/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 =
  trunc#(x) -> if#(x % 2 = 0, x, x - 1) 
  fNat#(true, I2, y) -> f#(I2 > y, trunc(I2), y + 1) 
  fNat#(true, I2, y) -> trunc#(I2) 
  f#(true, I3, I4) -> fNat#(I3 >= 0 && I4 >= 0, I3, I4) 
R = 
  if(false, u, v) -> v 
  if(true, I0, I1) -> I0 
  trunc(x) -> if(x % 2 = 0, x, x - 1) 
  fNat(true, I2, y) -> f(I2 > y, trunc(I2), y + 1) 
  f(true, I3, I4) -> fNat(I3 >= 0 && I4 >= 0, I3, I4) 

This problem is converted using chaining, where edges between chained DPs are removed.

DP problem for innermost termination.
P =
  trunc#(x) -> if#(x % 2 = 0, x, x - 1) 
  fNat#(true, I2, y) -> f#(I2 > y, trunc(I2), y + 1) 
  fNat#(true, I2, y) -> trunc#(I2) 
  f#(true, I3, I4) -> f_1#(I3, I4) 
  f_1#(I3, I4) -> fNat#(I3 >= 0 && I4 >= 0, I3, I4) 
  f_1#(I3, I4) -> trunc#(I3) [I3 >= 0 && I4 >= 0]
  f_1#(I3, I4) -> f#(I3 > I4, trunc(I3), I4 + 1) [I3 >= 0 && I4 >= 0]
  f_1#(I3, I4) -> f_1#(trunc(I3), I4 + 1) [I3 >= 0 && I4 >= 0, I3 > I4]
  fNat#(true, I2, y) -> f_1#(trunc(I2), y + 1) [I2 > y]
R = 
  if(false, u, v) -> v 
  if(true, I0, I1) -> I0 
  trunc(x) -> if(x % 2 = 0, x, x - 1) 
  fNat(true, I2, y) -> f(I2 > y, trunc(I2), y + 1) 
  f(true, I3, I4) -> fNat(I3 >= 0 && I4 >= 0, I3, I4) 

The dependency graph for this problem is:
  0 -> 
  1 -> 
  2 -> 0
  3 -> 7, 6, 5, 4
  4 -> 
  5 -> 0
  6 -> 
  7 -> 7, 4, 5, 6
  8 -> 4, 5, 6, 7
Where:
  0) trunc#(x) -> if#(x % 2 = 0, x, x - 1) 
  1) fNat#(true, I2, y) -> f#(I2 > y, trunc(I2), y + 1) 
  2) fNat#(true, I2, y) -> trunc#(I2) 
  3) f#(true, I3, I4) -> f_1#(I3, I4) 
  4) f_1#(I3, I4) -> fNat#(I3 >= 0 && I4 >= 0, I3, I4) 
  5) f_1#(I3, I4) -> trunc#(I3) [I3 >= 0 && I4 >= 0]
  6) f_1#(I3, I4) -> f#(I3 > I4, trunc(I3), I4 + 1) [I3 >= 0 && I4 >= 0]
  7) f_1#(I3, I4) -> f_1#(trunc(I3), I4 + 1) [I3 >= 0 && I4 >= 0, I3 > I4]
  8) fNat#(true, I2, y) -> f_1#(trunc(I2), y + 1) [I2 > y]

We have the following SCCs.
  { 7 }

DP problem for innermost termination.
P =
  f_1#(I3, I4) -> f_1#(trunc(I3), I4 + 1) [I3 >= 0 && I4 >= 0, I3 > I4]
R = 
  if(false, u, v) -> v 
  if(true, I0, I1) -> I0 
  trunc(x) -> if(x % 2 = 0, x, x - 1) 
  fNat(true, I2, y) -> f(I2 > y, trunc(I2), y + 1) 
  f(true, I3, I4) -> fNat(I3 >= 0 && I4 >= 0, I3, I4)