12.44/12.55 YES 12.44/12.55 12.44/12.55 DP problem for innermost termination. 12.44/12.55 P = 12.44/12.55 f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 12.44/12.55 f9#(I0, I1, I2, I3, I4) -> f7#(I0, I1, I2, I3, I4) 12.44/12.55 f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) 12.44/12.55 f7#(I10, I11, I12, I13, I14) -> f8#(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7#(I15, I16, I17, I18, I19) -> f5#(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) 12.44/12.55 f5#(I25, I26, I27, I28, I29) -> f6#(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) 12.44/12.55 f3#(I40, I41, I42, I43, I44) -> f4#(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3#(I45, I46, I47, I48, I49) -> f1#(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) 12.44/12.55 f1#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 R = 12.44/12.55 f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 12.44/12.55 f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) 12.44/12.55 f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) 12.44/12.55 f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) 12.44/12.55 f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) 12.44/12.55 f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) 12.44/12.55 f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 The dependency graph for this problem is: 12.44/12.55 0 -> 1 12.44/12.55 1 -> 3, 4 12.44/12.55 2 -> 3, 4 12.44/12.55 3 -> 2 12.44/12.55 4 -> 6, 7 12.44/12.55 5 -> 6, 7 12.44/12.55 6 -> 5 12.44/12.55 7 -> 9, 10 12.44/12.55 8 -> 9, 10 12.44/12.55 9 -> 8 12.44/12.55 10 -> 12 12.44/12.55 11 -> 12 12.44/12.55 12 -> 11 12.44/12.55 Where: 12.44/12.55 0) f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 12.44/12.55 1) f9#(I0, I1, I2, I3, I4) -> f7#(I0, I1, I2, I3, I4) 12.44/12.55 2) f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) 12.44/12.55 3) f7#(I10, I11, I12, I13, I14) -> f8#(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 4) f7#(I15, I16, I17, I18, I19) -> f5#(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 5) f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) 12.44/12.55 6) f5#(I25, I26, I27, I28, I29) -> f6#(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 7) f5#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 8) f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) 12.44/12.55 9) f3#(I40, I41, I42, I43, I44) -> f4#(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 10) f3#(I45, I46, I47, I48, I49) -> f1#(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 11) f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) 12.44/12.55 12) f1#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 We have the following SCCs. 12.44/12.55 { 2, 3 } 12.44/12.55 { 5, 6 } 12.44/12.55 { 8, 9 } 12.44/12.55 { 11, 12 } 12.44/12.55 12.44/12.55 DP problem for innermost termination. 12.44/12.55 P = 12.44/12.55 f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) 12.44/12.55 f1#(I55, I56, I57, I58, I59) -> f2#(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 R = 12.44/12.55 f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 12.44/12.55 f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) 12.44/12.55 f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) 12.44/12.55 f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) 12.44/12.55 f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) 12.44/12.55 f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) 12.44/12.55 f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 We use the basic value criterion with the projection function NU: 12.44/12.55 NU[f1#(z1,z2,z3,z4,z5)] = z4 12.44/12.55 NU[f2#(z1,z2,z3,z4,z5)] = z4 12.44/12.55 12.44/12.55 This gives the following inequalities: 12.44/12.55 ==> I53 (>! \union =) I53 12.44/12.55 1 <= I58 ==> I58 >! -1 + I58 12.44/12.55 12.44/12.55 We remove all the strictly oriented dependency pairs. 12.44/12.55 12.44/12.55 DP problem for innermost termination. 12.44/12.55 P = 12.44/12.55 f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) 12.44/12.55 R = 12.44/12.55 f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 12.44/12.55 f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) 12.44/12.55 f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) 12.44/12.55 f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) 12.44/12.55 f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) 12.44/12.55 f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) 12.44/12.55 f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 The dependency graph for this problem is: 12.44/12.55 11 -> 12.44/12.55 Where: 12.44/12.55 11) f2#(I50, I51, I52, I53, I54) -> f1#(I50, I51, I52, I53, I54) 12.44/12.55 12.44/12.55 We have the following SCCs. 12.44/12.55 12.44/12.55 12.44/12.55 DP problem for innermost termination. 12.44/12.55 P = 12.44/12.55 f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) 12.44/12.55 f3#(I40, I41, I42, I43, I44) -> f4#(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 R = 12.44/12.55 f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 12.44/12.55 f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) 12.44/12.55 f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) 12.44/12.55 f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) 12.44/12.55 f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) 12.44/12.55 f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) 12.44/12.55 f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 We use the reverse value criterion with the projection function NU: 12.44/12.55 NU[f3#(z1,z2,z3,z4,z5)] = z5 + -1 * (1 + z3) 12.44/12.55 NU[f4#(z1,z2,z3,z4,z5)] = z5 + -1 * (1 + z3) 12.44/12.55 12.44/12.55 This gives the following inequalities: 12.44/12.55 ==> I39 + -1 * (1 + I37) >= I39 + -1 * (1 + I37) 12.44/12.55 1 + I42 <= I44 ==> I44 + -1 * (1 + I42) > I44 + -1 * (1 + (1 + I42)) with I44 + -1 * (1 + I42) >= 0 12.44/12.55 12.44/12.55 We remove all the strictly oriented dependency pairs. 