NO

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

Initial linear ITS problem
   Start location: l3
      0: l0 -> l1 : __const_1000^0'=__const_1000^post_1, __const_200^0'=__const_200^post_1, x^0'=x^post_1, [ x^post_1==x^0+__const_1000^0 && 1+__const_200^0<=x^post_1 && __const_1000^0==__const_1000^post_1 && __const_200^0==__const_200^post_1 ], cost: 1
      1: l1 -> l0 : __const_1000^0'=__const_1000^post_2, __const_200^0'=__const_200^post_2, x^0'=x^post_2, [ __const_1000^0==__const_1000^post_2 && __const_200^0==__const_200^post_2 && x^0==x^post_2 ], cost: 1
      2: l2 -> l0 : __const_1000^0'=__const_1000^post_3, __const_200^0'=__const_200^post_3, x^0'=x^post_3, [ __const_1000^0==__const_1000^post_3 && __const_200^0==__const_200^post_3 && x^0==x^post_3 ], cost: 1
      3: l3 -> l2 : __const_1000^0'=__const_1000^post_4, __const_200^0'=__const_200^post_4, x^0'=x^post_4, [ __const_1000^0==__const_1000^post_4 && __const_200^0==__const_200^post_4 && x^0==x^post_4 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      3: l3 -> l2 : __const_1000^0'=__const_1000^post_4, __const_200^0'=__const_200^post_4, x^0'=x^post_4, [ __const_1000^0==__const_1000^post_4 && __const_200^0==__const_200^post_4 && x^0==x^post_4 ], cost: 1

Simplified all rules, resulting in:
   Start location: l3
      0: l0 -> l1 : x^0'=x^0+__const_1000^0, [ 1+__const_200^0<=x^0+__const_1000^0 ], cost: 1
      1: l1 -> l0 : [], cost: 1
      2: l2 -> l0 : [], cost: 1
      3: l3 -> l2 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l3
      5: l0 -> l0 : x^0'=x^0+__const_1000^0, [ 1+__const_200^0<=x^0+__const_1000^0 ], cost: 2
      4: l3 -> l0 : [], cost: 2

Accelerating simple loops of location 0.
   Accelerating the following rules:
      5: l0 -> l0 : x^0'=x^0+__const_1000^0, [ 1+__const_200^0<=x^0+__const_1000^0 ], cost: 2

[test] deduced pseudo-invariant -__const_1000^0<=0, also trying __const_1000^0<=-1
   Accelerated rule 5 with non-termination, yielding the new rule 6.
   Accelerated rule 5 with non-termination, yielding the new rule 7.
   Accelerated rule 5 with backward acceleration, yielding the new rule 8.
   Accelerated rule 5 with backward acceleration, yielding the new rule 9.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Also removing duplicate rules: 7.

Accelerated all simple loops using metering functions (where possible):
   Start location: l3
      5: l0 -> l0 : x^0'=x^0+__const_1000^0, [ 1+__const_200^0<=x^0+__const_1000^0 ], cost: 2
      6: l0 -> [4] : [ 1+__const_200^0<=x^0+__const_1000^0 && __const_200^0==0 && x^0==1 && __const_1000^0==0 ], cost: NONTERM
      8: l0 -> [4] : [ 1+__const_200^0<=x^0+__const_1000^0 && -__const_1000^0<=0 ], cost: NONTERM
      9: l0 -> l0 : x^0'=x^0+k*__const_1000^0, [ __const_1000^0<=-1 && k>=0 && 1+__const_200^0<=(-1+k)*__const_1000^0+x^0+__const_1000^0 ], cost: 2*k
      4: l3 -> l0 : [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l3
      4: l3 -> l0 : [], cost: 2
     10: l3 -> l0 : x^0'=x^0+__const_1000^0, [ 1+__const_200^0<=x^0+__const_1000^0 ], cost: 4
     11: l3 -> [4] : [ 1+__const_200^0<=x^0+__const_1000^0 && __const_200^0==0 && x^0==1 && __const_1000^0==0 ], cost: NONTERM
     12: l3 -> [4] : [ 1+__const_200^0<=x^0+__const_1000^0 && -__const_1000^0<=0 ], cost: NONTERM
     13: l3 -> l0 : x^0'=x^0+k*__const_1000^0, [ __const_1000^0<=-1 && k>=0 && 1+__const_200^0<=(-1+k)*__const_1000^0+x^0+__const_1000^0 ], cost: 2+2*k

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l3
     11: l3 -> [4] : [ 1+__const_200^0<=x^0+__const_1000^0 && __const_200^0==0 && x^0==1 && __const_1000^0==0 ], cost: NONTERM
     12: l3 -> [4] : [ 1+__const_200^0<=x^0+__const_1000^0 && -__const_1000^0<=0 ], cost: NONTERM
     13: l3 -> l0 : x^0'=x^0+k*__const_1000^0, [ __const_1000^0<=-1 && k>=0 && 1+__const_200^0<=(-1+k)*__const_1000^0+x^0+__const_1000^0 ], cost: 2+2*k

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l3
     11: l3 -> [4] : [ 1+__const_200^0<=x^0+__const_1000^0 && __const_200^0==0 && x^0==1 && __const_1000^0==0 ], cost: NONTERM
     12: l3 -> [4] : [ 1+__const_200^0<=x^0+__const_1000^0 && -__const_1000^0<=0 ], cost: NONTERM
     13: l3 -> l0 : x^0'=x^0+k*__const_1000^0, [ __const_1000^0<=-1 && k>=0 && 1+__const_200^0<=(-1+k)*__const_1000^0+x^0+__const_1000^0 ], cost: 2+2*k

Computing asymptotic complexity for rule 12
   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:  [ 1+__const_200^0<=x^0+__const_1000^0 && -__const_1000^0<=0 ]

NO