/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 =
  f5#(x1, x2) -> f1#(x1, x2) 
  f2#(I2, I3) -> f3#(I2, rnd2) [rnd2 = rnd2]
  f3#(I4, I5) -> f2#(1 + I4, I5) [1 <= I5]
  f3#(I6, I7) -> f2#(-1 + I6, I7) [I7 <= 0]
  f1#(I8, I9) -> f2#(2, I9) 
R = 
  f5(x1, x2) -> f1(x1, x2) 
  f2(I0, I1) -> f4(I0, I1) [2 <= 0]
  f2(I2, I3) -> f3(I2, rnd2) [rnd2 = rnd2]
  f3(I4, I5) -> f2(1 + I4, I5) [1 <= I5]
  f3(I6, I7) -> f2(-1 + I6, I7) [I7 <= 0]
  f1(I8, I9) -> f2(2, I9) 

The dependency graph for this problem is:
  0 -> 4
  1 -> 2, 3
  2 -> 1
  3 -> 1
  4 -> 1
Where:
  0) f5#(x1, x2) -> f1#(x1, x2) 
  1) f2#(I2, I3) -> f3#(I2, rnd2) [rnd2 = rnd2]
  2) f3#(I4, I5) -> f2#(1 + I4, I5) [1 <= I5]
  3) f3#(I6, I7) -> f2#(-1 + I6, I7) [I7 <= 0]
  4) f1#(I8, I9) -> f2#(2, I9) 

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

DP problem for innermost termination.
P =
  f2#(I2, I3) -> f3#(I2, rnd2) [rnd2 = rnd2]
  f3#(I4, I5) -> f2#(1 + I4, I5) [1 <= I5]
  f3#(I6, I7) -> f2#(-1 + I6, I7) [I7 <= 0]
R = 
  f5(x1, x2) -> f1(x1, x2) 
  f2(I0, I1) -> f4(I0, I1) [2 <= 0]
  f2(I2, I3) -> f3(I2, rnd2) [rnd2 = rnd2]
  f3(I4, I5) -> f2(1 + I4, I5) [1 <= I5]
  f3(I6, I7) -> f2(-1 + I6, I7) [I7 <= 0]
  f1(I8, I9) -> f2(2, I9)