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

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

Added constraints to the guards to ensure costs are nonnegative:
   Start location: f0
      0: f0 -> f : [], cost: 0
      1: f -> f : x'=1+x, [ 0<x && y>=0 ], cost: y

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, [ 0<x && y>=0 ], cost: y

   Accelerated rule 1 with NONTERM, 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 -> [2] : [ 0<x && y>=1 ], cost: NONTERM

Chained accelerated rules (with incoming rules):
   Start location: f0
      0: f0 -> f : [], cost: 0
      3: f0 -> [2] : [ 0<x && y>=1 ], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: f0
      3: f0 -> [2] : [ 0<x && y>=1 ], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: f0
      3: f0 -> [2] : [ 0<x && y>=1 ], cost: NONTERM

Computing asymptotic complexity for rule 3
   Guard is satisfiable, yielding nontermination
   Resulting cost NONTERM has complexity: Nonterm

Found new complexity Nonterm.

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

NO