NO

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

Initial linear ITS problem
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2==arg2P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 ], cost: 1
      3: f3 -> f4 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg1>0 && arg2>0 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      5: f3 -> f6 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1<=0 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      6: f3 -> f6 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg2<=0 && arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1P_3==10*arg2-2*arg1 && arg2==arg2P_3 ], cost: 1
      4: f5 -> f3 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

Removed unreachable and leaf rules:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2==arg2P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 ], cost: 1
      3: f3 -> f4 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg1>0 && arg2>0 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1P_3==10*arg2-2*arg1 && arg2==arg2P_3 ], cost: 1
      4: f5 -> f3 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1
      1: f2 -> f3 : arg2'=arg2P_2, [], cost: 1
      3: f3 -> f4 : [ arg1>0 && arg2>0 ], cost: 1
      2: f4 -> f5 : arg1'=10*arg2-2*arg1, [], cost: 1
      4: f5 -> f3 : [], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     11: f3 -> f3 : arg1'=10*arg2-2*arg1, [ arg1>0 && arg2>0 ], cost: 3
      9: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3

Accelerating simple loops of location 2.
   Accelerating the following rules:
     11: f3 -> f3 : arg1'=10*arg2-2*arg1, [ arg1>0 && arg2>0 ], cost: 3

   Accelerated rule 11 with non-termination, yielding the new rule 12.
   Accelerated rule 11 with backward acceleration, yielding the new rule 13.
[accelerate] Nesting with 1 inner and 0 outer candidates
   Removing the simple loops: 11.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     12: f3 -> [7] : [ 10*arg2-2*arg1>0 && arg2==3 && arg1==10 ], cost: NONTERM
     13: f3 -> f3 : arg1'=4^k*arg1+10/3*arg2-10/3*4^k*arg2, [ arg2>0 && k>=0 && 4^(-1+k)*arg1+10/3*arg2-10/3*arg2*4^(-1+k)>0 && -2*4^(-1+k)*arg1+10/3*arg2+20/3*arg2*4^(-1+k)>0 ], cost: 6*k
      9: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3

Chained accelerated rules (with incoming rules):
   Start location: __init
      9: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3
     14: __init -> [7] : [], cost: NONTERM
     15: __init -> f3 : arg1'=10/3*arg2P_2+4^k*arg1P_1-10/3*4^k*arg2P_2, arg2'=arg2P_2, [ arg2P_2>0 && k>=0 && 10/3*arg2P_2-10/3*arg2P_2*4^(-1+k)+arg1P_1*4^(-1+k)>0 && 10/3*arg2P_2+20/3*arg2P_2*4^(-1+k)-2*arg1P_1*4^(-1+k)>0 ], cost: 3+6*k

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     14: __init -> [7] : [], cost: NONTERM
     15: __init -> f3 : arg1'=10/3*arg2P_2+4^k*arg1P_1-10/3*4^k*arg2P_2, arg2'=arg2P_2, [ arg2P_2>0 && k>=0 && 10/3*arg2P_2-10/3*arg2P_2*4^(-1+k)+arg1P_1*4^(-1+k)>0 && 10/3*arg2P_2+20/3*arg2P_2*4^(-1+k)-2*arg1P_1*4^(-1+k)>0 ], cost: 3+6*k

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     14: __init -> [7] : [], cost: NONTERM
     15: __init -> f3 : arg1'=10/3*arg2P_2+4^k*arg1P_1-10/3*4^k*arg2P_2, arg2'=arg2P_2, [ arg2P_2>0 && k>=0 && 10/3*arg2P_2-10/3*arg2P_2*4^(-1+k)+arg1P_1*4^(-1+k)>0 && 10/3*arg2P_2+20/3*arg2P_2*4^(-1+k)-2*arg1P_1*4^(-1+k)>0 ], cost: 3+6*k

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:  []

NO