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


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MAYBE

DP problem for innermost termination.
P =
  init#(x1, x2) -> f1#(rnd1, rnd2) 
  f3#(I0, I1) -> f3#(I0 + 1, I1 - 2) [1 <= I1 - 1]
  f3#(I2, I3) -> f2#(I2, I4) [I3 <= 1]
  f2#(I5, I6) -> f3#(0, I5) [1 <= I5 - 1]
  f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f3(I0 + 1, I1 - 2) [1 <= I1 - 1]
  f3(I2, I3) -> f2(I2, I4) [I3 <= 1]
  f2(I5, I6) -> f3(0, I5) [1 <= I5 - 1]
  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]

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

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

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f3#(I0 + 1, I1 - 2) [1 <= I1 - 1]
  f3#(I2, I3) -> f2#(I2, I4) [I3 <= 1]
  f2#(I5, I6) -> f3#(0, I5) [1 <= I5 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f3(I0 + 1, I1 - 2) [1 <= I1 - 1]
  f3(I2, I3) -> f2(I2, I4) [I3 <= 1]
  f2(I5, I6) -> f3(0, I5) [1 <= I5 - 1]
  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]