WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: l6
      0: l0 -> l1 : oldX0^0'=oldX0^post_1, oldX1^0'=oldX1^post_1, x0^0'=x0^post_1, [ oldX0^post_1==x0^0 && oldX1^post_1==oldX1^post_1 && x0^post_1==oldX1^post_1 ], cost: 1
      1: l0 -> l2 : oldX0^0'=oldX0^post_2, oldX1^0'=oldX1^post_2, x0^0'=x0^post_2, [ oldX0^post_2==x0^0 && x0^post_2==-1+oldX0^post_2 && oldX1^0==oldX1^post_2 ], cost: 1
      5: l2 -> l4 : oldX0^0'=oldX0^post_6, oldX1^0'=oldX1^post_6, x0^0'=x0^post_6, [ oldX0^post_6==x0^0 && x0^post_6==oldX0^post_6 && oldX1^0==oldX1^post_6 ], cost: 1
      2: l3 -> l1 : oldX0^0'=oldX0^post_3, oldX1^0'=oldX1^post_3, x0^0'=x0^post_3, [ oldX0^post_3==x0^0 && oldX1^post_3==oldX1^post_3 && x0^post_3==oldX1^post_3 ], cost: 1
      3: l4 -> l0 : oldX0^0'=oldX0^post_4, oldX1^0'=oldX1^post_4, x0^0'=x0^post_4, [ oldX0^post_4==x0^0 && 0<=oldX0^post_4 && x0^post_4==oldX0^post_4 && oldX1^0==oldX1^post_4 ], cost: 1
      4: l4 -> l3 : oldX0^0'=oldX0^post_5, oldX1^0'=oldX1^post_5, x0^0'=x0^post_5, [ oldX0^post_5==x0^0 && 1+oldX0^post_5<=0 && x0^post_5==oldX0^post_5 && oldX1^0==oldX1^post_5 ], cost: 1
      6: l5 -> l1 : oldX0^0'=oldX0^post_7, oldX1^0'=oldX1^post_7, x0^0'=x0^post_7, [ oldX0^0==oldX0^post_7 && oldX1^0==oldX1^post_7 && x0^0==x0^post_7 ], cost: 1
      7: l5 -> l0 : oldX0^0'=oldX0^post_8, oldX1^0'=oldX1^post_8, x0^0'=x0^post_8, [ oldX0^0==oldX0^post_8 && oldX1^0==oldX1^post_8 && x0^0==x0^post_8 ], cost: 1
      8: l5 -> l3 : oldX0^0'=oldX0^post_9, oldX1^0'=oldX1^post_9, x0^0'=x0^post_9, [ oldX0^0==oldX0^post_9 && oldX1^0==oldX1^post_9 && x0^0==x0^post_9 ], cost: 1
      9: l5 -> l4 : oldX0^0'=oldX0^post_10, oldX1^0'=oldX1^post_10, x0^0'=x0^post_10, [ oldX0^0==oldX0^post_10 && oldX1^0==oldX1^post_10 && x0^0==x0^post_10 ], cost: 1
     10: l5 -> l2 : oldX0^0'=oldX0^post_11, oldX1^0'=oldX1^post_11, x0^0'=x0^post_11, [ oldX0^0==oldX0^post_11 && oldX1^0==oldX1^post_11 && x0^0==x0^post_11 ], cost: 1
     11: l6 -> l5 : oldX0^0'=oldX0^post_12, oldX1^0'=oldX1^post_12, x0^0'=x0^post_12, [ oldX0^0==oldX0^post_12 && oldX1^0==oldX1^post_12 && x0^0==x0^post_12 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
     11: l6 -> l5 : oldX0^0'=oldX0^post_12, oldX1^0'=oldX1^post_12, x0^0'=x0^post_12, [ oldX0^0==oldX0^post_12 && oldX1^0==oldX1^post_12 && x0^0==x0^post_12 ], cost: 1

Removed unreachable and leaf rules:
   Start location: l6
      1: l0 -> l2 : oldX0^0'=oldX0^post_2, oldX1^0'=oldX1^post_2, x0^0'=x0^post_2, [ oldX0^post_2==x0^0 && x0^post_2==-1+oldX0^post_2 && oldX1^0==oldX1^post_2 ], cost: 1
      5: l2 -> l4 : oldX0^0'=oldX0^post_6, oldX1^0'=oldX1^post_6, x0^0'=x0^post_6, [ oldX0^post_6==x0^0 && x0^post_6==oldX0^post_6 && oldX1^0==oldX1^post_6 ], cost: 1
      3: l4 -> l0 : oldX0^0'=oldX0^post_4, oldX1^0'=oldX1^post_4, x0^0'=x0^post_4, [ oldX0^post_4==x0^0 && 0<=oldX0^post_4 && x0^post_4==oldX0^post_4 && oldX1^0==oldX1^post_4 ], cost: 1
      7: l5 -> l0 : oldX0^0'=oldX0^post_8, oldX1^0'=oldX1^post_8, x0^0'=x0^post_8, [ oldX0^0==oldX0^post_8 && oldX1^0==oldX1^post_8 && x0^0==x0^post_8 ], cost: 1
      9: l5 -> l4 : oldX0^0'=oldX0^post_10, oldX1^0'=oldX1^post_10, x0^0'=x0^post_10, [ oldX0^0==oldX0^post_10 && oldX1^0==oldX1^post_10 && x0^0==x0^post_10 ], cost: 1
     10: l5 -> l2 : oldX0^0'=oldX0^post_11, oldX1^0'=oldX1^post_11, x0^0'=x0^post_11, [ oldX0^0==oldX0^post_11 && oldX1^0==oldX1^post_11 && x0^0==x0^post_11 ], cost: 1
     11: l6 -> l5 : oldX0^0'=oldX0^post_12, oldX1^0'=oldX1^post_12, x0^0'=x0^post_12, [ oldX0^0==oldX0^post_12 && oldX1^0==oldX1^post_12 && x0^0==x0^post_12 ], cost: 1

