18.42/18.16 YES 18.42/18.16 18.42/18.16 DP problem for innermost termination. 18.42/18.16 P = 18.42/18.16 f13#(x1, x2, x3, x4, x5) -> f12#(x1, x2, x3, x4, x5) 18.42/18.16 f12#(I0, I1, I2, I3, I4) -> f6#(0, I1, I2, I3, I4) 18.42/18.16 f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 18.42/18.16 f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 18.42/18.16 f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 18.42/18.16 f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 18.42/18.16 f8#(I25, I26, I27, I28, I29) -> f7#(I25, I26, I27, I28, I26) [1 + I26 <= 200] 18.42/18.16 f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 18.42/18.16 f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 18.42/18.16 f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 18.42/18.16 f3#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I50, I53, I54) [1 + I50 <= 100] 18.42/18.16 f3#(I55, I56, I57, I58, I59) -> f4#(I55, 100, I57, I58, I59) [100 <= I55] 18.42/18.16 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 18.42/18.16 R = 18.42/18.16 f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 18.42/18.16 f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 18.42/18.16 f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 18.42/18.16 f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 18.42/18.16 f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 18.42/18.16 f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 18.42/18.16 f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200] 18.42/18.16 f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31] 18.42/18.16 f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 18.42/18.16 f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 18.42/18.16 f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 18.42/18.16 f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100] 18.42/18.16 f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55] 18.42/18.16 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 18.42/18.16 18.42/18.16 The dependency graph for this problem is: 18.42/18.16 0 -> 1 18.42/18.16 1 -> 8 18.42/18.16 2 -> 3 18.42/18.16 3 -> 4 18.42/18.16 4 -> 8 18.42/18.16 5 -> 6 18.42/18.16 6 -> 7 18.42/18.16 7 -> 9 18.42/18.16 8 -> 10, 11 18.42/18.16 9 -> 5 18.42/18.16 10 -> 12 18.42/18.16 11 -> 5 18.42/18.16 12 -> 2 18.42/18.16 Where: 18.42/18.16 0) f13#(x1, x2, x3, x4, x5) -> f12#(x1, x2, x3, x4, x5) 18.42/18.16 1) f12#(I0, I1, I2, I3, I4) -> f6#(0, I1, I2, I3, I4) 18.42/18.16 2) f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 18.42/18.16 3) f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 18.42/18.16 4) f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 18.42/18.16 5) f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 18.42/18.16 6) f8#(I25, I26, I27, I28, I29) -> f7#(I25, I26, I27, I28, I26) [1 + I26 <= 200] 18.42/18.16 7) f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 18.42/18.16 8) f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 18.42/18.16 9) f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 18.42/18.16 10) f3#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I50, I53, I54) [1 + I50 <= 100] 18.42/18.16 11) f3#(I55, I56, I57, I58, I59) -> f4#(I55, 100, I57, I58, I59) [100 <= I55] 18.42/18.16 12) f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 18.42/18.16 18.42/18.16 We have the following SCCs. 18.42/18.16 { 2, 3, 4, 8, 10, 12 } 18.42/18.16 { 5, 6, 7, 9 } 18.42/18.16 18.42/18.16 DP problem for innermost termination. 18.42/18.16 P = 18.42/18.16 f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 18.42/18.16 f8#(I25, I26, I27, I28, I29) -> f7#(I25, I26, I27, I28, I26) [1 + I26 <= 200] 18.42/18.16 f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 18.42/18.16 f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 18.42/18.16 R = 18.42/18.16 f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 18.42/18.16 f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 18.42/18.16 f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 18.42/18.16 f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 18.42/18.16 f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 18.42/18.16 f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 18.42/18.16 f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200] 18.42/18.16 f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31] 18.42/18.16 f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 18.42/18.16 f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 18.42/18.16 f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 18.42/18.16 f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100] 18.42/18.16 f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55] 18.42/18.16 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 18.42/18.16 18.42/18.16 We use the extended value criterion with the projection function NU: 18.42/18.16 NU[f5#(x0,x1,x2,x3,x4)] = -x1 + 198 18.42/18.16 NU[f7#(x0,x1,x2,x3,x4)] = -x1 + 198 18.42/18.16 NU[f8#(x0,x1,x2,x3,x4)] = -x1 + 199 18.42/18.16 NU[f4#(x0,x1,x2,x3,x4)] = -x1 + 199 18.42/18.16 18.42/18.16 This gives the following inequalities: 18.42/18.16 ==> -I21 + 199 >= -I21 + 199 18.42/18.16 1 + I26 <= 200 ==> -I26 + 199 > -I26 + 198 with -I26 + 199 >= 0 18.42/18.16 ==> -I36 + 198 >= -I36 + 198 18.42/18.16 ==> -I46 + 198 >= -(1 + I46) + 199 18.42/18.16 18.42/18.16 We remove all the strictly oriented dependency pairs. 18.42/18.16 18.42/18.16 DP problem for innermost termination. 