NO

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

Initial linear ITS problem
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg3==arg3P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && 0==arg3P_3 ], cost: 1
      6: f4 -> f5 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1<0 && arg2<0 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      7: f4 -> f5 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1<0 && arg2>0 && arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1
      8: f4 -> f5 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg1>0 && arg2<0 && arg1==arg1P_9 && arg2==arg2P_9 && arg3==arg3P_9 ], cost: 1
      9: f4 -> f5 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg1>0 && arg2>0 && arg1==arg1P_10 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1
     11: f4 -> f9 : arg1'=arg1P_12, arg2'=arg2P_12, arg3'=arg3P_12, [ arg1==0 && arg1==arg1P_12 && arg2==arg2P_12 && arg3==arg3P_12 ], cost: 1
     12: f4 -> f9 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [ arg2==0 && arg1==arg1P_13 && arg2==arg2P_13 && arg3==arg3P_13 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1==arg1P_4 && arg2==arg2P_4 && arg1==arg3P_4 ], cost: 1
      4: f6 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      5: f7 -> f8 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg3==arg2P_6 && arg3==arg3P_6 ], cost: 1
     10: f8 -> f4 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], cost: 1

Removed unreachable and leaf rules:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg3==arg3P_2 ], cost: 1
      2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && 0==arg3P_3 ], cost: 1
      6: f4 -> f5 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1<0 && arg2<0 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      7: f4 -> f5 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1<0 && arg2>0 && arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1
      8: f4 -> f5 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg1>0 && arg2<0 && arg1==arg1P_9 && arg2==arg2P_9 && arg3==arg3P_9 ], cost: 1
      9: f4 -> f5 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg1>0 && arg2>0 && arg1==arg1P_10 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1==arg1P_4 && arg2==arg2P_4 && arg1==arg3P_4 ], cost: 1
      4: f6 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      5: f7 -> f8 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg3==arg2P_6 && arg3==arg3P_6 ], cost: 1
     10: f8 -> f4 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], 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
      2: f3 -> f4 : arg3'=0, [], cost: 1
      6: f4 -> f5 : [ arg1<0 && arg2<0 ], cost: 1
      7: f4 -> f5 : [ arg1<0 && arg2>0 ], cost: 1
      8: f4 -> f5 : [ arg1>0 && arg2<0 ], cost: 1
      9: f4 -> f5 : [ arg1>0 && arg2>0 ], cost: 1
      3: f5 -> f6 : arg3'=arg1, [], cost: 1
      4: f6 -> f7 : arg1'=arg2, [], cost: 1
      5: f7 -> f8 : arg2'=arg3, [], cost: 1
     10: f8 -> f4 : [], cost: 1
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
      6: f4 -> f5 : [ arg1<0 && arg2<0 ], cost: 1
      7: f4 -> f5 : [ arg1<0 && arg2>0 ], cost: 1
      8: f4 -> f5 : [ arg1>0 && arg2<0 ], cost: 1
      9: f4 -> f5 : [ arg1>0 && arg2>0 ], cost: 1
     19: f5 -> f4 : arg1'=arg2, arg2'=arg1, arg3'=arg1, [], cost: 4
     16: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=0, [], cost: 4

Eliminated locations (on tree-shaped paths):
   Start location: __init
     20: f4 -> f4 : arg1'=arg2, arg2'=arg1, arg3'=arg1, [ arg1<0 && arg2<0 ], cost: 5
     21: f4 -> f4 : arg1'=arg2, arg2'=arg1, arg3'=arg1, [ arg1<0 && arg2>0 ], cost: 5
     22: f4 -> f4 : arg1'=arg2, arg2'=arg1, arg3'=arg1, [ arg1>0 && arg2<0 ], cost: 5
     23: f4 -> f4 : arg1'=arg2, arg2'=arg1, arg3'=arg1, [ arg1>0 && arg2>0 ], cost: 5
     16: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=0, [], cost: 4

Accelerating simple loops of location 3.
   Accelerating the following rules:
     20: f4 -> f4 : arg1'=arg2, arg2'=arg1, arg3'=arg1, [ arg1<0 && arg2<0 ], cost: 5
     21: f4 -> f4 : arg1'=arg2, arg2'=arg1, arg3'=arg1, [ arg1<0 && arg2>0 ], cost: 5
     22: f4 -> f4 : arg1'=arg2, arg2'=arg1, arg3'=arg1, [ arg1>0 && arg2<0 ], cost: 5
     23: f4 -> f4 : arg1'=arg2, arg2'=arg1, arg3'=arg1, [ arg1>0 && arg2>0 ], cost: 5

   Accelerated rule 20 with non-termination, yielding the new rule 24.
   Failed to prove monotonicity of the guard of rule 21.
   Failed to prove monotonicity of the guard of rule 22.
   Accelerated rule 23 with non-termination, yielding the new rule 25.
[accelerate] Nesting with 2 inner and 2 outer candidates
   Nested simple loops 22 (outer loop) and 21 (inner loop) with Rule(3 | arg1<0, arg2>0, | NONTERM || 10 | ), resulting in the new rules: 26, 27.
   Nested simple loops 21 (outer loop) and 22 (inner loop) with Rule(3 | arg1>0, arg2<0, | NONTERM || 10 | ), resulting in the new rules: 28, 29.
   Removing the simple loops: 20 21 22 23.
   Also removing duplicate rules: 26 27.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     24: f4 -> [10] : [ arg1<0 && arg2<0 ], cost: NONTERM
     25: f4 -> [10] : [ arg1>0 && arg2>0 ], cost: NONTERM
     28: f4 -> [10] : [ arg1>0 && arg2<0 ], cost: NONTERM
     29: f4 -> [10] : [ arg1<0 && arg2>0 ], cost: NONTERM
     16: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=0, [], cost: 4

Chained accelerated rules (with incoming rules):
   Start location: __init
     16: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=0, [], cost: 4
     30: __init -> [10] : [], cost: NONTERM
     31: __init -> [10] : [], cost: NONTERM
     32: __init -> [10] : [], cost: NONTERM
     33: __init -> [10] : [], cost: NONTERM

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     30: __init -> [10] : [], cost: NONTERM
     31: __init -> [10] : [], cost: NONTERM
     32: __init -> [10] : [], cost: NONTERM
     33: __init -> [10] : [], cost: NONTERM

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     33: __init -> [10] : [], cost: NONTERM

Computing asymptotic complexity for rule 33
   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