/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) -> f3#(x1, x2, x3) 
  f3#(I0, I1, I2) -> f1#(I0, I1, I2) 
  f2#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f2#(I6, I7, I6 + I8) [1 + I7 <= I6 + I8]
R = 
  f4(x1, x2, x3) -> f3(x1, x2, x3) 
  f3(I0, I1, I2) -> f1(I0, I1, I2) 
  f2(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f2(I6, I7, I6 + I8) [1 + I7 <= I6 + I8]

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

We have the following SCCs.
  { 2, 3 }

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