/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, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  f3#(I0, I1, I2, I3) -> f3#(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
  f3#(I4, I5, I6, I7) -> f2#(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7]
  f2#(I9, I10, I11, I12) -> f3#(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1]
  f1#(I13, I14, I15, I16) -> f2#(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f3(I0, I1, I2, I3) -> f3(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
  f3(I4, I5, I6, I7) -> f2(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7]
  f2(I9, I10, I11, I12) -> f3(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1]
  f1(I13, I14, I15, I16) -> f2(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1]

The dependency graph for this problem is:
  0 -> 4
  1 -> 1, 2
  2 -> 
  3 -> 1
  4 -> 3
Where:
  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  1) f3#(I0, I1, I2, I3) -> f3#(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
  2) f3#(I4, I5, I6, I7) -> f2#(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7]
  3) f2#(I9, I10, I11, I12) -> f3#(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1]
  4) f1#(I13, I14, I15, I16) -> f2#(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1]

We have the following SCCs.
  { 1 }

DP problem for innermost termination.
P =
  f3#(I0, I1, I2, I3) -> f3#(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f3(I0, I1, I2, I3) -> f3(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1]
  f3(I4, I5, I6, I7) -> f2(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7]
  f2(I9, I10, I11, I12) -> f3(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1]
  f1(I13, I14, I15, I16) -> f2(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1]

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

This gives the following inequalities:
  I1 = I2 /\ I3 <= I1 - 1  ==>  I1 - 1 + -1 * I3 > I1 - 1 - 1 + -1 * I3  with  I1 - 1 + -1 * I3 >= 0

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