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


--------------------------------------------------------------------------------


MAYBE

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

The dependency graph for this problem is:
  0 -> 7
  1 -> 6
  2 -> 
  3 -> 
  4 -> 
  5 -> 6
  6 -> 1, 5
  7 -> 1, 3
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f2#(I0, I1) -> f2#(I0 + 1, -1 * I1 - 1) [0 - I0 <= I1 /\ 0 <= I1 - 1 /\ I0 <= I1 - 1]
  2) f2#(I2, I3) -> f2#(I2 + 1, -1 * I3 - 1) [0 - I2 <= I3 /\ I3 <= -1 /\ I2 <= I3 - 1]
  3) f2#(I4, I5) -> f2#(I4 + 1, 0) [0 - I4 <= I5 /\ 0 <= I5 - 1 /\ I5 <= I4]
  4) f2#(I6, I7) -> f2#(I6 + 1, 0) [0 - I6 <= I7 /\ I7 <= -1 /\ I7 <= I6]
  5) f2#(I8, I9) -> f2#(I8 + 1, -1 * I9 + 1) [I9 <= 0 - I8 - 1 /\ 0 <= I9 - 1]
  6) f2#(I10, I11) -> f2#(I10 + 1, -1 * I11 + 1) [I11 <= 0 - I10 - 1 /\ I11 <= -1]
  7) f1#(I12, I13) -> f2#(5, I13) [-1 <= I13 - 1 /\ 0 <= I12 - 1]

We have the following SCCs.
  { 1, 5, 6 }

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