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


--------------------------------------------------------------------------------


MAYBE

DP problem for innermost termination.
P =
  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
  f4#(I0, I1, I2) -> f1#(I1, I1, I2) 
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f3#(-1 + I6, -1 + I7, 1 + I8) 
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f1(I1, I1, I2) 
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(-1 + I6, -1 + I7, 1 + I8) 
  f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
  1) f4#(I0, I1, I2) -> f1#(I1, I1, I2) 
  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  3) f1#(I6, I7, I8) -> f3#(-1 + I6, -1 + I7, 1 + I8) 

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f3#(-1 + I6, -1 + I7, 1 + I8) 
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f1(I1, I1, I2) 
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(-1 + I6, -1 + I7, 1 + I8) 
  f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10]