/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, x3) -> f1#(rnd1, rnd2, rnd3) 
  f2#(I0, I1, I2) -> f2#(I3, I4, I2) [0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I4 <= I0 /\ I3 <= I1 /\ I3 <= I0]
  f2#(I5, I6, I7) -> f2#(I8, I9, I7) [0 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 <= I6 /\ I9 <= I5 /\ I8 <= I6 /\ I8 <= I5]
  f1#(I10, I11, I12) -> f2#(I13, I14, 1) [1 = I11 /\ 0 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 <= I10 /\ I13 <= I10]
  f1#(I15, I16, I17) -> f2#(I18, I19, 2) [2 = I16 /\ 0 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 <= I15 /\ I18 <= I15]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(I3, I4, I2) [0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I4 <= I0 /\ I3 <= I1 /\ I3 <= I0]
  f2(I5, I6, I7) -> f2(I8, I9, I7) [0 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 <= I6 /\ I9 <= I5 /\ I8 <= I6 /\ I8 <= I5]
  f1(I10, I11, I12) -> f2(I13, I14, 1) [1 = I11 /\ 0 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 <= I10 /\ I13 <= I10]
  f1(I15, I16, I17) -> f2(I18, I19, 2) [2 = I16 /\ 0 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 <= I15 /\ I18 <= I15]

The dependency graph for this problem is:
  0 -> 3, 4
  1 -> 1, 2
  2 -> 1, 2
  3 -> 1, 2
  4 -> 1, 2
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f2#(I0, I1, I2) -> f2#(I3, I4, I2) [0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I4 <= I0 /\ I3 <= I1 /\ I3 <= I0]
  2) f2#(I5, I6, I7) -> f2#(I8, I9, I7) [0 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 <= I6 /\ I9 <= I5 /\ I8 <= I6 /\ I8 <= I5]
  3) f1#(I10, I11, I12) -> f2#(I13, I14, 1) [1 = I11 /\ 0 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 <= I10 /\ I13 <= I10]
  4) f1#(I15, I16, I17) -> f2#(I18, I19, 2) [2 = I16 /\ 0 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 <= I15 /\ I18 <= I15]

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f2#(I0, I1, I2) -> f2#(I3, I4, I2) [0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I4 <= I0 /\ I3 <= I1 /\ I3 <= I0]
  f2#(I5, I6, I7) -> f2#(I8, I9, I7) [0 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 <= I6 /\ I9 <= I5 /\ I8 <= I6 /\ I8 <= I5]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(I3, I4, I2) [0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I4 <= I0 /\ I3 <= I1 /\ I3 <= I0]
  f2(I5, I6, I7) -> f2(I8, I9, I7) [0 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 <= I6 /\ I9 <= I5 /\ I8 <= I6 /\ I8 <= I5]
  f1(I10, I11, I12) -> f2(I13, I14, 1) [1 = I11 /\ 0 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 <= I10 /\ I13 <= I10]
  f1(I15, I16, I17) -> f2(I18, I19, 2) [2 = I16 /\ 0 <= I19 - 1 /\ 0 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 <= I15 /\ I18 <= I15]