/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#(-1 * I0 + 2, I2) [2 <= I0 - 1]
  f2#(I3, I4) -> f2#(-1 * I3 - 2, I5) [I3 <= -1 /\ I3 <= -1 - 1 /\ I3 <= -2 - 1]
  f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1]
  f2#(I9, I10) -> f2#(I9 + 2, I11) [-3 <= I9 - 1 /\ I9 <= -1]
  f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(-1 * I0 + 2, I2) [2 <= I0 - 1]
  f2(I3, I4) -> f2(-1 * I3 - 2, I5) [I3 <= -1 /\ I3 <= -1 - 1 /\ I3 <= -2 - 1]
  f2(I6, I7) -> f2(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1]
  f2(I9, I10) -> f2(I9 + 2, I11) [-3 <= I9 - 1 /\ I9 <= -1]
  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]

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

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

DP problem for innermost termination.
P =
  f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1]
  f2#(I9, I10) -> f2#(I9 + 2, I11) [-3 <= I9 - 1 /\ I9 <= -1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(-1 * I0 + 2, I2) [2 <= I0 - 1]
  f2(I3, I4) -> f2(-1 * I3 - 2, I5) [I3 <= -1 /\ I3 <= -1 - 1 /\ I3 <= -2 - 1]
  f2(I6, I7) -> f2(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1]
  f2(I9, I10) -> f2(I9 + 2, I11) [-3 <= I9 - 1 /\ I9 <= -1]
  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]