34.61/34.28 YES 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f13#(x1, x2, x3, x4) -> f12#(x1, x2, x3, x4) 34.61/34.28 f12#(I0, I1, I2, I3) -> f1#(I0, 1, I2, I3) 34.61/34.28 f2#(I4, I5, I6, I7) -> f3#(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2#(I8, I9, I10, I11) -> f5#(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4#(I12, I13, I14, I15) -> f3#(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4#(I16, I17, I18, I19) -> f1#(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 f6#(I24, I25, I26, I27) -> f7#(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f8#(I32, I33, I34, I35) -> f10#(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8#(I36, I37, I38, I39) -> f5#(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9#(I40, I41, I42, I43) -> f10#(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9#(I44, I45, I46, I47) -> f7#(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7#(I48, I49, I50, I51) -> f8#(I48, I49, I50, I51) 34.61/34.28 f5#(I52, I53, I54, I55) -> f6#(I52, I53, I54, I55) 34.61/34.28 f3#(I56, I57, I58, I59) -> f4#(I56, I57, I58, I59) 34.61/34.28 f1#(I60, I61, I62, I63) -> f2#(I60, I61, I62, I63) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 The dependency graph for this problem is: 34.61/34.28 0 -> 1 34.61/34.28 1 -> 15 34.61/34.28 2 -> 14 34.61/34.28 3 -> 13 34.61/34.28 4 -> 14 34.61/34.28 5 -> 15 34.61/34.28 6 -> 10, 11 34.61/34.28 7 -> 12 34.61/34.28 8 -> 6 34.61/34.28 9 -> 13 34.61/34.28 10 -> 6 34.61/34.28 11 -> 12 34.61/34.28 12 -> 8, 9 34.61/34.28 13 -> 7 34.61/34.28 14 -> 4, 5 34.61/34.28 15 -> 2, 3 34.61/34.28 Where: 34.61/34.28 0) f13#(x1, x2, x3, x4) -> f12#(x1, x2, x3, x4) 34.61/34.28 1) f12#(I0, I1, I2, I3) -> f1#(I0, 1, I2, I3) 34.61/34.28 2) f2#(I4, I5, I6, I7) -> f3#(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 3) f2#(I8, I9, I10, I11) -> f5#(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 4) f4#(I12, I13, I14, I15) -> f3#(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 5) f4#(I16, I17, I18, I19) -> f1#(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 6) f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 7) f6#(I24, I25, I26, I27) -> f7#(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 8) f8#(I32, I33, I34, I35) -> f10#(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 9) f8#(I36, I37, I38, I39) -> f5#(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 10) f9#(I40, I41, I42, I43) -> f10#(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 11) f9#(I44, I45, I46, I47) -> f7#(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 12) f7#(I48, I49, I50, I51) -> f8#(I48, I49, I50, I51) 34.61/34.28 13) f5#(I52, I53, I54, I55) -> f6#(I52, I53, I54, I55) 34.61/34.28 14) f3#(I56, I57, I58, I59) -> f4#(I56, I57, I58, I59) 34.61/34.28 15) f1#(I60, I61, I62, I63) -> f2#(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 We have the following SCCs. 34.61/34.28 { 2, 4, 5, 14, 15 } 34.61/34.28 { 6, 7, 8, 9, 10, 11, 12, 13 } 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 