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

The dependency graph for this problem is:
  0 -> 1
  1 -> 4
  2 -> 4
  3 -> 2
  4 -> 3
Where:
  0) f5#(x1, x2) -> f4#(x1, x2) 
  1) f4#(I0, I1) -> f1#(I0, I1) [1 <= I1]
  2) f3#(I2, I3) -> f1#(I2, I3) 
  3) f2#(I4, I5) -> f3#(I4, I5) [I4 = I4]
  4) f1#(I6, I7) -> f2#(I6, -1 + I7) 

We have the following SCCs.
  { 2, 3, 4 }

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