WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: l3
      0: l0 -> l1 : c^0'=c^post_1, p^0'=p^post_1, s^0'=s^post_1, [ 1+p^0<=c^0 && p^post_1==s^0 && s^post_1==1+s^0 && c^0==c^post_1 ], cost: 1
      1: l1 -> l0 : c^0'=c^post_2, p^0'=p^post_2, s^0'=s^post_2, [ c^0==c^post_2 && p^0==p^post_2 && s^0==s^post_2 ], cost: 1
      2: l2 -> l0 : c^0'=c^post_3, p^0'=p^post_3, s^0'=s^post_3, [ s^post_3==1 && p^post_3==1 && c^0==c^post_3 ], cost: 1
      3: l3 -> l2 : c^0'=c^post_4, p^0'=p^post_4, s^0'=s^post_4, [ c^0==c^post_4 && p^0==p^post_4 && s^0==s^post_4 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      3: l3 -> l2 : c^0'=c^post_4, p^0'=p^post_4, s^0'=s^post_4, [ c^0==c^post_4 && p^0==p^post_4 && s^0==s^post_4 ], cost: 1

Simplified all rules, resulting in:
   Start location: l3
      0: l0 -> l1 : p^0'=s^0, s^0'=1+s^0, [ 1+p^0<=c^0 ], cost: 1
      1: l1 -> l0 : [], cost: 1
      2: l2 -> l0 : p^0'=1, s^0'=1, [], cost: 1
      3: l3 -> l2 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l3
      5: l0 -> l0 : p^0'=s^0, s^0'=1+s^0, [ 1+p^0<=c^0 ], cost: 2
      4: l3 -> l0 : p^0'=1, s^0'=1, [], cost: 2

Accelerating simple loops of location 0.
   Accelerating the following rules:
      5: l0 -> l0 : p^0'=s^0, s^0'=1+s^0, [ 1+p^0<=c^0 ], cost: 2

[test] deduced invariant p^0-s^0<=0
   Accelerated rule 5 with backward acceleration, yielding the new rule 6.
[accelerate] Nesting with 1 inner and 1 outer candidates

Accelerated all simple loops using metering functions (where possible):
   Start location: l3
      5: l0 -> l0 : p^0'=s^0, s^0'=1+s^0, [ 1+p^0<=c^0 ], cost: 2
      6: l0 -> l0 : p^0'=c^0, s^0'=1+c^0, [ p^0-s^0<=0 && 1-s^0+c^0>=1 ], cost: 2-2*s^0+2*c^0
      4: l3 -> l0 : p^0'=1, s^0'=1, [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l3
      4: l3 -> l0 : p^0'=1, s^0'=1, [], cost: 2
      7: l3 -> l0 : p^0'=1, s^0'=2, [ 2<=c^0 ], cost: 4
      8: l3 -> l0 : p^0'=c^0, s^0'=1+c^0, [ c^0>=1 ], cost: 2+2*c^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l3
      8: l3 -> l0 : p^0'=c^0, s^0'=1+c^0, [ c^0>=1 ], cost: 2+2*c^0

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l3
      8: l3 -> l0 : p^0'=c^0, s^0'=1+c^0, [ c^0>=1 ], cost: 2+2*c^0

Computing asymptotic complexity for rule 8
   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:  [ c^0==c^post_4 && p^0==p^post_4 && s^0==s^post_4 ]

WORST_CASE(Omega(1),?)