NO

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

Initial linear ITS problem
   Start location: l5
      0: l0 -> l1 : x^0'=x^post_1, y^0'=y^post_1, [ x^0==x^post_1 && y^0==y^post_1 ], cost: 1
      1: l1 -> l3 : x^0'=x^post_2, y^0'=y^post_2, [ 1<=x^0 && x^0==x^post_2 && y^0==y^post_2 ], cost: 1
      2: l3 -> l4 : x^0'=x^post_3, y^0'=y^post_3, [ 1+y^0<=0 && x^0==x^post_3 && y^0==y^post_3 ], cost: 1
      3: l3 -> l4 : x^0'=x^post_4, y^0'=y^post_4, [ 1<=y^0 && x^0==x^post_4 && y^0==y^post_4 ], cost: 1
      4: l4 -> l2 : x^0'=x^post_5, y^0'=y^post_5, [ x^post_5==y^0+x^0 && y^0==y^post_5 ], cost: 1
      5: l2 -> l1 : x^0'=x^post_6, y^0'=y^post_6, [ x^0==x^post_6 && y^0==y^post_6 ], cost: 1
      6: l5 -> l0 : x^0'=x^post_7, y^0'=y^post_7, [ x^0==x^post_7 && y^0==y^post_7 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      6: l5 -> l0 : x^0'=x^post_7, y^0'=y^post_7, [ x^0==x^post_7 && y^0==y^post_7 ], cost: 1

Simplified all rules, resulting in:
   Start location: l5
      0: l0 -> l1 : [], cost: 1
      1: l1 -> l3 : [ 1<=x^0 ], cost: 1
      2: l3 -> l4 : [ 1+y^0<=0 ], cost: 1
      3: l3 -> l4 : [ 1<=y^0 ], cost: 1
      4: l4 -> l2 : x^0'=y^0+x^0, [], cost: 1
      5: l2 -> l1 : [], cost: 1
      6: l5 -> l0 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l5
      1: l1 -> l3 : [ 1<=x^0 ], cost: 1
      2: l3 -> l4 : [ 1+y^0<=0 ], cost: 1
      3: l3 -> l4 : [ 1<=y^0 ], cost: 1
      8: l4 -> l1 : x^0'=y^0+x^0, [], cost: 2
      7: l5 -> l1 : [], cost: 2

Eliminated locations (on tree-shaped paths):
   Start location: l5
      9: l1 -> l4 : [ 1<=x^0 && 1+y^0<=0 ], cost: 2
     10: l1 -> l4 : [ 1<=x^0 && 1<=y^0 ], cost: 2
      8: l4 -> l1 : x^0'=y^0+x^0, [], cost: 2
      7: l5 -> l1 : [], cost: 2

Eliminated locations (on tree-shaped paths):
   Start location: l5
     11: l1 -> l1 : x^0'=y^0+x^0, [ 1<=x^0 && 1+y^0<=0 ], cost: 4
     12: l1 -> l1 : x^0'=y^0+x^0, [ 1<=x^0 && 1<=y^0 ], cost: 4
      7: l5 -> l1 : [], cost: 2

Accelerating simple loops of location 1.
   Accelerating the following rules:
     11: l1 -> l1 : x^0'=y^0+x^0, [ 1<=x^0 && 1+y^0<=0 ], cost: 4
     12: l1 -> l1 : x^0'=y^0+x^0, [ 1<=x^0 && 1<=y^0 ], cost: 4

   Accelerated rule 11 with backward acceleration, yielding the new rule 13.
   Accelerated rule 12 with non-termination, yielding the new rule 14.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 11 12.

Accelerated all simple loops using metering functions (where possible):
   Start location: l5
     13: l1 -> l1 : x^0'=y^0*k+x^0, [ 1+y^0<=0 && k>=0 && 1<=y^0*(-1+k)+x^0 ], cost: 4*k
     14: l1 -> [6] : [ 1<=x^0 && 1<=y^0 ], cost: NONTERM
      7: l5 -> l1 : [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l5
      7: l5 -> l1 : [], cost: 2
     15: l5 -> l1 : x^0'=y^0*k+x^0, [ 1+y^0<=0 && k>=0 && 1<=y^0*(-1+k)+x^0 ], cost: 2+4*k
     16: l5 -> [6] : [ 1<=x^0 && 1<=y^0 ], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l5
     15: l5 -> l1 : x^0'=y^0*k+x^0, [ 1+y^0<=0 && k>=0 && 1<=y^0*(-1+k)+x^0 ], cost: 2+4*k
     16: l5 -> [6] : [ 1<=x^0 && 1<=y^0 ], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l5
     15: l5 -> l1 : x^0'=y^0*k+x^0, [ 1+y^0<=0 && k>=0 && 1<=y^0*(-1+k)+x^0 ], cost: 2+4*k
     16: l5 -> [6] : [ 1<=x^0 && 1<=y^0 ], cost: NONTERM

Computing asymptotic complexity for rule 16
   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 && 1<=y^0 ]

NO