NO

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

Initial linear ITS problem
   Start location: l4
      0: l0 -> l1 : x^0'=x^post_1, y^0'=y^post_1, [ 1+x^0<=0 && x^0==x^post_1 && y^0==y^post_1 ], cost: 1
      1: l0 -> l2 : x^0'=x^post_2, y^0'=y^post_2, [ 0<=x^0 && x^post_2==-y^0+x^0 && y^0==y^post_2 ], cost: 1
      2: l2 -> l0 : x^0'=x^post_3, y^0'=y^post_3, [ x^0==x^post_3 && y^0==y^post_3 ], cost: 1
      3: l3 -> l0 : x^0'=x^post_4, y^0'=y^post_4, [ 1+y^0+x^0<=0 && 1<=x^0 && x^0==x^post_4 && y^0==y^post_4 ], cost: 1
      4: l4 -> l3 : 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: l4 -> l3 : x^0'=x^post_5, y^0'=y^post_5, [ x^0==x^post_5 && y^0==y^post_5 ], cost: 1

Removed unreachable and leaf rules:
   Start location: l4
      1: l0 -> l2 : x^0'=x^post_2, y^0'=y^post_2, [ 0<=x^0 && x^post_2==-y^0+x^0 && y^0==y^post_2 ], cost: 1
      2: l2 -> l0 : x^0'=x^post_3, y^0'=y^post_3, [ x^0==x^post_3 && y^0==y^post_3 ], cost: 1
      3: l3 -> l0 : x^0'=x^post_4, y^0'=y^post_4, [ 1+y^0+x^0<=0 && 1<=x^0 && x^0==x^post_4 && y^0==y^post_4 ], cost: 1
      4: l4 -> l3 : 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: l4
      1: l0 -> l2 : x^0'=-y^0+x^0, [ 0<=x^0 ], cost: 1
      2: l2 -> l0 : [], cost: 1
      3: l3 -> l0 : [ 1+y^0+x^0<=0 && 1<=x^0 ], cost: 1
      4: l4 -> l3 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l4
      6: l0 -> l0 : x^0'=-y^0+x^0, [ 0<=x^0 ], cost: 2
      5: l4 -> l0 : [ 1+y^0+x^0<=0 && 1<=x^0 ], cost: 2

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

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

Accelerated all simple loops using metering functions (where possible):
   Start location: l4
      6: l0 -> l0 : x^0'=-y^0+x^0, [ 0<=x^0 ], cost: 2
      7: l0 -> [5] : [ y^0==0 && x^0==0 ], cost: NONTERM
      9: l0 -> [5] : [ 0<=x^0 && y^0<=0 ], cost: NONTERM
      5: l4 -> l0 : [ 1+y^0+x^0<=0 && 1<=x^0 ], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l4
      5: l4 -> l0 : [ 1+y^0+x^0<=0 && 1<=x^0 ], cost: 2
     10: l4 -> l0 : x^0'=-y^0+x^0, [ 1+y^0+x^0<=0 && 1<=x^0 ], cost: 4
     11: l4 -> [5] : [ 1+y^0+x^0<=0 && 1<=x^0 && y^0<=0 ], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l4
     11: l4 -> [5] : [ 1+y^0+x^0<=0 && 1<=x^0 && y^0<=0 ], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l4
     11: l4 -> [5] : [ 1+y^0+x^0<=0 && 1<=x^0 && y^0<=0 ], cost: NONTERM

Computing asymptotic complexity for rule 11
   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+y^0+x^0<=0 && 1<=x^0 ]

NO