f6#(I24, I25, I26, I27) -> f7#(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f8#(I32, I33, I34, I35) -> f10#(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8#(I36, I37, I38, I39) -> f5#(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9#(I40, I41, I42, I43) -> f10#(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9#(I44, I45, I46, I47) -> f7#(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7#(I48, I49, I50, I51) -> f8#(I48, I49, I50, I51) 34.61/34.28 f5#(I52, I53, I54, I55) -> f6#(I52, I53, I54, I55) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 We use the extended value criterion with the projection function NU: 34.61/34.28 NU[f5#(x0,x1,x2,x3)] = x0 - x1 34.61/34.28 NU[f8#(x0,x1,x2,x3)] = x0 - x1 - 1 34.61/34.28 NU[f7#(x0,x1,x2,x3)] = x0 - x1 - 1 34.61/34.28 NU[f6#(x0,x1,x2,x3)] = x0 - x1 34.61/34.28 NU[f9#(x0,x1,x2,x3)] = x0 - x1 - 1 34.61/34.28 NU[f10#(x0,x1,x2,x3)] = x0 - x1 - 1 34.61/34.28 34.61/34.28 This gives the following inequalities: 34.61/34.28 ==> I20 - I21 - 1 >= I20 - I21 - 1 34.61/34.28 I25 <= I24 ==> I24 - I25 > I24 - I25 - 1 with I24 - I25 >= 0 34.61/34.28 I34 <= I32 ==> I32 - I33 - 1 >= I32 - I33 - 1 34.61/34.28 1 + I36 <= I38 ==> I36 - I37 - 1 >= I36 - (1 + I37) 34.61/34.28 I43 <= I40 ==> I40 - I41 - 1 >= I40 - I41 - 1 34.61/34.28 1 + I44 <= I47 ==> I44 - I45 - 1 >= I44 - I45 - 1 34.61/34.28 ==> I48 - I49 - 1 >= I48 - I49 - 1 34.61/34.28 ==> I52 - I53 >= I52 - I53 34.61/34.28 34.61/34.28 We remove all the strictly oriented dependency pairs. 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 f8#(I32, I33, I34, I35) -> f10#(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8#(I36, I37, I38, I39) -> f5#(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9#(I40, I41, I42, I43) -> f10#(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9#(I44, I45, I46, I47) -> f7#(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7#(I48, I49, I50, I51) -> f8#(I48, I49, I50, I51) 34.61/34.28 f5#(I52, I53, I54, I55) -> f6#(I52, I53, I54, I55) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 The dependency graph for this problem is: 34.61/34.28 6 -> 10, 11 34.61/34.28 8 -> 6 34.61/34.28 9 -> 13 34.61/34.28 10 -> 6 34.61/34.28 11 -> 12 34.61/34.28 12 -> 8, 9 34.61/34.28 13 -> 34.61/34.28 Where: 34.61/34.28 6) f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 8) f8#(I32, I33, I34, I35) -> f10#(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 9) f8#(I36, I37, I38, I39) -> f5#(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 10) f9#(I40, I41, I42, I43) -> f10#(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 11) f9#(I44, I45, I46, I47) -> f7#(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 12) f7#(I48, I49, I50, I51) -> f8#(I48, I49, I50, I51) 34.61/34.28 13) f5#(I52, I53, I54, I55) -> f6#(I52, I53, I54, I55) 34.61/34.28 34.61/34.28 We have the following SCCs. 34.61/34.28 { 6, 8, 10, 11, 12 } 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 f8#(I32, I33, I34, I35) -> f10#(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f9#(I40, I41, I42, I43) -> f10#(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9#(I44, I45, I46, I47) -> f7#(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7#(I48, I49, I50, I51) -> f8#(I48, I49, I50, I51) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 We use the extended value criterion with the projection function NU: 34.61/34.28 NU[f7#(x0,x1,x2,x3)] = x0 - x2 34.61/34.28 NU[f8#(x0,x1,x2,x3)] = x0 - x2 34.61/34.28 