/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) -> f4#(x1, x2, x3, x4, x5) 
  f4#(I0, I1, I2, I3, I4) -> f3#(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  f1#(I10, I11, I12, I13, I14) -> f3#(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
R = 
  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f3(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f1(I10, I11, I12, I13, I14) -> f3(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) [I16 <= I15]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
  1) f4#(I0, I1, I2, I3, I4) -> f3#(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  3) f1#(I10, I11, I12, I13, I14) -> f3#(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  f1#(I10, I11, I12, I13, I14) -> f3#(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
R = 
  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f3(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f1(I10, I11, I12, I13, I14) -> f3(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) [I16 <= I15]

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

This gives the following inequalities:
    ==>  I6 + -1 * (1 + I5) >= I6 + -1 * (1 + I5)
  rnd5 = rnd5 /\ 1 + I10 <= I11  ==>  I11 + -1 * (1 + I10) > I11 + -1 * (1 + (1 + I10))  with  I11 + -1 * (1 + I10) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
R = 
  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f3(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f1(I10, I11, I12, I13, I14) -> f3(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11]
  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) [I16 <= I15]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 

We have the following SCCs.