12.44/12.55 12.44/12.55 DP problem for innermost termination. 12.44/12.55 P = 12.44/12.55 f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) 12.44/12.55 R = 12.44/12.55 f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 12.44/12.55 f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) 12.44/12.55 f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) 12.44/12.55 f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) 12.44/12.55 f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) 12.44/12.55 f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) 12.44/12.55 f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 The dependency graph for this problem is: 12.44/12.55 8 -> 12.44/12.55 Where: 12.44/12.55 8) f4#(I35, I36, I37, I38, I39) -> f3#(I35, I36, I37, I38, I39) 12.44/12.55 12.44/12.55 We have the following SCCs. 12.44/12.55 12.44/12.55 12.44/12.55 DP problem for innermost termination. 12.44/12.55 P = 12.44/12.55 f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) 12.44/12.55 f5#(I25, I26, I27, I28, I29) -> f6#(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 R = 12.44/12.55 f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 12.44/12.55 f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) 12.44/12.55 f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) 12.44/12.55 f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) 12.44/12.55 f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) 12.44/12.55 f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) 12.44/12.55 f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 We use the basic value criterion with the projection function NU: 12.44/12.55 NU[f5#(z1,z2,z3,z4,z5)] = z2 12.44/12.55 NU[f6#(z1,z2,z3,z4,z5)] = z2 12.44/12.55 12.44/12.55 This gives the following inequalities: 12.44/12.55 ==> I21 (>! \union =) I21 12.44/12.55 1 <= I26 ==> I26 >! -1 + I26 12.44/12.55 12.44/12.55 We remove all the strictly oriented dependency pairs. 12.44/12.55 12.44/12.55 DP problem for innermost termination. 12.44/12.55 P = 12.44/12.55 f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) 12.44/12.55 R = 12.44/12.55 f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 12.44/12.55 f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) 12.44/12.55 f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) 12.44/12.55 f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) 12.44/12.55 f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) 12.44/12.55 f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) 12.44/12.55 f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 The dependency graph for this problem is: 12.44/12.55 5 -> 12.44/12.55 Where: 12.44/12.55 5) f6#(I20, I21, I22, I23, I24) -> f5#(I20, I21, I22, I23, I24) 12.44/12.55 12.44/12.55 We have the following SCCs. 12.44/12.55 12.44/12.55 12.44/12.55 DP problem for innermost termination. 12.44/12.55 P = 12.44/12.55 f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) 12.44/12.55 f7#(I10, I11, I12, I13, I14) -> f8#(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 R = 12.44/12.55 f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 12.44/12.55 f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) 12.44/12.55 f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) 12.44/12.55 f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) 12.44/12.55 f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) 12.44/12.55 f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) 12.44/12.55 f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 We use the reverse value criterion with the projection function NU: 12.44/12.55 NU[f7#(z1,z2,z3,z4,z5)] = z5 + -1 * (1 + z1) 12.44/12.55 NU[f8#(z1,z2,z3,z4,z5)] = z5 + -1 * (1 + z1) 12.44/12.55 12.44/12.55 This gives the following inequalities: 12.44/12.55 ==> I9 + -1 * (1 + I5) >= I9 + -1 * (1 + I5) 12.44/12.55 1 + I10 <= I14 ==> I14 + -1 * (1 + I10) > I14 + -1 * (1 + (1 + I10)) with I14 + -1 * (1 + I10) >= 0 12.44/12.55 12.44/12.55 We remove all the strictly oriented dependency pairs. 12.44/12.55 12.44/12.55 DP problem for innermost termination. 12.44/12.55 P = 12.44/12.55 f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) 12.44/12.55 R = 12.44/12.55 f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 12.44/12.55 f9(I0, I1, I2, I3, I4) -> f7(I0, I1, I2, I3, I4) 12.44/12.55 f8(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) 12.44/12.55 f7(I10, I11, I12, I13, I14) -> f8(1 + I10, I11, I12, I13, I14) [1 + I10 <= I14] 12.44/12.55 f7(I15, I16, I17, I18, I19) -> f5(I15, I15, I17, I18, I19) [I19 <= I15] 12.44/12.55 f6(I20, I21, I22, I23, I24) -> f5(I20, I21, I22, I23, I24) 12.44/12.55 f5(I25, I26, I27, I28, I29) -> f6(I25, -1 + I26, I27, I28, I29) [1 <= I26] 12.44/12.55 f5(I30, I31, I32, I33, I34) -> f3(I30, I31, I31, I33, I34) [I31 <= 0] 12.44/12.55 f4(I35, I36, I37, I38, I39) -> f3(I35, I36, I37, I38, I39) 12.44/12.55 f3(I40, I41, I42, I43, I44) -> f4(I40, I41, 1 + I42, I43, I44) [1 + I42 <= I44] 12.44/12.55 f3(I45, I46, I47, I48, I49) -> f1(I45, I46, I47, I47, I49) [I49 <= I47] 12.44/12.55 f2(I50, I51, I52, I53, I54) -> f1(I50, I51, I52, I53, I54) 12.44/12.55 f1(I55, I56, I57, I58, I59) -> f2(I55, I56, I57, -1 + I58, I59) [1 <= I58] 12.44/12.55 12.44/12.55 The dependency graph for this problem is: 12.44/12.55 2 -> 12.44/12.55 Where: 12.44/12.55 2) f8#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) 12.44/12.55 12.44/12.55 We have the following SCCs. 12.44/12.55 12.44/15.53 EOF