/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f15#(x1, x2, x3, x4, x5, x6, x7) -> f14#(x1, x2, x3, x4, x5, x6, x7) f14#(I0, I1, I2, I3, I4, I5, I6) -> f1#(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) f7#(I29, I30, I31, I32, I33, I34, I35) -> f6#(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7#(I36, I37, I38, I39, I40, I41, I42) -> f10#(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9#(I43, I44, I45, I46, I47, I48, I49) -> f8#(I43, I44, I45, I46, I47, I48, I49) f11#(I50, I51, I52, I53, I54, I55, I56) -> f10#(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11#(I57, I58, I59, I60, I61, I62, I63) -> f13#(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13#(I65, I66, I67, I68, I69, I70, I71) -> f12#(I65, I66, I67, I68, I69, I70, I71) f12#(I72, I73, I74, I75, I76, I77, I78) -> f13#(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12#(I79, I80, I81, I82, I83, I84, I85) -> f9#(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10#(I87, I88, I89, I90, I91, I92, I93) -> f11#(I87, I88, I89, I90, I91, I92, I93) f8#(I94, I95, I96, I97, I98, I99, I100) -> f9#(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8#(I101, I102, I103, I104, I105, I106, I107) -> f5#(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6#(I108, I109, I110, I111, I112, I113, I114) -> f7#(I108, I109, I110, I111, I112, I113, I114) f3#(I115, I116, I117, I118, I119, I120, I121) -> f5#(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f1#(I129, I130, I131, I132, I133, I134, I135) -> f2#(I129, I130, I131, I132, I133, I134, I135) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) The dependency graph for this problem is: 0 -> 1 1 -> 18 2 -> 18 3 -> 16 4 -> 17 5 -> 16 6 -> 13 7 -> 14, 15 8 -> 13 9 -> 10 10 -> 11, 12 11 -> 10 12 -> 7 13 -> 8, 9 14 -> 7 15 -> 4 16 -> 5, 6 17 -> 4 18 -> 2, 3 Where: 0) f15#(x1, x2, x3, x4, x5, x6, x7) -> f14#(x1, x2, x3, x4, x5, x6, x7) 1) f14#(I0, I1, I2, I3, I4, I5, I6) -> f1#(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] 2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] 3) f2#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] 4) f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) 5) f7#(I29, I30, I31, I32, I33, I34, I35) -> f6#(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] 6) f7#(I36, I37, I38, I39, I40, I41, I42) -> f10#(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] 7) f9#(I43, I44, I45, I46, I47, I48, I49) -> f8#(I43, I44, I45, I46, I47, I48, I49) 8) f11#(I50, I51, I52, I53, I54, I55, I56) -> f10#(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] 9) f11#(I57, I58, I59, I60, I61, I62, I63) -> f13#(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] 10) f13#(I65, I66, I67, I68, I69, I70, I71) -> f12#(I65, I66, I67, I68, I69, I70, I71) 11) f12#(I72, I73, I74, I75, I76, I77, I78) -> f13#(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] 12) f12#(I79, I80, I81, I82, I83, I84, I85) -> f9#(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] 13) f10#(I87, I88, I89, I90, I91, I92, I93) -> f11#(I87, I88, I89, I90, I91, I92, I93) 14) f8#(I94, I95, I96, I97, I98, I99, I100) -> f9#(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] 15) f8#(I101, I102, I103, I104, I105, I106, I107) -> f5#(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] 16) f6#(I108, I109, I110, I111, I112, I113, I114) -> f7#(I108, I109, I110, I111, I112, I113, I114) 17) f3#(I115, I116, I117, I118, I119, I120, I121) -> f5#(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] 18) f1#(I129, I130, I131, I132, I133, I134, I135) -> f2#(I129, I130, I131, I132, I133, I134, I135) We have the following SCCs. { 2, 18 } { 5, 16 } { 8, 13 } { 10, 11 } { 7, 14 } { 4, 17 } DP problem for innermost termination. P = f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) f3#(I115, I116, I117, I118, I119, I120, I121) -> f5#(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z5) NU[f5#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z5) This gives the following inequalities: ==> 50 + -1 * (1 + I26) >= 50 + -1 * (1 + I26) 1 + I119 <= 50 ==> 50 + -1 * (1 + I119) > 50 + -1 * (1 + (1 + I119)) with 50 + -1 * (1 + I119) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) The dependency graph for this problem is: 4 -> Where: 4) f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) We have the following SCCs. DP problem for innermost termination. P = f9#(I43, I44, I45, I46, I47, I48, I49) -> f8#(I43, I44, I45, I46, I47, I48, I49) f8#(I94, I95, I96, I97, I98, I99, I100) -> f9#(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) We use the reverse value criterion with the projection function NU: NU[f8#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z2) NU[f9#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z2) This gives the following inequalities: ==> 50 + -1 * (1 + I44) >= 50 + -1 * (1 + I44) 1 + I95 <= 50 ==> 50 + -1 * (1 + I95) > 50 + -1 * (1 + (1 + I95)) with 50 + -1 * (1 + I95) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f9#(I43, I44, I45, I46, I47, I48, I49) -> f8#(I43, I44, I45, I46, I47, I48, I49) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) The dependency graph for this problem is: 7 -> Where: 7) f9#(I43, I44, I45, I46, I47, I48, I49) -> f8#(I43, I44, I45, I46, I47, I48, I49) We have the following SCCs. DP problem for innermost termination. P = f13#(I65, I66, I67, I68, I69, I70, I71) -> f12#(I65, I66, I67, I68, I69, I70, I71) f12#(I72, I73, I74, I75, I76, I77, I78) -> f13#(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) We use the reverse value criterion with the projection function NU: NU[f12#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z1) NU[f13#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z1) This gives the following inequalities: ==> 50 + -1 * (1 + I65) >= 50 + -1 * (1 + I65) 1 + I72 <= 50 ==> 50 + -1 * (1 + I72) > 50 + -1 * (1 + (1 + I72)) with 50 + -1 * (1 + I72) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f13#(I65, I66, I67, I68, I69, I70, I71) -> f12#(I65, I66, I67, I68, I69, I70, I71) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) The dependency graph for this problem is: 10 -> Where: 10) f13#(I65, I66, I67, I68, I69, I70, I71) -> f12#(I65, I66, I67, I68, I69, I70, I71) We have the following SCCs. DP problem for innermost termination. P = f11#(I50, I51, I52, I53, I54, I55, I56) -> f10#(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f10#(I87, I88, I89, I90, I91, I92, I93) -> f11#(I87, I88, I89, I90, I91, I92, I93) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) We use the reverse value criterion with the projection function NU: NU[f10#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z5) NU[f11#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z5) This gives the following inequalities: 1 + I54 <= 50 ==> 50 + -1 * (1 + I54) > 50 + -1 * (1 + (1 + I54)) with 50 + -1 * (1 + I54) >= 0 ==> 50 + -1 * (1 + I91) >= 50 + -1 * (1 + I91) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f10#(I87, I88, I89, I90, I91, I92, I93) -> f11#(I87, I88, I89, I90, I91, I92, I93) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) The dependency graph for this problem is: 13 -> Where: 13) f10#(I87, I88, I89, I90, I91, I92, I93) -> f11#(I87, I88, I89, I90, I91, I92, I93) We have the following SCCs. DP problem for innermost termination. P = f7#(I29, I30, I31, I32, I33, I34, I35) -> f6#(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f6#(I108, I109, I110, I111, I112, I113, I114) -> f7#(I108, I109, I110, I111, I112, I113, I114) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) We use the reverse value criterion with the projection function NU: NU[f6#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z4) NU[f7#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z4) This gives the following inequalities: 1 + I32 <= 50 ==> 50 + -1 * (1 + I32) > 50 + -1 * (1 + (1 + I32)) with 50 + -1 * (1 + I32) >= 0 ==> 50 + -1 * (1 + I111) >= 50 + -1 * (1 + I111) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I108, I109, I110, I111, I112, I113, I114) -> f7#(I108, I109, I110, I111, I112, I113, I114) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) The dependency graph for this problem is: 16 -> Where: 16) f6#(I108, I109, I110, I111, I112, I113, I114) -> f7#(I108, I109, I110, I111, I112, I113, I114) We have the following SCCs. DP problem for innermost termination. P = f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f1#(I129, I130, I131, I132, I133, I134, I135) -> f2#(I129, I130, I131, I132, I133, I134, I135) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z3) NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = 50 + -1 * (1 + z3) This gives the following inequalities: 1 + I9 <= 50 ==> 50 + -1 * (1 + I9) > 50 + -1 * (1 + (1 + I9)) with 50 + -1 * (1 + I9) >= 0 ==> 50 + -1 * (1 + I131) >= 50 + -1 * (1 + I131) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I129, I130, I131, I132, I133, I134, I135) -> f2#(I129, I130, I131, I132, I133, I134, I135) R = f15(x1, x2, x3, x4, x5, x6, x7) -> f14(x1, x2, x3, x4, x5, x6, x7) f14(I0, I1, I2, I3, I4, I5, I6) -> f1(I0, I1, 0, I3, 0, rnd6, rnd7) [rnd6 = rnd6 /\ rnd7 = rnd7 /\ y1 = 0] f2(I7, I8, I9, I10, I11, I12, I13) -> f1(I7, I8, 1 + I9, I10, I11, I12, I13) [1 + I9 <= 50] f2(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, 0, I18, I19, I20) [50 <= I16 /\ I21 = 0] f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) f7(I29, I30, I31, I32, I33, I34, I35) -> f6(I29, I30, I31, 1 + I32, I33, I34, I35) [1 + I32 <= 50] f7(I36, I37, I38, I39, I40, I41, I42) -> f10(I36, I37, I38, I39, 0, I41, I42) [50 <= I39] f9(I43, I44, I45, I46, I47, I48, I49) -> f8(I43, I44, I45, I46, I47, I48, I49) f11(I50, I51, I52, I53, I54, I55, I56) -> f10(I50, I51, I52, I53, 1 + I54, I55, I56) [1 + I54 <= 50] f11(I57, I58, I59, I60, I61, I62, I63) -> f13(0, I58, I59, I60, I61, I62, I63) [50 <= I61 /\ I64 = 0] f13(I65, I66, I67, I68, I69, I70, I71) -> f12(I65, I66, I67, I68, I69, I70, I71) f12(I72, I73, I74, I75, I76, I77, I78) -> f13(1 + I72, I73, I74, I75, I76, I77, I78) [1 + I72 <= 50] f12(I79, I80, I81, I82, I83, I84, I85) -> f9(I79, 0, I81, I82, I83, I84, I85) [50 <= I79 /\ I86 = 0] f10(I87, I88, I89, I90, I91, I92, I93) -> f11(I87, I88, I89, I90, I91, I92, I93) f8(I94, I95, I96, I97, I98, I99, I100) -> f9(I94, 1 + I95, I96, I97, I98, I99, I100) [1 + I95 <= 50] f8(I101, I102, I103, I104, I105, I106, I107) -> f5(I101, I102, I103, I104, 0, I106, I107) [50 <= I102] f6(I108, I109, I110, I111, I112, I113, I114) -> f7(I108, I109, I110, I111, I112, I113, I114) f3(I115, I116, I117, I118, I119, I120, I121) -> f5(I115, I116, I117, I118, 1 + I119, I120, I121) [1 + I119 <= 50] f3(I122, I123, I124, I125, I126, I127, I128) -> f4(I122, I123, I124, I125, I126, I127, I128) [50 <= I126] f1(I129, I130, I131, I132, I133, I134, I135) -> f2(I129, I130, I131, I132, I133, I134, I135) The dependency graph for this problem is: 18 -> Where: 18) f1#(I129, I130, I131, I132, I133, I134, I135) -> f2#(I129, I130, I131, I132, I133, I134, I135) We have the following SCCs.