/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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MAYBE

DP problem for innermost termination.
P =
  init#(x1, x2) -> f1#(rnd1, rnd2) 
  f2#(I0, I1) -> f2#(20, I2) [30 <= I0 - 1]
  f2#(I3, I4) -> f2#(29, I5) [25 = I3]
  f2#(I6, I7) -> f2#(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30]
  f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
  f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(20, I2) [30 <= I0 - 1]
  f2(I3, I4) -> f2(29, I5) [25 = I3]
  f2(I6, I7) -> f2(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30]
  f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]

The dependency graph for this problem is:
  0 -> 5
  1 -> 4
  2 -> 3
  3 -> 2, 3
  4 -> 4
  5 -> 1, 2, 3, 4
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f2#(I0, I1) -> f2#(20, I2) [30 <= I0 - 1]
  2) f2#(I3, I4) -> f2#(29, I5) [25 = I3]
  3) f2#(I6, I7) -> f2#(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30]
  4) f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
  5) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]

We have the following SCCs.
  { 2, 3 }
  { 4 }

DP problem for innermost termination.
P =
  f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(20, I2) [30 <= I0 - 1]
  f2(I3, I4) -> f2(29, I5) [25 = I3]
  f2(I6, I7) -> f2(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30]
  f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]

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

This gives the following inequalities:
  10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30  ==>  I9 >! I9 - 1

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

DP problem for innermost termination.
P =
  f2#(I3, I4) -> f2#(29, I5) [25 = I3]
  f2#(I6, I7) -> f2#(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(20, I2) [30 <= I0 - 1]
  f2(I3, I4) -> f2(29, I5) [25 = I3]
  f2(I6, I7) -> f2(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30]
  f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]