NO

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

Initial linear ITS problem
   Start location: l5
      0: l0 -> l1 : p^0'=p^post_1, y_1^0'=y_1^post_1, [ p^0==p^post_1 && y_1^0==y_1^post_1 ], cost: 1
      1: l1 -> l0 : p^0'=p^post_2, y_1^0'=y_1^post_2, [ p^0==p^post_2 && y_1^0==y_1^post_2 ], cost: 1
      2: l2 -> l3 : p^0'=p^post_3, y_1^0'=y_1^post_3, [ p^post_3==0 && y_1^0==y_1^post_3 ], cost: 1
      3: l3 -> l4 : p^0'=p^post_4, y_1^0'=y_1^post_4, [ 1<=y_1^0 && y_1^post_4==-1+y_1^0 && p^0==p^post_4 ], cost: 1
      5: l3 -> l0 : p^0'=p^post_6, y_1^0'=y_1^post_6, [ y_1^0<=0 && p^post_6==-1 && y_1^0==y_1^post_6 ], cost: 1
      4: l4 -> l3 : p^0'=p^post_5, y_1^0'=y_1^post_5, [ p^0==p^post_5 && y_1^0==y_1^post_5 ], cost: 1
      6: l5 -> l2 : p^0'=p^post_7, y_1^0'=y_1^post_7, [ p^0==p^post_7 && y_1^0==y_1^post_7 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      6: l5 -> l2 : p^0'=p^post_7, y_1^0'=y_1^post_7, [ p^0==p^post_7 && y_1^0==y_1^post_7 ], cost: 1

Simplified all rules, resulting in:
   Start location: l5
      0: l0 -> l1 : [], cost: 1
      1: l1 -> l0 : [], cost: 1
      2: l2 -> l3 : p^0'=0, [], cost: 1
      3: l3 -> l4 : y_1^0'=-1+y_1^0, [ 1<=y_1^0 ], cost: 1
      5: l3 -> l0 : p^0'=-1, [ y_1^0<=0 ], cost: 1
      4: l4 -> l3 : [], cost: 1
      6: l5 -> l2 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l5
      9: l0 -> l0 : [], cost: 2
      5: l3 -> l0 : p^0'=-1, [ y_1^0<=0 ], cost: 1
      8: l3 -> l3 : y_1^0'=-1+y_1^0, [ 1<=y_1^0 ], cost: 2
      7: l5 -> l3 : p^0'=0, [], cost: 2

Accelerating simple loops of location 0.
   Accelerating the following rules:
      9: l0 -> l0 : [], cost: 2

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

Accelerating simple loops of location 3.
   Accelerating the following rules:
      8: l3 -> l3 : y_1^0'=-1+y_1^0, [ 1<=y_1^0 ], cost: 2

   Accelerated rule 8 with backward acceleration, yielding the new rule 11.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 8.

Accelerated all simple loops using metering functions (where possible):
   Start location: l5
     10: l0 -> [6] : [], cost: NONTERM
      5: l3 -> l0 : p^0'=-1, [ y_1^0<=0 ], cost: 1
     11: l3 -> l3 : y_1^0'=0, [ y_1^0>=0 ], cost: 2*y_1^0
      7: l5 -> l3 : p^0'=0, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l5
      5: l3 -> l0 : p^0'=-1, [ y_1^0<=0 ], cost: 1
     12: l3 -> [6] : [ y_1^0<=0 ], cost: NONTERM
      7: l5 -> l3 : p^0'=0, [], cost: 2
     13: l5 -> l3 : p^0'=0, y_1^0'=0, [ y_1^0>=0 ], cost: 2+2*y_1^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l5
     12: l3 -> [6] : [ y_1^0<=0 ], cost: NONTERM
      7: l5 -> l3 : p^0'=0, [], cost: 2
     13: l5 -> l3 : p^0'=0, y_1^0'=0, [ y_1^0>=0 ], cost: 2+2*y_1^0

Eliminated locations (on tree-shaped paths):
   Start location: l5
     14: l5 -> [6] : [ y_1^0<=0 ], cost: NONTERM
     15: l5 -> [6] : [ y_1^0>=0 ], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l5
     14: l5 -> [6] : [ y_1^0<=0 ], cost: NONTERM
     15: l5 -> [6] : [ y_1^0>=0 ], cost: NONTERM

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:  [ y_1^0<=0 ]

NO