### Pre-processing the ITS problem ###

Initial linear ITS problem
   Start location: f0
      0: f0 -> f : [], cost: 0
      1: f -> f : x'=-1+x, f : x'=x-y, [ x>0 && y>x ], cost: 1

Checking for constant complexity:
   Could not prove constant complexity.

### Simplification by acceleration and chaining ###

Accelerating simple loops of location 1.
   Accelerating the following rules:
      1: f -> f : x'=-1+x, f : x'=x-y, [ x>0 && y>x ], cost: 1

   Accelerated rule 1 with metering function x (after partial deletion), yielding the new rule 2.
   Removing the simple loops: 1.

Accelerated all simple loops using metering functions (where possible):
   Start location: f0
      0: f0 -> f : [], cost: 0
      2: f -> f : x'=0, [ x>0 && y>x ], cost: x

Chained accelerated rules (with incoming rules):
   Start location: f0
      0: f0 -> f : [], cost: 0
      3: f0 -> f : x'=0, [ x>0 && y>x ], cost: x

Removed unreachable locations (and leaf rules with constant cost):
   Start location: f0
      3: f0 -> f : x'=0, [ x>0 && y>x ], cost: x

### Obtained a tail recursive problem, continuing simplification ###

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: f0
      3: f0 -> f : x'=0, [ x>0 && y>x ], cost: x

Computing asymptotic complexity for rule 3
   Solved the limit problem by the following transformations:
      Created initial limit problem:
      x (+), -x+y (+/+!) [not solved]

      removing all constraints (solved by SMT)
      resulting limit problem: [solved]

      applying transformation rule (C) using substitution {x==n,y==2*n}
      resulting limit problem:
      [solved]

   Solution:
      x / n
      y / 2*n
   Resulting cost n has complexity: Poly(n^1)

Found new complexity Poly(n^1).

Obtained the following overall complexity (w.r.t. the length of the input n):
   Complexity:  Poly(n^1)
   Cpx degree:  1
   Solved cost: n
   Rule cost:   x
   Rule guard:  [ x>0 && y>x ]

WORST_CASE(Omega(n^1),?)