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

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

We have the following SCCs.
  { 1, 2 }

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

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z3 - 1 + -1 * z2

This gives the following inequalities:
  I1 <= I0 - 1 /\ I1 <= I2 - 1  ==>  I2 - 1 + -1 * I1 > I2 - 1 + -1 * (I1 + 1)  with  I2 - 1 + -1 * I1 >= 0
  I3 <= I4 /\ I4 <= I5 - 1  ==>  I5 - 1 + -1 * I4 >= I5 - 1 + -1 * I4

We remove all the strictly oriented dependency pairs.

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

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z2 + -1 * z1

This gives the following inequalities:
  I3 <= I4 /\ I4 <= I5 - 1  ==>  I4 + -1 * I3 > I4 + -1 * (I3 + 1)  with  I4 + -1 * I3 >= 0

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