Simplified all rules, resulting in:
   Start location: l6
      1: l0 -> l2 : oldX0^0'=x0^0, x0^0'=-1+x0^0, [], cost: 1
      5: l2 -> l4 : oldX0^0'=x0^0, [], cost: 1
      3: l4 -> l0 : oldX0^0'=x0^0, [ 0<=x0^0 ], cost: 1
      7: l5 -> l0 : [], cost: 1
      9: l5 -> l4 : [], cost: 1
     10: l5 -> l2 : [], cost: 1
     11: l6 -> l5 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on tree-shaped paths):
   Start location: l6
      1: l0 -> l2 : oldX0^0'=x0^0, x0^0'=-1+x0^0, [], cost: 1
      5: l2 -> l4 : oldX0^0'=x0^0, [], cost: 1
      3: l4 -> l0 : oldX0^0'=x0^0, [ 0<=x0^0 ], cost: 1
     12: l6 -> l0 : [], cost: 2
     13: l6 -> l4 : [], cost: 2
     14: l6 -> l2 : [], cost: 2

Eliminated location l0 (as a last resort):
   Start location: l6
      5: l2 -> l4 : oldX0^0'=x0^0, [], cost: 1
     15: l4 -> l2 : oldX0^0'=x0^0, x0^0'=-1+x0^0, [ 0<=x0^0 ], cost: 2
     13: l6 -> l4 : [], cost: 2
     14: l6 -> l2 : [], cost: 2
     16: l6 -> l2 : oldX0^0'=x0^0, x0^0'=-1+x0^0, [], cost: 3

Eliminated location l2 (as a last resort):
   Start location: l6
     18: l4 -> l4 : oldX0^0'=-1+x0^0, x0^0'=-1+x0^0, [ 0<=x0^0 ], cost: 3
     13: l6 -> l4 : [], cost: 2
     17: l6 -> l4 : oldX0^0'=x0^0, [], cost: 3
     19: l6 -> l4 : oldX0^0'=-1+x0^0, x0^0'=-1+x0^0, [], cost: 4

Accelerating simple loops of location 4.
   Accelerating the following rules:
     18: l4 -> l4 : oldX0^0'=-1+x0^0, x0^0'=-1+x0^0, [ 0<=x0^0 ], cost: 3

   Accelerated rule 18 with backward acceleration, yielding the new rule 20.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 18.

Accelerated all simple loops using metering functions (where possible):
   Start location: l6
     20: l4 -> l4 : oldX0^0'=-1, x0^0'=-1, [ 1+x0^0>=1 ], cost: 3+3*x0^0
     13: l6 -> l4 : [], cost: 2
     17: l6 -> l4 : oldX0^0'=x0^0, [], cost: 3
     19: l6 -> l4 : oldX0^0'=-1+x0^0, x0^0'=-1+x0^0, [], cost: 4

Chained accelerated rules (with incoming rules):
   Start location: l6
     13: l6 -> l4 : [], cost: 2
     17: l6 -> l4 : oldX0^0'=x0^0, [], cost: 3
     19: l6 -> l4 : oldX0^0'=-1+x0^0, x0^0'=-1+x0^0, [], cost: 4
     21: l6 -> l4 : oldX0^0'=-1, x0^0'=-1, [ 1+x0^0>=1 ], cost: 5+3*x0^0
     22: l6 -> l4 : oldX0^0'=-1, x0^0'=-1, [ 1+x0^0>=1 ], cost: 6+3*x0^0
     23: l6 -> l4 : oldX0^0'=-1, x0^0'=-1, [ x0^0>=1 ], cost: 4+3*x0^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l6
     21: l6 -> l4 : oldX0^0'=-1, x0^0'=-1, [ 1+x0^0>=1 ], cost: 5+3*x0^0
     22: l6 -> l4 : oldX0^0'=-1, x0^0'=-1, [ 1+x0^0>=1 ], cost: 6+3*x0^0
     23: l6 -> l4 : oldX0^0'=-1, x0^0'=-1, [ x0^0>=1 ], cost: 4+3*x0^0

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l6
     22: l6 -> l4 : oldX0^0'=-1, x0^0'=-1, [ 1+x0^0>=1 ], cost: 6+3*x0^0
     23: l6 -> l4 : oldX0^0'=-1, x0^0'=-1, [ x0^0>=1 ], cost: 4+3*x0^0

Computing asymptotic complexity for rule 22
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 23
   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:  [ oldX0^0==oldX0^post_12 && oldX1^0==oldX1^post_12 && x0^0==x0^post_12 ]

WORST_CASE(Omega(1),?)