YES DP problem for innermost termination. P = f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) f8#(I0, I1, I2, I3, I4, I5, I6) -> f1#(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7] f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] f2#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15] f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) f7#(I28, I29, I30, I31, I32, I33, I34) -> f6#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] f7#(I35, I36, I37, I38, I39, I40, I41) -> f5#(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36] f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) f3#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) R = f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36] f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57] f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) The dependency graph for this problem is: 0 -> 1 1 -> 9 2 -> 9 3 -> 7 4 -> 8 5 -> 7 6 -> 4 7 -> 5, 6 8 -> 4 9 -> 2, 3 Where: 0) f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) 1) f8#(I0, I1, I2, I3, I4, I5, I6) -> f1#(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7] 2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] 3) f2#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15] 4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 5) f7#(I28, I29, I30, I31, I32, I33, I34) -> f6#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] 6) f7#(I35, I36, I37, I38, I39, I40, I41) -> f5#(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36] 7) f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) 8) f3#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] 9) f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) We have the following SCCs. { 2, 9 } { 5, 7 } { 4, 8 } DP problem for innermost termination. P = f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) f3#(I49, I50, I51, I52, I53, I54, I55) -> f5#(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] R = f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36] f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57] f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2) NU[f5#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2) This gives the following inequalities: ==> I24 + -1 * (1 + I22) >= I24 + -1 * (1 + I22) 1 + I50 <= I52 ==> I52 + -1 * (1 + I50) > I52 + -1 * (1 + (1 + I50)) with I52 + -1 * (1 + I50) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) R = f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36] f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57] f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) The dependency graph for this problem is: 4 -> Where: 4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) We have the following SCCs. DP problem for innermost termination. P = f7#(I28, I29, I30, I31, I32, I33, I34) -> f6#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) R = f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36] f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57] f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) We use the reverse value criterion with the projection function NU: NU[f6#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2) NU[f7#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2) This gives the following inequalities: 1 + I29 <= I31 ==> I31 + -1 * (1 + I29) > I31 + -1 * (1 + (1 + I29)) with I31 + -1 * (1 + I29) >= 0 ==> I45 + -1 * (1 + I43) >= I45 + -1 * (1 + I43) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) R = f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36] f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57] f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) The dependency graph for this problem is: 7 -> Where: 7) f6#(I42, I43, I44, I45, I46, I47, I48) -> f7#(I42, I43, I44, I45, I46, I47, I48) We have the following SCCs. DP problem for innermost termination. P = f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) R = f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36] f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57] f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2) NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z2) This gives the following inequalities: 1 + I8 <= I10 ==> I10 + -1 * (1 + I8) > I10 + -1 * (1 + (1 + I8)) with I10 + -1 * (1 + I8) >= 0 ==> I66 + -1 * (1 + I64) >= I66 + -1 * (1 + I64) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) R = f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) f8(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, 0, rnd3, I0, I0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd3 = I0 /\ rnd7 = rnd7] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, 1 + I8, I9, I10, I11, I12, I13) [1 + I8 <= I10] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, 0, I16, I17, I18, I19, I20) [I17 <= I15] f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) f7(I28, I29, I30, I31, I32, I33, I34) -> f6(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I31] f7(I35, I36, I37, I38, I39, I40, I41) -> f5(I35, 0, I37, I38, I39, I40, I41) [I38 <= I36] f6(I42, I43, I44, I45, I46, I47, I48) -> f7(I42, I43, I44, I45, I46, I47, I48) f3(I49, I50, I51, I52, I53, I54, I55) -> f5(I49, 1 + I50, I51, I52, I53, I54, I55) [1 + I50 <= I52] f3(I56, I57, I58, I59, I60, I61, I62) -> f4(I56, I57, I58, I59, I60, I61, I62) [I59 <= I57] f1(I63, I64, I65, I66, I67, I68, I69) -> f2(I63, I64, I65, I66, I67, I68, I69) The dependency graph for this problem is: 9 -> Where: 9) f1#(I63, I64, I65, I66, I67, I68, I69) -> f2#(I63, I64, I65, I66, I67, I68, I69) We have the following SCCs.