WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: l2
      0: l0 -> l1 : a^0'=a^post_1, ret_returnOne2^0'=ret_returnOne2^post_1, [ a^1_1==-1 && ret_returnOne2^post_1==1 && a^post_1==ret_returnOne2^post_1 ], cost: 1
      1: l2 -> l0 : a^0'=a^post_2, ret_returnOne2^0'=ret_returnOne2^post_2, [ a^0==a^post_2 && ret_returnOne2^0==ret_returnOne2^post_2 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      1: l2 -> l0 : a^0'=a^post_2, ret_returnOne2^0'=ret_returnOne2^post_2, [ a^0==a^post_2 && ret_returnOne2^0==ret_returnOne2^post_2 ], cost: 1

Removed unreachable and leaf rules:
   Start location: l2
     <empty>

Empty problem, aborting

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_2 && ret_returnOne2^0==ret_returnOne2^post_2 ]

WORST_CASE(Omega(1),?)