/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) 
  f3#(I0, I1) -> f2#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
  f3#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0]
  f2#(I6, I7) -> f3#(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1]
  f2#(I9, I10) -> f3#(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0]
  f1#(I12, I13) -> f2#(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
  f3(I3, I4) -> f2(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0]
  f2(I6, I7) -> f3(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1]
  f2(I9, I10) -> f3(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0]
  f1(I12, I13) -> f2(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1]

The dependency graph for this problem is:
  0 -> 5
  1 -> 3
  2 -> 4
  3 -> 1
  4 -> 2
  5 -> 3, 4
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f3#(I0, I1) -> f2#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
  2) f3#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0]
  3) f2#(I6, I7) -> f3#(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1]
  4) f2#(I9, I10) -> f3#(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0]
  5) f1#(I12, I13) -> f2#(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1]

We have the following SCCs.
  { 2, 4 }
  { 1, 3 }

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f2#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
  f2#(I6, I7) -> f3#(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
  f3(I3, I4) -> f2(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0]
  f2(I6, I7) -> f3(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1]
  f2(I9, I10) -> f3(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0]
  f1(I12, I13) -> f2(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2)] = z1
  NU[f3#(z1,z2)] = z1

This gives the following inequalities:
  I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1  ==>  I0 >! I0 - 1
  I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1  ==>  I6 >! I6 - 1

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

DP problem for innermost termination.
P =
  f3#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0]
  f2#(I9, I10) -> f3#(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
  f3(I3, I4) -> f2(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0]
  f2(I6, I7) -> f3(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1]
  f2(I9, I10) -> f3(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0]
  f1(I12, I13) -> f2(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1]