NU[f9#(x0,x1,x2,x3)] = x0 - x2 - 1 34.61/34.28 NU[f10#(x0,x1,x2,x3)] = x0 - x2 - 1 34.61/34.28 34.61/34.28 This gives the following inequalities: 34.61/34.28 ==> I20 - I22 - 1 >= I20 - I22 - 1 34.61/34.28 I34 <= I32 ==> I32 - I34 > I32 - I34 - 1 with I32 - I34 >= 0 34.61/34.28 I43 <= I40 ==> I40 - I42 - 1 >= I40 - I42 - 1 34.61/34.28 1 + I44 <= I47 ==> I44 - I46 - 1 >= I44 - (1 + I46) 34.61/34.28 ==> I48 - I50 >= I48 - I50 34.61/34.28 34.61/34.28 We remove all the strictly oriented dependency pairs. 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 f9#(I40, I41, I42, I43) -> f10#(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9#(I44, I45, I46, I47) -> f7#(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7#(I48, I49, I50, I51) -> f8#(I48, I49, I50, I51) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 The dependency graph for this problem is: 34.61/34.28 6 -> 10, 11 34.61/34.28 10 -> 6 34.61/34.28 11 -> 12 34.61/34.28 12 -> 34.61/34.28 Where: 34.61/34.28 6) f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 10) f9#(I40, I41, I42, I43) -> f10#(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 11) f9#(I44, I45, I46, I47) -> f7#(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 12) f7#(I48, I49, I50, I51) -> f8#(I48, I49, I50, I51) 34.61/34.28 34.61/34.28 We have the following SCCs. 34.61/34.28 { 6, 10 } 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 f9#(I40, I41, I42, I43) -> f10#(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 We use the reverse value criterion with the projection function NU: 34.61/34.28 NU[f9#(z1,z2,z3,z4)] = z1 + -1 * z4 34.61/34.28 NU[f10#(z1,z2,z3,z4)] = z1 + -1 * z4 34.61/34.28 34.61/34.28 This gives the following inequalities: 34.61/34.28 ==> I20 + -1 * I23 >= I20 + -1 * I23 34.61/34.28 I43 <= I40 ==> I40 + -1 * I43 > I40 + -1 * (1 + I43) with I40 + -1 * I43 >= 0 34.61/34.28 34.61/34.28 We remove all the strictly oriented dependency pairs. 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 The dependency graph for this problem is: 34.61/34.28 6 -> 34.61/34.28 Where: 34.61/34.28 6) f10#(I20, I21, I22, I23) -> f9#(I20, I21, I22, I23) 34.61/34.28 34.61/34.28 We have the following SCCs. 34.61/34.28 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f2#(I4, I5, I6, I7) -> f3#(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f4#(I12, I13, I14, I15) -> f3#(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4#(I16, I17, I18, I19) -> f1#(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f3#(I56, I57, I58, I59) -> f4#(I56, I57, I58, I59) 34.61/34.28 f1#(I60, I61, I62, I63) -> f2#(I60, I61, I62, I63) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 We use the extended value criterion with the projection function NU: 34.61/34.28 NU[f1#(x0,x1,x2,x3)] = x0 - x1 34.61/34.28 NU[f4#(x0,x1,x2,x3)] = x0 - x1 - 1 34.61/34.28 NU[f3#(x0,x1,x2,x3)] = x0 - x1 - 1 34.61/34.28 NU[f2#(x0,x1,x2,x3)] = x0 - x1 34.61/34.28 34.61/34.28 This gives the following inequalities: 34.61/34.28 I5 <= I4 ==> I4 - I5 > I4 - I5 - 1 with I4 - I5 >= 0 34.61/34.28 I14 <= I12 ==> I12 - I13 - 1 >= I12 - I13 - 1 34.61/34.28 1 + I16 <= I18 ==> I16 - I17 - 1 >= I16 - (1 + I17) 34.61/34.28 ==> I56 - I57 - 1 >= I56 - I57 - 1 34.61/34.28 ==> I60 - I61 >= I60 - I61 34.61/34.28 34.61/34.28 We remove all the strictly oriented dependency pairs. 