/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f11#(x1, x2, x3, x4, x5, x6, x7, x8) -> f10#(x1, x2, x3, x4, x5, x6, x7, x8) f10#(I0, I1, I2, I3, I4, I5, I6, I7) -> f4#(I0, rnd2, rnd3, 35, 0, 10, I6, 285) [rnd2 = rnd3 /\ rnd3 = rnd3] f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f8#(rnd1, I9, I10, I11, I12, I13, 1, I15) [rnd1 = rnd1 /\ 1 + I12 <= I13] f8#(I24, I25, I26, I27, I28, I29, I30, I31) -> f7#(I24, I25, I26, I27, I28, I29, I30, I31) f7#(I32, I33, I34, I35, I36, I37, I38, I39) -> f8#(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] f7#(I41, I42, I43, I44, I45, I46, I47, I48) -> f5#(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] f4#(I49, I50, I51, I52, I53, I54, I55, I56) -> f6#(I49, I50, I51, I52, I53, I54, I55, I56) f5#(I57, I58, I59, I60, I61, I62, I63, I64) -> f3#(I57, I58, I59, I60, I61, I62, I63, I64) f5#(I65, I66, I67, I68, I69, I70, I71, I72) -> f2#(I65, -1 + I66, I67, I68, I69, I70, I71, I72) f5#(I73, I74, I75, I76, I77, I78, I79, I80) -> f3#(I73, I74, I75, I76, I77, I78, I79, I80) f2#(I81, I82, I83, I84, I85, I86, I87, I88) -> f4#(I81, I82, I83, I84, 1 + I85, I86, I87, I88) f3#(I89, I90, I91, I92, I93, I94, I95, I96) -> f1#(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] f3#(I97, I98, I99, I100, I101, I102, I103, I104) -> f1#(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] f1#(I105, I106, I107, I108, I109, I110, I111, I112) -> f2#(I105, I106, I107, I108, I109, I110, I111, I112) R = f11(x1, x2, x3, x4, x5, x6, x7, x8) -> f10(x1, x2, x3, x4, x5, x6, x7, x8) f10(I0, I1, I2, I3, I4, I5, I6, I7) -> f4(I0, rnd2, rnd3, 35, 0, 10, I6, 285) [rnd2 = rnd3 /\ rnd3 = rnd3] f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f8(rnd1, I9, I10, I11, I12, I13, 1, I15) [rnd1 = rnd1 /\ 1 + I12 <= I13] f6(I16, I17, I18, I19, I20, I21, I22, I23) -> f9(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= I20] f8(I24, I25, I26, I27, I28, I29, I30, I31) -> f7(I24, I25, I26, I27, I28, I29, I30, I31) f7(I32, I33, I34, I35, I36, I37, I38, I39) -> f8(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] f7(I41, I42, I43, I44, I45, I46, I47, I48) -> f5(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] f4(I49, I50, I51, I52, I53, I54, I55, I56) -> f6(I49, I50, I51, I52, I53, I54, I55, I56) f5(I57, I58, I59, I60, I61, I62, I63, I64) -> f3(I57, I58, I59, I60, I61, I62, I63, I64) f5(I65, I66, I67, I68, I69, I70, I71, I72) -> f2(I65, -1 + I66, I67, I68, I69, I70, I71, I72) f5(I73, I74, I75, I76, I77, I78, I79, I80) -> f3(I73, I74, I75, I76, I77, I78, I79, I80) f2(I81, I82, I83, I84, I85, I86, I87, I88) -> f4(I81, I82, I83, I84, 1 + I85, I86, I87, I88) f3(I89, I90, I91, I92, I93, I94, I95, I96) -> f1(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] f3(I97, I98, I99, I100, I101, I102, I103, I104) -> f1(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] f1(I105, I106, I107, I108, I109, I110, I111, I112) -> f2(I105, I106, I107, I108, I109, I110, I111, I112) The dependency graph for this problem is: 0 -> 1 1 -> 6 2 -> 3 3 -> 4, 5 4 -> 3 5 -> 7, 8, 9 6 -> 2 7 -> 11, 12 8 -> 10 9 -> 11, 12 10 -> 6 11 -> 13 12 -> 13 13 -> 10 Where: 0) f11#(x1, x2, x3, x4, x5, x6, x7, x8) -> f10#(x1, x2, x3, x4, x5, x6, x7, x8) 1) f10#(I0, I1, I2, I3, I4, I5, I6, I7) -> f4#(I0, rnd2, rnd3, 