18.42/18.16 P = 18.42/18.16 f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 18.42/18.16 f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 18.42/18.16 f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 18.42/18.16 R = 18.42/18.16 f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 18.42/18.16 f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 18.42/18.16 f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 18.42/18.16 f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 18.42/18.16 f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 18.42/18.16 f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 18.42/18.16 f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200] 18.42/18.16 f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31] 18.42/18.16 f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 18.42/18.16 f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 18.42/18.16 f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 18.42/18.16 f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100] 18.42/18.16 f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55] 18.42/18.16 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 18.42/18.16 18.42/18.16 The dependency graph for this problem is: 18.42/18.16 5 -> 18.42/18.16 7 -> 9 18.42/18.16 9 -> 5 18.42/18.16 Where: 18.42/18.16 5) f4#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 18.42/18.16 7) f7#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) 18.42/18.16 9) f5#(I45, I46, I47, I48, I49) -> f4#(I45, 1 + I46, I47, I48, I49) 18.42/18.16 18.42/18.16 We have the following SCCs. 18.42/18.16 18.42/18.16 18.42/18.16 DP problem for innermost termination. 18.42/18.16 P = 18.42/18.16 f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 18.42/18.16 f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 18.42/18.16 f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 18.42/18.16 f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 18.42/18.16 f3#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I50, I53, I54) [1 + I50 <= 100] 18.42/18.16 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 18.42/18.16 R = 18.42/18.16 f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 18.42/18.16 f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 18.42/18.16 f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 18.42/18.16 f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 18.42/18.16 f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 18.42/18.16 f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 18.42/18.16 f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200] 18.42/18.16 f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31] 18.42/18.16 f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 18.42/18.16 f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 18.42/18.16 f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 18.42/18.16 f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100] 18.42/18.16 f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55] 18.42/18.16 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 18.42/18.16 18.42/18.16 We use the extended value criterion with the projection function NU: 18.42/18.16 NU[f1#(x0,x1,x2,x3,x4)] = -x0 + 98 18.42/18.16 NU[f3#(x0,x1,x2,x3,x4)] = -x0 + 99 18.42/18.16 NU[f6#(x0,x1,x2,x3,x4)] = -x0 + 99 18.42/18.16 NU[f10#(x0,x1,x2,x3,x4)] = -x0 + 98 18.42/18.16 NU[f11#(x0,x1,x2,x3,x4)] = -x0 + 98 18.42/18.16 NU[f2#(x0,x1,x2,x3,x4)] = -x0 + 98 18.42/18.16 18.42/18.16 This gives the following inequalities: 18.42/18.16 ==> -I5 + 98 >= -I5 + 98 18.42/18.16 ==> -I10 + 98 >= -I10 + 98 18.42/18.16 ==> -I15 + 98 >= -(1 + I15) + 99 18.42/18.16 ==> -I40 + 99 >= -I40 + 99 18.42/18.16 1 + I50 <= 100 ==> -I50 + 99 > -I50 + 98 with -I50 + 99 >= 0 18.42/18.16 ==> -I60 + 98 >= -I60 + 98 18.42/18.16 18.42/18.16 We remove all the strictly oriented dependency pairs. 18.42/18.16 18.42/18.16 DP problem for innermost termination. 18.42/18.16 P = 18.42/18.16 f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 18.42/18.16 f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 18.42/18.16 f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 18.42/18.16 f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 18.42/18.16 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 18.42/18.16 R = 18.42/18.16 f13(x1, x2, x3, x4, x5) -> f12(x1, x2, x3, x4, x5) 18.42/18.16 f12(I0, I1, I2, I3, I4) -> f6(0, I1, I2, I3, I4) 18.42/18.16 f2(I5, I6, I7, I8, I9) -> f11(I5, I6, I7, I5, I9) 18.42/18.16 f11(I10, I11, I12, I13, I14) -> f10(I10, I11, I12, I13, I14) 18.42/18.16 f10(I15, I16, I17, I18, I19) -> f6(1 + I15, I16, I17, I18, I19) 18.42/18.16 f4(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 18.42/18.16 f8(I25, I26, I27, I28, I29) -> f7(I25, I26, I27, I28, I26) [1 + I26 <= 200] 18.42/18.16 f8(I30, I31, I32, I33, I34) -> f9(I30, I31, I32, I33, I34) [200 <= I31] 18.42/18.16 f7(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) 18.42/18.16 f6(I40, I41, I42, I43, I44) -> f3(I40, I41, I42, I43, I44) 18.42/18.16 f5(I45, I46, I47, I48, I49) -> f4(I45, 1 + I46, I47, I48, I49) 18.42/18.16 f3(I50, I51, I52, I53, I54) -> f1(I50, I51, I50, I53, I54) [1 + I50 <= 100] 18.42/18.16 f3(I55, I56, I57, I58, I59) -> f4(I55, 100, I57, I58, I59) [100 <= I55] 18.42/18.16 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 18.42/18.16 18.42/18.16 The dependency graph for this problem is: 18.42/18.16 2 -> 3 18.42/18.16 3 -> 4 18.42/18.16 4 -> 8 18.42/18.16 8 -> 18.42/18.16 12 -> 2 18.42/18.16 Where: 18.42/18.16 2) f2#(I5, I6, I7, I8, I9) -> f11#(I5, I6, I7, I5, I9) 18.42/18.16 3) f11#(I10, I11, I12, I13, I14) -> f10#(I10, I11, I12, I13, I14) 18.42/18.16 4) f10#(I15, I16, I17, I18, I19) -> f6#(1 + I15, I16, I17, I18, I19) 18.42/18.16 8) f6#(I40, I41, I42, I43, I44) -> f3#(I40, I41, I42, I43, I44) 18.42/18.16 12) f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 18.42/18.16 18.42/18.16 We have the following SCCs. 18.42/18.16 18.45/21.14 EOF