairs. 25.02/24.67 25.02/24.67 DP problem for innermost termination. 25.02/24.67 P = 25.02/24.67 f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) 25.02/24.67 f9#(I30, I31, I32, I33, I34) -> f6#(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 R = 25.02/24.67 f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 25.02/24.67 f10(I0, I1, I2, I3, I4) -> f1(4, 0, I2, I3, I4) 25.02/24.67 f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 f8(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8(I20, I21, I22, I23, I24) -> f1(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) 25.02/24.67 f9(I30, I31, I32, I33, I34) -> f6(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7(I35, I36, I37, I38, I39) -> f8(I35, I36, I37, I38, I39) 25.02/24.67 f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f5(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) 25.02/24.67 f5(I50, I51, I52, I53, I54) -> f3(I50, I51, I52, I53, I54) 25.02/24.67 f3(I55, I56, I57, I58, I59) -> f4(I55, I56, I57, I58, I59) 25.02/24.67 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 The dependency graph for this problem is: 25.02/24.67 2 -> 8 25.02/24.67 5 -> 12 25.02/24.67 6 -> 9 25.02/24.67 7 -> 9 25.02/24.67 8 -> 5 25.02/24.67 9 -> 8 25.02/24.67 12 -> 2 25.02/24.67 Where: 25.02/24.67 2) f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 5) f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 6) f9#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) 25.02/24.67 7) f9#(I30, I31, I32, I33, I34) -> f6#(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 8) f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 9) f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) 25.02/24.67 12) f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 We have the following SCCs. 25.02/24.67 { 2, 5, 8, 12 } 25.02/24.67 25.02/24.67 DP problem for innermost termination. 25.02/24.67 P = 25.02/24.67 f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 R = 25.02/24.67 f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 25.02/24.67 f10(I0, I1, I2, I3, I4) -> f1(4, 0, I2, I3, I4) 25.02/24.67 f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 f8(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8(I20, I21, I22, I23, I24) -> f1(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) 25.02/24.67 f9(I30, I31, I32, I33, I34) -> f6(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7(I35, I36, I37, I38, I39) -> f8(I35, I36, I37, I38, I39) 25.02/24.67 f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f5(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) 25.02/24.67 f5(I50, I51, I52, I53, I54) -> f3(I50, I51, I52, I53, I54) 25.02/24.67 f3(I55, I56, I57, I58, I59) -> f4(I55, I56, I57, I58, I59) 25.02/24.67 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 We use the extended value criterion with the projection function NU: 25.02/24.67 NU[f1#(x0,x1,x2,x3,x4)] = x0 25.02/24.67 NU[f8#(x0,x1,x2,x3,x4)] = x0 - 1 25.02/24.67 NU[f7#(x0,x1,x2,x3,x4)] = x0 - 1 25.02/24.67 NU[f2#(x0,x1,x2,x3,x4)] = x0 25.02/24.67 25.02/24.67 This gives the following inequalities: 25.02/24.67 0 <= I5 ==> I5 > I5 - 1 with I5 >= 0 25.02/24.67 I20 <= I21 ==> I20 - 1 >= (-1 + I20) 25.02/24.67 ==> I35 - 1 >= I35 - 1 25.02/24.67 ==> I60 >= I60 25.02/24.67 25.02/24.67 We remove all the strictly oriented dependency pairs. 25.02/24.67 25.02/24.67 DP problem for innermost termination. 25.02/24.67 P = 25.02/24.67 f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 R = 25.02/24.67 f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 25.02/24.67 f10(I0, I1, I2, I3, I4) -> f1(4, 0, I2, I3, I4) 25.02/24.67 f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 f8(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8(I20, I21, I22, I23, I24) -> f1(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) 25.02/24.67 f9(I30, I31, I32, I33, I34) -> f6(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7(I35, I36, I37, I38, I39) -> f8(I35, I36, I37, I38, I39) 25.02/24.67 f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f5(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) 25.02/24.67 f5(I50, I51, I52, I53, I54) -> f3(I50, I51, I52, I53, I54) 25.02/24.67 f3(I55, I56, I57, I58, I59) -> f4(I55, I56, I57, I58, I59) 25.02/24.67 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 The dependency graph for this problem is: 25.02/24.67 5 -> 12 25.02/24.67 8 -> 5 25.02/24.67 12 -> 25.02/24.67 Where: 25.02/24.67 5) f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 8) f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 12) f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 We have the following SCCs. 25.02/24.67 25.02/27.65 EOF