NO

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

Initial linear ITS problem
   Start location: l4
      0: l0 -> l1 : x_5^0'=x_5^post_1, [ x_5^0==x_5^post_1 ], cost: 1
      1: l1 -> l2 : x_5^0'=x_5^post_2, [ x_5^post_2==-1+x_5^0 && x_5^post_2<=0 ], cost: 1
      3: l1 -> l3 : x_5^0'=x_5^post_4, [ x_5^post_4==-1+x_5^0 && 0<=-1+x_5^post_4 ], cost: 1
      2: l2 -> l1 : x_5^0'=x_5^post_3, [ x_5^0==x_5^post_3 ], cost: 1
      4: l3 -> l1 : x_5^0'=x_5^post_5, [ x_5^0==x_5^post_5 ], cost: 1
      5: l4 -> l0 : x_5^0'=x_5^post_6, [ x_5^0==x_5^post_6 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      5: l4 -> l0 : x_5^0'=x_5^post_6, [ x_5^0==x_5^post_6 ], cost: 1

Simplified all rules, resulting in:
   Start location: l4
      0: l0 -> l1 : [], cost: 1
      1: l1 -> l2 : x_5^0'=-1+x_5^0, [ -1+x_5^0<=0 ], cost: 1
      3: l1 -> l3 : x_5^0'=-1+x_5^0, [ 0<=-2+x_5^0 ], cost: 1
      2: l2 -> l1 : [], cost: 1
      4: l3 -> l1 : [], cost: 1
      5: l4 -> l0 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l4
      7: l1 -> l1 : x_5^0'=-1+x_5^0, [ -1+x_5^0<=0 ], cost: 2
      8: l1 -> l1 : x_5^0'=-1+x_5^0, [ 0<=-2+x_5^0 ], cost: 2
      6: l4 -> l1 : [], cost: 2

Accelerating simple loops of location 1.
   Accelerating the following rules:
      7: l1 -> l1 : x_5^0'=-1+x_5^0, [ -1+x_5^0<=0 ], cost: 2
      8: l1 -> l1 : x_5^0'=-1+x_5^0, [ 0<=-2+x_5^0 ], cost: 2

   Accelerated rule 7 with non-termination, yielding the new rule 9.
   Accelerated rule 8 with backward acceleration, yielding the new rule 10.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 7 8.

Accelerated all simple loops using metering functions (where possible):
   Start location: l4
      9: l1 -> [5] : [ -1+x_5^0<=0 ], cost: NONTERM
     10: l1 -> l1 : x_5^0'=1, [ -1+x_5^0>=0 ], cost: -2+2*x_5^0
      6: l4 -> l1 : [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l4
      6: l4 -> l1 : [], cost: 2
     11: l4 -> [5] : [ -1+x_5^0<=0 ], cost: NONTERM
     12: l4 -> l1 : x_5^0'=1, [ -1+x_5^0>=0 ], cost: 2*x_5^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l4
     11: l4 -> [5] : [ -1+x_5^0<=0 ], cost: NONTERM
     12: l4 -> l1 : x_5^0'=1, [ -1+x_5^0>=0 ], cost: 2*x_5^0

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l4
     11: l4 -> [5] : [ -1+x_5^0<=0 ], cost: NONTERM
     12: l4 -> l1 : x_5^0'=1, [ -1+x_5^0>=0 ], cost: 2*x_5^0

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+x_5^0<=0 ]

NO