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

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2) -> f4#(x1, x2) 
  1) f4#(I0, I1) -> f1#(I0, I1) [1 <= I0 /\ 1 + I0 + I1 <= 0]
  2) f3#(I2, I3) -> f1#(I2, I3) 
  3) f1#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I4]

We have the following SCCs.
  { 2, 3 }

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