35, 0, 10, I6, 285) [rnd2 = rnd3 /\ rnd3 = rnd3] 2) f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f8#(rnd1, I9, I10, I11, I12, I13, 1, I15) [rnd1 = rnd1 /\ 1 + I12 <= I13] 3) f8#(I24, I25, I26, I27, I28, I29, I30, I31) -> f7#(I24, I25, I26, I27, I28, I29, I30, I31) 4) f7#(I32, I33, I34, I35, I36, I37, I38, I39) -> f8#(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] 5) f7#(I41, I42, I43, I44, I45, I46, I47, I48) -> f5#(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] 6) f4#(I49, I50, I51, I52, I53, I54, I55, I56) -> f6#(I49, I50, I51, I52, I53, I54, I55, I56) 7) f5#(I57, I58, I59, I60, I61, I62, I63, I64) -> f3#(I57, I58, I59, I60, I61, I62, I63, I64) 8) f5#(I65, I66, I67, I68, I69, I70, I71, I72) -> f2#(I65, -1 + I66, I67, I68, I69, I70, I71, I72) 9) f5#(I73, I74, I75, I76, I77, I78, I79, I80) -> f3#(I73, I74, I75, I76, I77, I78, I79, I80) 10) f2#(I81, I82, I83, I84, I85, I86, I87, I88) -> f4#(I81, I82, I83, I84, 1 + I85, I86, I87, I88) 11) f3#(I89, I90, I91, I92, I93, I94, I95, I96) -> f1#(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] 12) f3#(I97, I98, I99, I100, I101, I102, I103, I104) -> f1#(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] 13) f1#(I105, I106, I107, I108, I109, I110, I111, I112) -> f2#(I105, I106, I107, I108, I109, I110, I111, I112) We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 } DP problem for innermost termination. P = f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f8#(rnd1, I9, I10, I11, I12, I13, 1, I15) [rnd1 = rnd1 /\ 1 + I12 <= I13] f8#(I24, I25, I26, I27, I28, I29, I30, I31) -> f7#(I24, I25, I26, I27, I28, I29, I30, I31) f7#(I32, I33, I34, I35, I36, I37, I38, I39) -> f8#(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] f7#(I41, I42, I43, I44, I45, I46, I47, I48) -> f5#(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] f4#(I49, I50, I51, I52, I53, I54, I55, I56) -> f6#(I49, I50, I51, I52, I53, I54, I55, I56) f5#(I57, I58, I59, I60, I61, I62, I63, I64) -> f3#(I57, I58, I59, I60, I61, I62, I63, I64) f5#(I65, I66, I67, I68, I69, I70, I71, I72) -> f2#(I65, -1 + I66, I67, I68, I69, I70, I71, I72) f5#(I73, I74, I75, I76, I77, I78, I79, I80) -> f3#(I73, I74, I75, I76, I77, I78, I79, I80) f2#(I81, I82, I83, I84, I85, I86, I87, I88) -> f4#(I81, I82, I83, I84, 1 + I85, I86, I87, I88) f3#(I89, I90, I91, I92, I93, I94, I95, I96) -> f1#(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] f3#(I97, I98, I99, I100, I101, I102, I103, I104) -> f1#(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] f1#(I105, I106, I107, I108, I109, I110, I111, I112) -> f2#(I105, I106, I107, I108, I109, I110, I111, I112) R = f11(x1, x2, x3, x4, x5, x6, x7, x8) -> f10(x1, x2, x3, x4, x5, x6, x7, x8) f10(I0, I1, I2, I3, I4, I5, I6, I7) -> f4(I0, rnd2, rnd3, 35, 0, 10, I6, 285) [rnd2 = rnd3 /\ rnd3 = rnd3] f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f8(rnd1, I9, I10, I11, I12, I13, 1, I15) [rnd1 = rnd1 /\ 1 + I12 <= I13] f6(I16, I17, I18, I19, I20, I21, I22, I23) -> f9(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= I20] f8(I24, I25, I26, I27, I28, I29, I30, I31) -> f7(I24, I25, I26, I27, I28, I29, I30, I31) f7(I32, I33, I34, I35, I36, I37, I38, I39) -> f8(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] f7(I41, I42, I43, I44, I45, I46, I47, I48) -> f5(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] f4(I49, I50, I51, I52, I53, I54, I55, I56) -> f6(I49, I50, I51, I52, I53, I54, I55, I56) f5(I57, I58, I59, I60, I61, I62, I63, I64) -> f3(I57, I58, I59, I60, I61, I62, I63, I64) f5(I65, I66, I67, I68, I69, I70, I71, I72) -> f2(I65, -1 + I66, I67, I68, I69, I70, I71, I72) f5(I73, I74, I75, I76, I77, I78, I79, I80) -> f3(I73, I74, I75, I76, I77, I78, I79, I80) f2(I81, I82, I83, I84, I85, I86, I87, I88) -> f4(I81, I82, I83, I84, 1 + I85, I86, I87, I88) f3(I89, I90, I91, I92, I93, I94, I95, I96) -> f1(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] f3(I97, I98, I99, I100, I101, I102, I103, I104) -> f1(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] f1(I105, I106, I107, I108, I109, I110, I111, I112) -> f2(I105, I106, I107, I108, I109, I110, I111, I112) We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3,x4,x5,x6,x7)] = -x4 + x5 NU[f2#(x0,x1,x2,x3,x4,x5,x6,x7)] = -x4 + x5 NU[f3#(x0,x1,x2,x3,x4,x5,x6,x7)] = -x4 + x5 NU[f4#(x0,x1,x2,x3,x4,x5,x6,x7)] = -x4 + x5 + 1 NU[f5#(x0,x1,x2,x3,x4,x5,x6,x7)] = -x4 + x5 NU[f7#(x0,x1,x2,x3,x4,x5,x6,x7)] = -x4 + x5 NU[f8#(x0,x1,x2,x3,x4,x5,x6,x7)] = -x4 + x5 NU[f6#(x0,x1,x2,x3,x4,x5,x6,x7)] = -x4 + x5 + 1 This gives the following inequalities: rnd1 = rnd1 /\ 1 + I12 <= I13 ==> -I12 + I13 + 1 > -I12 + I13 with -I12 + I13 + 1 >= 0 ==> -I28 + I29 >= -I28 + I29 I40 = I40 /\ 1 + I38 <= I33 ==> -I36 + I37 >= -I36 + I37 I42 <= I47 ==> -I45 + I46 >= -I45 + I46 ==> -I53 + I54 + 1 >= -I53 + I54 + 1 ==> -I61 + I62 >= -I61 + I62 ==> -I69 + I70 >= -I69 + I70 ==> -I77 + I78 >= -I77 + I78 ==> -I85 + I86 >= -(1 + I85) + I86 + 1 1 + I90 <= I92 ==> -I93 + I94 >= -I93 + I94 I100 <= I98 ==> -I101 + I102 >= -I101 + I102 ==> -I109 + I110 >= -I109 + I110 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f8#(I24, I25, I26, I27, I28, I29, I30, I31) -> f7#(I24, I25, I26, I27, I28, I29, I30, I31) f7#(I32, I33, I34, I35, I36, I37, I38, I39) -> f8#(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] f7#(I41, I42, I43, I44, I45, I46, I47, I48) -> f5#(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] f4#(I49, I50, I51, I52, I53, I54, I55, I56) -> f6#(I49, I50, I51, I52, I53, I54, I55, I56) f5#(I57, I58, I59, I60, I61, I62, I63, I64) -> f3#(I57, I58, I59, I60, I61, I62, I63, I64) f5#(I65, I66, I67, I68, I69, I70, I71, I72) -> f2#(I65, -1 + I66, I67, I68, I69, I70, I71, I72) f5#(I73, I74, I75, I76, I77, I78, I79, I80) -> f3#(I73, I74, I75, I76, I77, I78, I79, I80) f2#(I81, I82, I83, I84, I85, I86, I87, I88) -> f4#(I81, I82, I83, I84, 1 + I85, I86, I87, I88) f3#(I89, I90, I91, I92, I93, I94, I95, I96) -> f1#(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] f3#(I97, I98, I99, I100, I101, I102, I103, I104) -> f1#(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] f1#(I105, I106, I107, I108, I109, I110, I111, I112) -> f2#(I105, I106, I107, I108, I109, I110, I111, I112) R = f11(x1, x2, x3, x4, x5, x6, x7, x8) -> f10(x1, x2, x3, x4, x5, x6, x7, x8) f10(I0, I1, I2, I3, I4, I5, I6, I7) -> f4(I0, rnd2, rnd3, 35, 0, 10, I6, 285) [rnd2 = rnd3 /\ rnd3 = rnd3] f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f8(rnd1, I9, I10, I11, I12, I13, 