WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: l4
      0: l0 -> l1 : e^0'=e^post_1, n^0'=n^post_1, [ 1<=e^0 && n^0<=100 && n^post_1==11+n^0 && e^post_1==1+e^0 ], cost: 1
      2: l0 -> l2 : e^0'=e^post_3, n^0'=n^post_3, [ 1<=e^0 && 101<=n^0 && n^post_3==-10+n^0 && e^post_3==-1+e^0 ], cost: 1
      1: l1 -> l0 : e^0'=e^post_2, n^0'=n^post_2, [ e^0==e^post_2 && n^0==n^post_2 ], cost: 1
      3: l2 -> l0 : e^0'=e^post_4, n^0'=n^post_4, [ e^0==e^post_4 && n^0==n^post_4 ], cost: 1
      4: l3 -> l0 : e^0'=e^post_5, n^0'=n^post_5, [ n^post_5==n^post_5 && e^post_5==1 ], cost: 1
      5: l4 -> l3 : e^0'=e^post_6, n^0'=n^post_6, [ e^0==e^post_6 && n^0==n^post_6 ], cost: 1

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

Simplified all rules, resulting in:
   Start location: l4
      0: l0 -> l1 : e^0'=1+e^0, n^0'=11+n^0, [ 1<=e^0 && n^0<=100 ], cost: 1
      2: l0 -> l2 : e^0'=-1+e^0, n^0'=-10+n^0, [ 1<=e^0 && 101<=n^0 ], cost: 1
      1: l1 -> l0 : [], cost: 1
      3: l2 -> l0 : [], cost: 1
      4: l3 -> l0 : e^0'=1, n^0'=n^post_5, [], cost: 1
      5: l4 -> l3 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l4
      7: l0 -> l0 : e^0'=1+e^0, n^0'=11+n^0, [ 1<=e^0 && n^0<=100 ], cost: 2
      8: l0 -> l0 : e^0'=-1+e^0, n^0'=-10+n^0, [ 1<=e^0 && 101<=n^0 ], cost: 2
      6: l4 -> l0 : e^0'=1, n^0'=n^post_5, [], cost: 2

Accelerating simple loops of location 0.
   Accelerating the following rules:
      7: l0 -> l0 : e^0'=1+e^0, n^0'=11+n^0, [ 1<=e^0 && n^0<=100 ], cost: 2
      8: l0 -> l0 : e^0'=-1+e^0, n^0'=-10+n^0, [ 1<=e^0 && 101<=n^0 ], cost: 2

   Accelerated rule 7 with backward acceleration, yielding the new rule 9.
   Accelerated rule 8 with backward acceleration, yielding the new rule 10.
[accelerate] Nesting with 2 inner and 2 outer candidates
   Removing the simple loops: 7 8.

Accelerated all simple loops using metering functions (where possible):
   Start location: l4
      9: l0 -> l0 : e^0'=k+e^0, n^0'=n^0+11*k, [ 1<=e^0 && k>=0 && -11+n^0+11*k<=100 ], cost: 2*k
     10: l0 -> l0 : e^0'=-k_1+e^0, n^0'=n^0-10*k_1, [ k_1>=0 && 1<=1-k_1+e^0 && 101<=10+n^0-10*k_1 ], cost: 2*k_1
      6: l4 -> l0 : e^0'=1, n^0'=n^post_5, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l4
      6: l4 -> l0 : e^0'=1, n^0'=n^post_5, [], cost: 2
     11: l4 -> l0 : e^0'=1+k, n^0'=11*k+n^post_5, [ k>=0 && -11+11*k+n^post_5<=100 ], cost: 2+2*k
     12: l4 -> l0 : e^0'=1-k_1, n^0'=-10*k_1+n^post_5, [ k_1>=0 && 1<=2-k_1 && 101<=10-10*k_1+n^post_5 ], cost: 2+2*k_1

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l4
     11: l4 -> l0 : e^0'=1+k, n^0'=11*k+n^post_5, [ k>=0 && -11+11*k+n^post_5<=100 ], cost: 2+2*k
     12: l4 -> l0 : e^0'=1-k_1, n^0'=-10*k_1+n^post_5, [ k_1>=0 && 1<=2-k_1 && 101<=10-10*k_1+n^post_5 ], cost: 2+2*k_1

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l4
     11: l4 -> l0 : e^0'=1+k, n^0'=11*k+n^post_5, [ k>=0 && -11+11*k+n^post_5<=100 ], cost: 2+2*k
     12: l4 -> l0 : e^0'=1-k_1, n^0'=-10*k_1+n^post_5, [ k_1>=0 && 1<=2-k_1 && 101<=10-10*k_1+n^post_5 ], cost: 2+2*k_1

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

Computing asymptotic complexity for rule 12
   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:  [ e^0==e^post_6 && n^0==n^post_6 ]

WORST_CASE(Omega(1),?)