WORST_CASE(Omega(1),?)

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

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

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      5: l4 -> l3 : a^0'=a^post_6, x^0'=x^post_6, y^0'=y^post_6, z^0'=z^post_6, [ a^0==a^post_6 && x^0==x^post_6 && y^0==y^post_6 && z^0==z^post_6 ], cost: 1

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

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l4
      7: l0 -> l0 : x^0'=-1+x^0, [ 1<=x^0 ], cost: 2
      8: l0 -> l0 : a^0'=-1+a^0, x^0'=x^0+y^0, y^0'=y^0+z^0, z^0'=z^0+a^0, [ 1<=x^0 ], cost: 2
      6: l4 -> l0 : [], cost: 2

Accelerating simple loops of location 0.
   Accelerating the following rules:
      7: l0 -> l0 : x^0'=-1+x^0, [ 1<=x^0 ], cost: 2
      8: l0 -> l0 : a^0'=-1+a^0, x^0'=x^0+y^0, y^0'=y^0+z^0, z^0'=z^0+a^0, [ 1<=x^0 ], cost: 2

   Accelerated rule 7 with backward acceleration, yielding the new rule 9.
   Failed to prove monotonicity of the guard of rule 8.
[accelerate] Nesting with 2 inner and 2 outer candidates
   Removing the simple loops: 7.

Accelerated all simple loops using metering functions (where possible):
   Start location: l4
      8: l0 -> l0 : a^0'=-1+a^0, x^0'=x^0+y^0, y^0'=y^0+z^0, z^0'=z^0+a^0, [ 1<=x^0 ], cost: 2
      9: l0 -> l0 : x^0'=0, [ x^0>=0 ], cost: 2*x^0
      6: l4 -> l0 : [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l4
      6: l4 -> l0 : [], cost: 2
     10: l4 -> l0 : a^0'=-1+a^0, x^0'=x^0+y^0, y^0'=y^0+z^0, z^0'=z^0+a^0, [ 1<=x^0 ], cost: 4
     11: l4 -> l0 : x^0'=0, [ x^0>=0 ], cost: 2+2*x^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l4
     11: l4 -> l0 : x^0'=0, [ x^0>=0 ], cost: 2+2*x^0

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l4
     11: l4 -> l0 : x^0'=0, [ x^0>=0 ], cost: 2+2*x^0

Computing asymptotic complexity for rule 11
   Resulting cost 0 has complexity: Unknown

Obtained the following overall complexity (w.r.t. the length of the input n):
   Complexity:  Constant
   Cpx degree:  0
   Solved cost: 1
   Rule cost:   1
   Rule guard:  [ a^0==a^post_6 && x^0==x^post_6 && y^0==y^post_6 && z^0==z^post_6 ]

WORST_CASE(Omega(1),?)