/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files


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MAYBE

DP problem for innermost termination.
P =
  min#(I2, I3) -> if#(I2 < I3, I2, I3) 
  cond#(true, x, y) -> minus#(x, y + 1) 
  minus#(I4, I5) -> cond#(min(I4, I5) = I5, I4, I5) 
  minus#(I4, I5) -> min#(I4, I5) 
R = 
  if(false, u, v) -> v 
  if(true, I0, I1) -> I0 
  min(I2, I3) -> if(I2 < I3, I2, I3) 
  cond(true, x, y) -> 1 + minus(x, y + 1) 
  minus(I4, I5) -> cond(min(I4, I5) = I5, I4, I5) 
  minus(I6, I6) -> 0 

The dependency graph for this problem is:
  0 -> 
  1 -> 2, 3
  2 -> 1
  3 -> 0
Where:
  0) min#(I2, I3) -> if#(I2 < I3, I2, I3) 
  1) cond#(true, x, y) -> minus#(x, y + 1) 
  2) minus#(I4, I5) -> cond#(min(I4, I5) = I5, I4, I5) 
  3) minus#(I4, I5) -> min#(I4, I5) 

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  cond#(true, x, y) -> minus#(x, y + 1) 
  minus#(I4, I5) -> cond#(min(I4, I5) = I5, I4, I5) 
R = 
  if(false, u, v) -> v 
  if(true, I0, I1) -> I0 
  min(I2, I3) -> if(I2 < I3, I2, I3) 
  cond(true, x, y) -> 1 + minus(x, y + 1) 
  minus(I4, I5) -> cond(min(I4, I5) = I5, I4, I5) 
  minus(I6, I6) -> 0