NO

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

Initial linear ITS problem
   Start location: l3
      0: l0 -> l1 : x^0'=x^post_1, y^0'=y^post_1, [ 1<=x^0 && x^0==x^post_1 && y^0==y^post_1 ], cost: 1
      1: l1 -> l0 : x^0'=x^post_2, y^0'=y^post_2, [ y^post_2==-1+y^0 && x^post_2==x^post_2 ], cost: 1
      2: l1 -> l0 : x^0'=x^post_3, y^0'=y^post_3, [ x^post_3==-1+x^0 && y^0==y^post_3 ], cost: 1
      3: l2 -> l0 : x^0'=x^post_4, y^0'=y^post_4, [ x^0==x^post_4 && y^0==y^post_4 ], cost: 1
      4: l3 -> l2 : x^0'=x^post_5, y^0'=y^post_5, [ x^0==x^post_5 && y^0==y^post_5 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      4: l3 -> l2 : x^0'=x^post_5, y^0'=y^post_5, [ x^0==x^post_5 && y^0==y^post_5 ], cost: 1

Simplified all rules, resulting in:
   Start location: l3
      0: l0 -> l1 : [ 1<=x^0 ], cost: 1
      1: l1 -> l0 : x^0'=x^post_2, y^0'=-1+y^0, [], cost: 1
      2: l1 -> l0 : x^0'=-1+x^0, [], cost: 1
      3: l2 -> l0 : [], cost: 1
      4: l3 -> l2 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l3
      0: l0 -> l1 : [ 1<=x^0 ], cost: 1
      1: l1 -> l0 : x^0'=x^post_2, y^0'=-1+y^0, [], cost: 1
      2: l1 -> l0 : x^0'=-1+x^0, [], cost: 1
      5: l3 -> l0 : [], cost: 2

Eliminated locations (on tree-shaped paths):
   Start location: l3
      6: l0 -> l0 : x^0'=x^post_2, y^0'=-1+y^0, [ 1<=x^0 ], cost: 2
      7: l0 -> l0 : x^0'=-1+x^0, [ 1<=x^0 ], cost: 2
      5: l3 -> l0 : [], cost: 2

Accelerating simple loops of location 0.
   Accelerating the following rules:
      6: l0 -> l0 : x^0'=x^post_2, y^0'=-1+y^0, [ 1<=x^0 ], cost: 2
      7: l0 -> l0 : x^0'=-1+x^0, [ 1<=x^0 ], cost: 2

[test] deduced pseudo-invariant -x^post_2+x^0<=0, also trying x^post_2-x^0<=-1
   Accelerated rule 6 with non-termination, yielding the new rule 8.
   Accelerated rule 6 with non-termination, yielding the new rule 9.
   Accelerated rule 6 with backward acceleration, yielding the new rule 10.
   Accelerated rule 7 with backward acceleration, yielding the new rule 11.
[accelerate] Nesting with 1 inner and 2 outer candidates
   Removing the simple loops: 7.
   Also removing duplicate rules: 9.

Accelerated all simple loops using metering functions (where possible):
   Start location: l3
      6: l0 -> l0 : x^0'=x^post_2, y^0'=-1+y^0, [ 1<=x^0 ], cost: 2
      8: l0 -> [4] : [ 1<=x^0 && 1<=x^post_2 ], cost: NONTERM
     10: l0 -> [4] : [ 1<=x^0 && -x^post_2+x^0<=0 ], cost: NONTERM
     11: l0 -> l0 : x^0'=0, [ x^0>=0 ], cost: 2*x^0
      5: l3 -> l0 : [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l3
      5: l3 -> l0 : [], cost: 2
     12: l3 -> l0 : x^0'=x^post_2, y^0'=-1+y^0, [ 1<=x^0 ], cost: 4
     13: l3 -> [4] : [ 1<=x^0 ], cost: NONTERM
     14: l3 -> [4] : [ 1<=x^0 ], cost: NONTERM
     15: l3 -> l0 : x^0'=0, [ x^0>=0 ], cost: 2+2*x^0

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

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l3
     14: l3 -> [4] : [ 1<=x^0 ], cost: NONTERM
     15: l3 -> l0 : x^0'=0, [ x^0>=0 ], cost: 2+2*x^0

Computing asymptotic complexity for rule 14
   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<=x^0 ]

NO