/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 =
  f4#(x1, x2, x3, x4) -> f3#(x1, x2, x3, x4) 
  f3#(I0, I1, I2, I3) -> f1#(0, I1, rnd3, I3) [rnd3 = rnd3]
  f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
  f1#(I8, I9, I10, I11) -> f2#(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]
R = 
  f4(x1, x2, x3, x4) -> f3(x1, x2, x3, x4) 
  f3(I0, I1, I2, I3) -> f1(0, I1, rnd3, I3) [rnd3 = rnd3]
  f2(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
  f1(I8, I9, I10, I11) -> f2(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f4#(x1, x2, x3, x4) -> f3#(x1, x2, x3, x4) 
  1) f3#(I0, I1, I2, I3) -> f1#(0, I1, rnd3, I3) [rnd3 = rnd3]
  2) f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
  3) f1#(I8, I9, I10, I11) -> f2#(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f2#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
  f1#(I8, I9, I10, I11) -> f2#(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]
R = 
  f4(x1, x2, x3, x4) -> f3(x1, x2, x3, x4) 
  f3(I0, I1, I2, I3) -> f1(0, I1, rnd3, I3) [rnd3 = rnd3]
  f2(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
  f1(I8, I9, I10, I11) -> f2(1 + I8, rnd2, I10, I8 + I10) [rnd2 = I9 - (I8 + I10) /\ 1 <= I9]