/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) 
  f2#(I0, I1, I2) -> f2#(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
  f2#(I4, I5, I6) -> f2#(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
  f1#(I7, I8, I9) -> f2#(I10, I11, 2) [0 <= I7 - 1 /\ -1 <= I11 - 1 /\ -1 <= I8 - 1 /\ -1 <= I10 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
  f2(I4, I5, I6) -> f2(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
  f1(I7, I8, I9) -> f2(I10, I11, 2) [0 <= I7 - 1 /\ -1 <= I11 - 1 /\ -1 <= I8 - 1 /\ -1 <= I10 - 1]

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

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

DP problem for innermost termination.
P =
  f2#(I4, I5, I6) -> f2#(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
  f2(I4, I5, I6) -> f2(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
  f1(I7, I8, I9) -> f2(I10, I11, 2) [0 <= I7 - 1 /\ -1 <= I11 - 1 /\ -1 <= I8 - 1 /\ -1 <= I10 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z1

This gives the following inequalities:
  -1 = I5 /\ -1 <= I4 - 1  ==>  I4 >! I4 - 1

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

DP problem for innermost termination.
P =
  f2#(I0, I1, I2) -> f2#(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
  f2(I4, I5, I6) -> f2(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
  f1(I7, I8, I9) -> f2(I10, I11, 2) [0 <= I7 - 1 /\ -1 <= I11 - 1 /\ -1 <= I8 - 1 /\ -1 <= I10 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z2

This gives the following inequalities:
  -1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1  ==>  I1 >! I1 - 1

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