/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files


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YES

DP problem for innermost termination.
P =
  f8#(x1, x2, x3, x4, x5, x6, x7, x8) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8) 
  f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7) 
  f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f7#(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  f5#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  f2#(I24, I25, I26, I27, I28, I29, I30, I31) -> f5#(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32]
  f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 
  f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
  f1#(I61, I62, I63, I64, I65, I66, I67, I68) -> f2#(I61, I62, I63, I64, I65, I66, I67, I68) 
R = 
  f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f2(I0, I1, I2, I3, I4, I5, I6, I7) 
  f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f7(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  f5(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  f2(I24, I25, I26, I27, I28, I29, I30, I31) -> f5(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32]
  f4(I34, I35, I36, I37, I38, I39, I40, I41) -> f2(I34, I35, I36, I37, I38, I39, I40, I41) 
  f2(I42, I43, I44, I45, I46, I47, I48, I49) -> f4(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
  f2(I53, I54, I55, I56, I57, I58, I59, I60) -> f3(I53, I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
  f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I61, I62, I63, I64, I65, I66, I67, I68) 

The dependency graph for this problem is:
  0 -> 7
  1 -> 4, 6
  2 -> 1
  3 -> 2
  4 -> 3
  5 -> 4, 6
  6 -> 5
  7 -> 4, 6
Where:
  0) f8#(x1, x2, x3, x4, x5, x6, x7, x8) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8) 
  1) f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7) 
  2) f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f7#(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  3) f5#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  4) f2#(I24, I25, I26, I27, I28, I29, I30, I31) -> f5#(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32]
  5) f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 
  6) f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
  7) f1#(I61, I62, I63, I64, I65, I66, I67, I68) -> f2#(I61, I62, I63, I64, I65, I66, I67, I68) 

We have the following SCCs.
  { 1, 2, 3, 4, 5, 6 }

DP problem for innermost termination.
P =
  f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7) 
  f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f7#(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  f5#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  f2#(I24, I25, I26, I27, I28, I29, I30, I31) -> f5#(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32]
  f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 
  f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
R = 
  f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f2(I0, I1, I2, I3, I4, I5, I6, I7) 
  f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f7(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  f5(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  f2(I24, I25, I26, I27, I28, I29, I30, I31) -> f5(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32]
  f4(I34, I35, I36, I37, I38, I39, I40, I41) -> f2(I34, I35, I36, I37, I38, I39, I40, I41) 
  f2(I42, I43, I44, I45, I46, I47, I48, I49) -> f4(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
  f2(I53, I54, I55, I56, I57, I58, I59, I60) -> f3(I53, I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
  f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I61, I62, I63, I64, I65, I66, I67, I68) 

We use the extended value criterion with the projection function NU:
  NU[f4#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 1
  NU[f5#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 2
  NU[f6#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 2
  NU[f2#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 1
  NU[f7#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 1

This gives the following inequalities:
    ==>  I6 - 1 >= I6 - 1
  y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2  ==>  I14 - 2 >= (-1 + I14) - 1
  I16 = I16  ==>  I22 - 2 >= I22 - 2
  1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32  ==>  I30 - 1 > I30 - 2  with  I30 - 1 >= 0
    ==>  I40 - 1 >= I40 - 1
  1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49  ==>  I48 - 1 >= I48 - 1

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7) 
  f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f7#(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  f5#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 
  f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
R = 
  f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f2(I0, I1, I2, I3, I4, I5, I6, I7) 
  f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f7(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  f5(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  f2(I24, I25, I26, I27, I28, I29, I30, I31) -> f5(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32]
  f4(I34, I35, I36, I37, I38, I39, I40, I41) -> f2(I34, I35, I36, I37, I38, I39, I40, I41) 
  f2(I42, I43, I44, I45, I46, I47, I48, I49) -> f4(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
  f2(I53, I54, I55, I56, I57, I58, I59, I60) -> f3(I53, I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
  f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I61, I62, I63, I64, I65, I66, I67, I68) 

The dependency graph for this problem is:
  1 -> 6
  2 -> 1
  3 -> 2
  5 -> 6
  6 -> 5
Where:
  1) f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7) 
  2) f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f7#(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  3) f5#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  5) f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 
  6) f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]

We have the following SCCs.
  { 5, 6 }

DP problem for innermost termination.
P =
  f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 
  f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
R = 
  f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f2(I0, I1, I2, I3, I4, I5, I6, I7) 
  f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f7(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  f5(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  f2(I24, I25, I26, I27, I28, I29, I30, I31) -> f5(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32]
  f4(I34, I35, I36, I37, I38, I39, I40, I41) -> f2(I34, I35, I36, I37, I38, I39, I40, I41) 
  f2(I42, I43, I44, I45, I46, I47, I48, I49) -> f4(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
  f2(I53, I54, I55, I56, I57, I58, I59, I60) -> f3(I53, I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
  f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I61, I62, I63, I64, I65, I66, I67, I68) 

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3,z4,z5,z6,z7,z8)] = z8
  NU[f4#(z1,z2,z3,z4,z5,z6,z7,z8)] = z8

This gives the following inequalities:
    ==>  I41 (>! \union =) I41
  1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49  ==>  I49 >! -1 + I49

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 
R = 
  f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f2(I0, I1, I2, I3, I4, I5, I6, I7) 
  f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f7(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2]
  f5(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16]
  f2(I24, I25, I26, I27, I28, I29, I30, I31) -> f5(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32]
  f4(I34, I35, I36, I37, I38, I39, I40, I41) -> f2(I34, I35, I36, I37, I38, I39, I40, I41) 
  f2(I42, I43, I44, I45, I46, I47, I48, I49) -> f4(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49]
  f2(I53, I54, I55, I56, I57, I58, I59, I60) -> f3(I53, I54, I57, I56, I57, I58, I59, I60) [I59 <= 0]
  f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I61, I62, I63, I64, I65, I66, I67, I68) 

The dependency graph for this problem is:
  5 -> 
Where:
  5) f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 

We have the following SCCs.