/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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YES

DP problem for innermost termination.
P =
  f5#(x1, x2, x3, x4, x5, x6, x7, x8) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8) 
  f4#(I0, I1, I2, I3, I4, I5, I6, I7) -> f1#(I0, I1, I2, I3, I4, I5, I2, I7) 
  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 
  f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f3#(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
R = 
  f5(x1, x2, x3, x4, x5, x6, x7, x8) -> f4(x1, x2, x3, x4, x5, x6, x7, x8) 
  f4(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, I3, I4, I5, I2, I7) 
  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, I11, I12, I13, I14, I15) 
  f1(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
  f1(I24, I25, I26, I27, I28, I29, I30, I31) -> f2(rnd1, rnd2, I26, I27, I28, 0, I30, I31) [rnd1 = rnd2 /\ rnd2 = 0 /\ 1 - I27 + I30 <= 0 /\ 0 <= I28 - I31]
  f1(I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I40, I41, I34, I35, I36, 0, I38, I39) [I40 = I41 /\ I41 = 0 /\ 1 + I36 - I39 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2, x3, x4, x5, x6, x7, x8) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8) 
  1) f4#(I0, I1, I2, I3, I4, I5, I6, I7) -> f1#(I0, I1, I2, I3, I4, I5, I2, I7) 
  2) f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 
  3) f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f3#(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 
  f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f3#(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
R = 
  f5(x1, x2, x3, x4, x5, x6, x7, x8) -> f4(x1, x2, x3, x4, x5, x6, x7, x8) 
  f4(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, I3, I4, I5, I2, I7) 
  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, I11, I12, I13, I14, I15) 
  f1(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
  f1(I24, I25, I26, I27, I28, I29, I30, I31) -> f2(rnd1, rnd2, I26, I27, I28, 0, I30, I31) [rnd1 = rnd2 /\ rnd2 = 0 /\ 1 - I27 + I30 <= 0 /\ 0 <= I28 - I31]
  f1(I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I40, I41, I34, I35, I36, 0, I38, I39) [I40 = I41 /\ I41 = 0 /\ 1 + I36 - I39 <= 0]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8)] = -1 * z4 + z7 + -1 * 0
  NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8)] = -1 * z4 + z7 + -1 * 0

This gives the following inequalities:
    ==>  -1 * I11 + I14 + -1 * 0 >= -1 * I11 + I14 + -1 * 0
  0 <= -1 * I19 + I22 /\ 0 <= I20 - I23  ==>  -1 * I19 + I22 + -1 * 0 > -1 * I19 + (-1 + I22) + -1 * 0  with  -1 * I19 + I22 + -1 * 0 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 
R = 
  f5(x1, x2, x3, x4, x5, x6, x7, x8) -> f4(x1, x2, x3, x4, x5, x6, x7, x8) 
  f4(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, I3, I4, I5, I2, I7) 
  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, I11, I12, I13, I14, I15) 
  f1(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I16, I17, I18, I19, I20, I21, -1 + I22, I23) [0 <= -1 * I19 + I22 /\ 0 <= I20 - I23]
  f1(I24, I25, I26, I27, I28, I29, I30, I31) -> f2(rnd1, rnd2, I26, I27, I28, 0, I30, I31) [rnd1 = rnd2 /\ rnd2 = 0 /\ 1 - I27 + I30 <= 0 /\ 0 <= I28 - I31]
  f1(I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I40, I41, I34, I35, I36, 0, I38, I39) [I40 = I41 /\ I41 = 0 /\ 1 + I36 - I39 <= 0]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 

We have the following SCCs.