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f4#(I12, I13, I14, I15) -> f3#(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4#(I16, I17, I18, I19) -> f1#(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f3#(I56, I57, I58, I59) -> f4#(I56, I57, I58, I59) 34.61/34.28 f1#(I60, I61, I62, I63) -> f2#(I60, I61, I62, I63) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 The dependency graph for this problem is: 34.61/34.28 4 -> 14 34.61/34.28 5 -> 15 34.61/34.28 14 -> 4, 5 34.61/34.28 15 -> 34.61/34.28 Where: 34.61/34.28 4) f4#(I12, I13, I14, I15) -> f3#(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 5) f4#(I16, I17, I18, I19) -> f1#(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 14) f3#(I56, I57, I58, I59) -> f4#(I56, I57, I58, I59) 34.61/34.28 15) f1#(I60, I61, I62, I63) -> f2#(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 We have the following SCCs. 34.61/34.28 { 4, 14 } 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f4#(I12, I13, I14, I15) -> f3#(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f3#(I56, I57, I58, I59) -> f4#(I56, I57, I58, I59) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 We use the reverse value criterion with the projection function NU: 34.61/34.28 NU[f3#(z1,z2,z3,z4)] = z1 + -1 * z3 34.61/34.28 NU[f4#(z1,z2,z3,z4)] = z1 + -1 * z3 34.61/34.28 34.61/34.28 This gives the following inequalities: 34.61/34.28 I14 <= I12 ==> I12 + -1 * I14 > I12 + -1 * (1 + I14) with I12 + -1 * I14 >= 0 34.61/34.28 ==> I56 + -1 * I58 >= I56 + -1 * I58 34.61/34.28 34.61/34.28 We remove all the strictly oriented dependency pairs. 34.61/34.28 34.61/34.28 DP problem for innermost termination. 34.61/34.28 P = 34.61/34.28 f3#(I56, I57, I58, I59) -> f4#(I56, I57, I58, I59) 34.61/34.28 R = 34.61/34.28 f13(x1, x2, x3, x4) -> f12(x1, x2, x3, x4) 34.61/34.28 f12(I0, I1, I2, I3) -> f1(I0, 1, I2, I3) 34.61/34.28 f2(I4, I5, I6, I7) -> f3(I4, I5, 1, I7) [I5 <= I4] 34.61/34.28 f2(I8, I9, I10, I11) -> f5(I8, 1, I10, I11) [1 + I8 <= I9] 34.61/34.28 f4(I12, I13, I14, I15) -> f3(I12, I13, 1 + I14, I15) [I14 <= I12] 34.61/34.28 f4(I16, I17, I18, I19) -> f1(I16, 1 + I17, I18, I19) [1 + I16 <= I18] 34.61/34.28 f10(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 34.61/34.28 f6(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [I25 <= I24] 34.61/34.28 f6(I28, I29, I30, I31) -> f11(I28, I29, I30, I31) [1 + I28 <= I29] 34.61/34.28 f8(I32, I33, I34, I35) -> f10(I32, I33, I34, 1) [I34 <= I32] 34.61/34.28 f8(I36, I37, I38, I39) -> f5(I36, 1 + I37, I38, I39) [1 + I36 <= I38] 34.61/34.28 f9(I40, I41, I42, I43) -> f10(I40, I41, I42, 1 + I43) [I43 <= I40] 34.61/34.28 f9(I44, I45, I46, I47) -> f7(I44, I45, 1 + I46, I47) [1 + I44 <= I47] 34.61/34.28 f7(I48, I49, I50, I51) -> f8(I48, I49, I50, I51) 34.61/34.28 f5(I52, I53, I54, I55) -> f6(I52, I53, I54, I55) 34.61/34.28 f3(I56, I57, I58, I59) -> f4(I56, I57, I58, I59) 34.61/34.28 f1(I60, I61, I62, I63) -> f2(I60, I61, I62, I63) 34.61/34.28 34.61/34.28 The dependency graph for this problem is: 34.61/34.28 14 -> 34.61/34.28 Where: 34.61/34.28 14) f3#(I56, I57, I58, I59) -> f4#(I56, I57, I58, I59) 34.61/34.28 34.61/34.28 We have the following SCCs. 34.61/34.28 34.61/37.26 EOF