1, I15) [rnd1 = rnd1 /\ 1 + I12 <= I13] f6(I16, I17, I18, I19, I20, I21, I22, I23) -> f9(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= I20] f8(I24, I25, I26, I27, I28, I29, I30, I31) -> f7(I24, I25, I26, I27, I28, I29, I30, I31) f7(I32, I33, I34, I35, I36, I37, I38, I39) -> f8(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] f7(I41, I42, I43, I44, I45, I46, I47, I48) -> f5(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] f4(I49, I50, I51, I52, I53, I54, I55, I56) -> f6(I49, I50, I51, I52, I53, I54, I55, I56) f5(I57, I58, I59, I60, I61, I62, I63, I64) -> f3(I57, I58, I59, I60, I61, I62, I63, I64) f5(I65, I66, I67, I68, I69, I70, I71, I72) -> f2(I65, -1 + I66, I67, I68, I69, I70, I71, I72) f5(I73, I74, I75, I76, I77, I78, I79, I80) -> f3(I73, I74, I75, I76, I77, I78, I79, I80) f2(I81, I82, I83, I84, I85, I86, I87, I88) -> f4(I81, I82, I83, I84, 1 + I85, I86, I87, I88) f3(I89, I90, I91, I92, I93, I94, I95, I96) -> f1(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] f3(I97, I98, I99, I100, I101, I102, I103, I104) -> f1(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] f1(I105, I106, I107, I108, I109, I110, I111, I112) -> f2(I105, I106, I107, I108, I109, I110, I111, I112) The dependency graph for this problem is: 3 -> 4, 5 4 -> 3 5 -> 7, 8, 9 6 -> 7 -> 11, 12 8 -> 10 9 -> 11, 12 10 -> 6 11 -> 13 12 -> 13 13 -> 10 Where: 3) f8#(I24, I25, I26, I27, I28, I29, I30, I31) -> f7#(I24, I25, I26, I27, I28, I29, I30, I31) 4) f7#(I32, I33, I34, I35, I36, I37, I38, I39) -> f8#(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] 5) f7#(I41, I42, I43, I44, I45, I46, I47, I48) -> f5#(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] 6) f4#(I49, I50, I51, I52, I53, I54, I55, I56) -> f6#(I49, I50, I51, I52, I53, I54, I55, I56) 7) f5#(I57, I58, I59, I60, I61, I62, I63, I64) -> f3#(I57, I58, I59, I60, I61, I62, I63, I64) 8) f5#(I65, I66, I67, I68, I69, I70, I71, I72) -> f2#(I65, -1 + I66, I67, I68, I69, I70, I71, I72) 9) f5#(I73, I74, I75, I76, I77, I78, I79, I80) -> f3#(I73, I74, I75, I76, I77, I78, I79, I80) 10) f2#(I81, I82, I83, I84, I85, I86, I87, I88) -> f4#(I81, I82, I83, I84, 1 + I85, I86, I87, I88) 11) f3#(I89, I90, I91, I92, I93, I94, I95, I96) -> f1#(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] 12) f3#(I97, I98, I99, I100, I101, I102, I103, I104) -> f1#(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] 13) f1#(I105, I106, I107, I108, I109, I110, I111, I112) -> f2#(I105, I106, I107, I108, I109, I110, I111, I112) We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f8#(I24, I25, I26, I27, I28, I29, I30, I31) -> f7#(I24, I25, I26, I27, I28, I29, I30, I31) f7#(I32, I33, I34, I35, I36, I37, I38, I39) -> f8#(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] R = f11(x1, x2, x3, x4, x5, x6, x7, x8) -> f10(x1, x2, x3, x4, x5, x6, x7, x8) f10(I0, I1, I2, I3, I4, I5, I6, I7) -> f4(I0, rnd2, rnd3, 35, 0, 10, I6, 285) [rnd2 = rnd3 /\ rnd3 = rnd3] f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f8(rnd1, I9, I10, I11, I12, I13, 1, I15) [rnd1 = rnd1 /\ 1 + I12 <= I13] f6(I16, I17, I18, I19, I20, I21, I22, I23) -> f9(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= I20] f8(I24, I25, I26, I27, I28, I29, I30, I31) -> f7(I24, I25, I26, I27, I28, I29, I30, I31) f7(I32, I33, I34, I35, I36, I37, I38, I39) -> f8(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] f7(I41, I42, I43, I44, I45, I46, I47, I48) -> f5(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] f4(I49, I50, I51, I52, I53, I54, I55, I56) -> f6(I49, I50, I51, I52, I53, I54, I55, I56) f5(I57, I58, I59, I60, I61, I62, I63, I64) -> f3(I57, I58, I59, I60, I61, I62, I63, I64) f5(I65, I66, I67, I68, I69, I70, I71, I72) -> f2(I65, -1 + I66, I67, I68, I69, I70, I71, I72) f5(I73, I74, I75, I76, I77, I78, I79, I80) -> f3(I73, I74, I75, I76, I77, I78, I79, I80) f2(I81, I82, I83, I84, I85, I86, I87, I88) -> f4(I81, I82, I83, I84, 1 + I85, I86, I87, I88) f3(I89, I90, I91, I92, I93, I94, I95, I96) -> f1(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] f3(I97, I98, I99, I100, I101, I102, I103, I104) -> f1(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] f1(I105, I106, I107, I108, I109, I110, I111, I112) -> f2(I105, I106, I107, I108, I109, I110, I111, I112) We use the reverse value criterion with the projection function NU: NU[f7#(z1,z2,z3,z4,z5,z6,z7,z8)] = z2 + -1 * (1 + z7) NU[f8#(z1,z2,z3,z4,z5,z6,z7,z8)] = z2 + -1 * (1 + z7) This gives the following inequalities: ==> I25 + -1 * (1 + I30) >= I25 + -1 * (1 + I30) I40 = I40 /\ 1 + I38 <= I33 ==> I33 + -1 * (1 + I38) > I33 + -1 * (1 + (1 + I38)) with I33 + -1 * (1 + I38) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f8#(I24, I25, I26, I27, I28, I29, I30, I31) -> f7#(I24, I25, I26, I27, I28, I29, I30, I31) R = f11(x1, x2, x3, x4, x5, x6, x7, x8) -> f10(x1, x2, x3, x4, x5, x6, x7, x8) f10(I0, I1, I2, I3, I4, I5, I6, I7) -> f4(I0, rnd2, rnd3, 35, 0, 10, I6, 285) [rnd2 = rnd3 /\ rnd3 = rnd3] f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f8(rnd1, I9, I10, I11, I12, I13, 1, I15) [rnd1 = rnd1 /\ 1 + I12 <= I13] f6(I16, I17, I18, I19, I20, I21, I22, I23) -> f9(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= I20] f8(I24, I25, I26, I27, I28, I29, I30, I31) -> f7(I24, I25, I26, I27, I28, I29, I30, I31) f7(I32, I33, I34, I35, I36, I37, I38, I39) -> f8(I40, I33, I34, I35, I36, I37, 1 + I38, I39) [I40 = I40 /\ 1 + I38 <= I33] f7(I41, I42, I43, I44, I45, I46, I47, I48) -> f5(I41, I42, I43, I44, I45, I46, I47, I48) [I42 <= I47] f4(I49, I50, I51, I52, I53, I54, I55, I56) -> f6(I49, I50, I51, I52, I53, I54, I55, I56) f5(I57, I58, I59, I60, I61, I62, I63, I64) -> f3(I57, I58, I59, I60, I61, I62, I63, I64) f5(I65, I66, I67, I68, I69, I70, I71, I72) -> f2(I65, -1 + I66, I67, I68, I69, I70, I71, I72) f5(I73, I74, I75, I76, I77, I78, I79, I80) -> f3(I73, I74, I75, I76, I77, I78, I79, I80) f2(I81, I82, I83, I84, I85, I86, I87, I88) -> f4(I81, I82, I83, I84, 1 + I85, I86, I87, I88) f3(I89, I90, I91, I92, I93, I94, I95, I96) -> f1(I89, 1 + I90, I91, I92, I93, I94, I95, I96) [1 + I90 <= I92] f3(I97, I98, I99, I100, I101, I102, I103, I104) -> f1(I97, I98, I99, I100, I101, I102, I103, I104) [I100 <= I98] f1(I105, I106, I107, I108, I109, I110, I111, I112) -> f2(I105, I106, I107, I108, I109, I110, I111, I112) The dependency graph for this problem is: 3 -> Where: 3) f8#(I24, I25, I26, I27, I28, I29, I30, I31) -> f7#(I24, I25, I26, I27, I28, I29, I30, I31) We have the following SCCs.