WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: l4
      0: l0 -> l1 : __patmp1^0'=__patmp1^post_1, __patmp2^0'=__patmp2^post_1, k_208^0'=k_208^post_1, k_243^0'=k_243^post_1, len_263^0'=len_263^post_1, [ 0<=k_208^0 && __patmp1^post_1==1 && __patmp2^post_1==k_208^0 && len_263^post_1==__patmp1^post_1 && k_243^post_1==__patmp2^post_1 && 0<=-1+k_243^post_1 && k_208^0==k_208^post_1 ], cost: 1
      1: l2 -> l3 : __patmp1^0'=__patmp1^post_2, __patmp2^0'=__patmp2^post_2, k_208^0'=k_208^post_2, k_243^0'=k_243^post_2, len_263^0'=len_263^post_2, [ __patmp1^0==__patmp1^post_2 && __patmp2^0==__patmp2^post_2 && k_208^0==k_208^post_2 && k_243^0==k_243^post_2 && len_263^0==len_263^post_2 ], cost: 1
      2: l4 -> l0 : __patmp1^0'=__patmp1^post_3, __patmp2^0'=__patmp2^post_3, k_208^0'=k_208^post_3, k_243^0'=k_243^post_3, len_263^0'=len_263^post_3, [ __patmp1^0==__patmp1^post_3 && __patmp2^0==__patmp2^post_3 && k_208^0==k_208^post_3 && k_243^0==k_243^post_3 && len_263^0==len_263^post_3 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      2: l4 -> l0 : __patmp1^0'=__patmp1^post_3, __patmp2^0'=__patmp2^post_3, k_208^0'=k_208^post_3, k_243^0'=k_243^post_3, len_263^0'=len_263^post_3, [ __patmp1^0==__patmp1^post_3 && __patmp2^0==__patmp2^post_3 && k_208^0==k_208^post_3 && k_243^0==k_243^post_3 && len_263^0==len_263^post_3 ], cost: 1

Removed unreachable and leaf rules:
   Start location: l4
     <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:  [ __patmp1^0==__patmp1^post_3 && __patmp2^0==__patmp2^post_3 && k_208^0==k_208^post_3 && k_243^0==k_243^post_3 && len_263^0==len_263^post_3 ]

WORST_CASE(Omega(1),?)