NO

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

Initial linear ITS problem
   Start location: [0]
      0: [0] -> [1] : z'=1+z, [ z>=4 ], cost: 1
      1: [1] -> [2] : z'=1+z, [ x>=0 ], cost: 1
      2: [1] -> [2] : z'=-1+z, [ x<0 ], cost: 1
      3: [2] -> [3] : [ y>=2 ], cost: 1
      4: [3] -> [4] : [ x>=0 ], cost: 1
      7: [3] -> [6] : [ x<0 ], cost: 1
      5: [4] -> [5] : a'=z^2, b'=z*y, [ 0<=0 ], cost: 1
      6: [5] -> [3] : x'=-b+a, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      0: [0] -> [1] : z'=1+z, [ z>=4 ], cost: 1

Removed unreachable and leaf rules:
   Start location: [0]
      0: [0] -> [1] : z'=1+z, [ z>=4 ], cost: 1
      1: [1] -> [2] : z'=1+z, [ x>=0 ], cost: 1
      2: [1] -> [2] : z'=-1+z, [ x<0 ], cost: 1
      3: [2] -> [3] : [ y>=2 ], cost: 1
      4: [3] -> [4] : [ x>=0 ], cost: 1
      5: [4] -> [5] : a'=z^2, b'=z*y, [ 0<=0 ], cost: 1
      6: [5] -> [3] : x'=-b+a, [], cost: 1

Simplified all rules, resulting in:
   Start location: [0]
      0: [0] -> [1] : z'=1+z, [ z>=4 ], cost: 1
      1: [1] -> [2] : z'=1+z, [ x>=0 ], cost: 1
      2: [1] -> [2] : z'=-1+z, [ x<0 ], cost: 1
      3: [2] -> [3] : [ y>=2 ], cost: 1
      4: [3] -> [4] : [ x>=0 ], cost: 1
      5: [4] -> [5] : a'=z^2, b'=z*y, [], cost: 1
      6: [5] -> [3] : x'=-b+a, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: [0]
      0: [0] -> [1] : z'=1+z, [ z>=4 ], cost: 1
      1: [1] -> [2] : z'=1+z, [ x>=0 ], cost: 1
      2: [1] -> [2] : z'=-1+z, [ x<0 ], cost: 1
      3: [2] -> [3] : [ y>=2 ], cost: 1
      9: [3] -> [3] : x'=z^2-z*y, a'=z^2, b'=z*y, [ x>=0 ], cost: 3

Accelerating simple loops of location 3.
   Accelerating the following rules:
      9: [3] -> [3] : x'=z^2-z*y, a'=z^2, b'=z*y, [ x>=0 ], cost: 3

   Accelerated rule 9 with non-termination, yielding the new rule 10.
[accelerate] Nesting with 0 inner and 1 outer candidates

Accelerated all simple loops using metering functions (where possible):
   Start location: [0]
      0: [0] -> [1] : z'=1+z, [ z>=4 ], cost: 1
      1: [1] -> [2] : z'=1+z, [ x>=0 ], cost: 1
      2: [1] -> [2] : z'=-1+z, [ x<0 ], cost: 1
      3: [2] -> [3] : [ y>=2 ], cost: 1
      9: [3] -> [3] : x'=z^2-z*y, a'=z^2, b'=z*y, [ x>=0 ], cost: 3
     10: [3] -> [7] : [ x>=0 && z^2-z*y>=0 ], cost: NONTERM

Chained accelerated rules (with incoming rules):
   Start location: [0]
      0: [0] -> [1] : z'=1+z, [ z>=4 ], cost: 1
      1: [1] -> [2] : z'=1+z, [ x>=0 ], cost: 1
      2: [1] -> [2] : z'=-1+z, [ x<0 ], cost: 1
      3: [2] -> [3] : [ y>=2 ], cost: 1
     11: [2] -> [3] : x'=z^2-z*y, a'=z^2, b'=z*y, [ y>=2 && x>=0 ], cost: 4
     12: [2] -> [7] : [ y>=2 && x>=0 && z^2-z*y>=0 ], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: [0]
      0: [0] -> [1] : z'=1+z, [ z>=4 ], cost: 1
      1: [1] -> [2] : z'=1+z, [ x>=0 ], cost: 1
      2: [1] -> [2] : z'=-1+z, [ x<0 ], cost: 1
     12: [2] -> [7] : [ y>=2 && x>=0 && z^2-z*y>=0 ], cost: NONTERM

Eliminated locations (on tree-shaped paths):
   Start location: [0]
     13: [0] -> [2] : z'=2+z, [ z>=4 && x>=0 ], cost: 2
     14: [0] -> [2] : z'=z, [ z>=4 && x<0 ], cost: 2
     12: [2] -> [7] : [ y>=2 && x>=0 && z^2-z*y>=0 ], cost: NONTERM

Eliminated locations (on tree-shaped paths):
   Start location: [0]
     15: [0] -> [7] : [ z>=4 && x>=0 && y>=2 && x>=0 && (2+z)^2-y*(2+z)>=0 ], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: [0]
     15: [0] -> [7] : [ z>=4 && x>=0 && y>=2 && x>=0 && (2+z)^2-y*(2+z)>=0 ], cost: NONTERM

Computing asymptotic complexity for rule 15
   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:  [ z>=4 && x>=0 && y>=2 && x>=0 && (2+z)^2-y*(2+z)>=0 ]

NO