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

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

We have the following SCCs.
  { 1 }

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

We use the basic value criterion with the projection function NU:
  NU[f3#(z1,z2,z3)] = z3

This gives the following inequalities:
  I1 = I2 /\ 0 <= I1 - 1  ==>  I2 >! I1 - 1

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