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),?)