/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 =
  init#(x1, x2) -> f1#(rnd1, rnd2) 
  f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
  f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]
  f2#(I5, I6) -> f3#(I5, 3) [I5 <= 9]
  f1#(I7, I8) -> f2#(0, I9) 
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11]
  f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1]
  f2(I5, I6) -> f3(I5, 3) [I5 <= 9]
  f1(I7, I8) -> f2(0, I9) 

The dependency graph for this problem is:
  0 -> 4
  1 -> 1, 2
  2 -> 3
  3 -> 1
  4 -> 3
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
  2) f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]
  3) f2#(I5, I6) -> f3#(I5, 3) [I5 <= 9]
  4) f1#(I7, I8) -> f2#(0, I9) 

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

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
  f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]
  f2#(I5, I6) -> f3#(I5, 3) [I5 <= 9]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11]
  f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1]
  f2(I5, I6) -> f3(I5, 3) [I5 <= 9]
  f1(I7, I8) -> f2(0, I9) 

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2)] = 9 + -1 * z1
  NU[f3#(z1,z2)] = 9 + -1 * (z1 + 1)

This gives the following inequalities:
  I1 <= 11  ==>  9 + -1 * (I0 + 1) >= 9 + -1 * (I0 + 1)
  11 <= I3 - 1  ==>  9 + -1 * (I2 + 1) >= 9 + -1 * (I2 + 1)
  I5 <= 9  ==>  9 + -1 * I5 > 9 + -1 * (I5 + 1)  with  9 + -1 * I5 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
  f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11]
  f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1]
  f2(I5, I6) -> f3(I5, 3) [I5 <= 9]
  f1(I7, I8) -> f2(0, I9) 

The dependency graph for this problem is:
  1 -> 1, 2
  2 -> 
Where:
  1) f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
  2) f3#(I2, I3) -> f2#(I2 + 1, I4) [11 <= I3 - 1]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f3#(I0, I1 + 1) [I1 <= 11]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f3(I0, I1 + 1) [I1 <= 11]
  f3(I2, I3) -> f2(I2 + 1, I4) [11 <= I3 - 1]
  f2(I5, I6) -> f3(I5, 3) [I5 <= 9]
  f1(I7, I8) -> f2(0, I9) 

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1,z2)] = 11 + -1 * z2

This gives the following inequalities:
  I1 <= 11  ==>  11 + -1 * I1 > 11 + -1 * (I1 + 1)  with  11 + -1 * I1 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.