NO

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

Initial linear ITS problem
   Start location: l4
      0: l0 -> l1 : i^0'=i^post_1, j^0'=j^post_1, ret_pair5^0'=ret_pair5^post_1, [ ret_pair5^post_1==ret_pair5^post_1 && i^post_1==1+i^0 && j^post_1==2+j^0 ], cost: 1
      1: l0 -> l2 : i^0'=i^post_2, j^0'=j^post_2, ret_pair5^0'=ret_pair5^post_2, [ i^0==i^post_2 && j^0==j^post_2 && ret_pair5^0==ret_pair5^post_2 ], cost: 1
      2: l1 -> l0 : i^0'=i^post_3, j^0'=j^post_3, ret_pair5^0'=ret_pair5^post_3, [ i^0==i^post_3 && j^0==j^post_3 && ret_pair5^0==ret_pair5^post_3 ], cost: 1
      3: l3 -> l1 : i^0'=i^post_4, j^0'=j^post_4, ret_pair5^0'=ret_pair5^post_4, [ i^post_4==0 && j^post_4==0 && ret_pair5^0==ret_pair5^post_4 ], cost: 1
      4: l4 -> l3 : i^0'=i^post_5, j^0'=j^post_5, ret_pair5^0'=ret_pair5^post_5, [ i^0==i^post_5 && j^0==j^post_5 && ret_pair5^0==ret_pair5^post_5 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      4: l4 -> l3 : i^0'=i^post_5, j^0'=j^post_5, ret_pair5^0'=ret_pair5^post_5, [ i^0==i^post_5 && j^0==j^post_5 && ret_pair5^0==ret_pair5^post_5 ], cost: 1

Removed unreachable and leaf rules:
   Start location: l4
      0: l0 -> l1 : i^0'=i^post_1, j^0'=j^post_1, ret_pair5^0'=ret_pair5^post_1, [ ret_pair5^post_1==ret_pair5^post_1 && i^post_1==1+i^0 && j^post_1==2+j^0 ], cost: 1
      2: l1 -> l0 : i^0'=i^post_3, j^0'=j^post_3, ret_pair5^0'=ret_pair5^post_3, [ i^0==i^post_3 && j^0==j^post_3 && ret_pair5^0==ret_pair5^post_3 ], cost: 1
      3: l3 -> l1 : i^0'=i^post_4, j^0'=j^post_4, ret_pair5^0'=ret_pair5^post_4, [ i^post_4==0 && j^post_4==0 && ret_pair5^0==ret_pair5^post_4 ], cost: 1
      4: l4 -> l3 : i^0'=i^post_5, j^0'=j^post_5, ret_pair5^0'=ret_pair5^post_5, [ i^0==i^post_5 && j^0==j^post_5 && ret_pair5^0==ret_pair5^post_5 ], cost: 1

Simplified all rules, resulting in:
   Start location: l4
      0: l0 -> l1 : i^0'=1+i^0, j^0'=2+j^0, ret_pair5^0'=ret_pair5^post_1, [], cost: 1
      2: l1 -> l0 : [], cost: 1
      3: l3 -> l1 : i^0'=0, j^0'=0, [], cost: 1
      4: l4 -> l3 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l4
      6: l1 -> l1 : i^0'=1+i^0, j^0'=2+j^0, ret_pair5^0'=ret_pair5^post_1, [], cost: 2
      5: l4 -> l1 : i^0'=0, j^0'=0, [], cost: 2

Accelerating simple loops of location 1.
   Accelerating the following rules:
      6: l1 -> l1 : i^0'=1+i^0, j^0'=2+j^0, ret_pair5^0'=ret_pair5^post_1, [], cost: 2

   Accelerated rule 6 with non-termination, yielding the new rule 7.
[accelerate] Nesting with 0 inner and 0 outer candidates
   Removing the simple loops: 6.

Accelerated all simple loops using metering functions (where possible):
   Start location: l4
      7: l1 -> [5] : [], cost: NONTERM
      5: l4 -> l1 : i^0'=0, j^0'=0, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l4
      5: l4 -> l1 : i^0'=0, j^0'=0, [], cost: 2
      8: l4 -> [5] : [], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l4
      8: l4 -> [5] : [], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l4
      8: l4 -> [5] : [], cost: NONTERM

Computing asymptotic complexity for rule 8
   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:  []

NO