WORST_CASE(Omega(1),?)

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

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

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

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

### Simplification by acceleration and chaining ###

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

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

[test] deduced pseudo-invariant -y^0<=0, also trying y^0<=-1
   Accelerated rule 5 with backward acceleration, yielding the new rule 6.
   Accelerated rule 5 with backward acceleration, yielding the new rule 7.
[accelerate] Nesting with 2 inner and 1 outer candidates

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

Chained accelerated rules (with incoming rules):
   Start location: l3
      4: l3 -> l0 : [], cost: 2
      8: l3 -> l0 : x^0'=-y^0+x^0, y^0'=1+y^0, [ 1<=x^0 ], cost: 4
      9: l3 -> l0 : x^0'=-y^0*k+x^0+1/2*k-1/2*k^2, y^0'=y^0+k, [ -y^0<=0 && k>=0 && 1<=-1/2-y^0*(-1+k)-1/2*(-1+k)^2+x^0+1/2*k ], cost: 2+2*k
     10: l3 -> l0 : x^0'=-1/2*y^0+1/2*y^0^2+x^0, y^0'=0, [ 1<=x^0 && -y^0>=0 ], cost: 2-2*y^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l3
      9: l3 -> l0 : x^0'=-y^0*k+x^0+1/2*k-1/2*k^2, y^0'=y^0+k, [ -y^0<=0 && k>=0 && 1<=-1/2-y^0*(-1+k)-1/2*(-1+k)^2+x^0+1/2*k ], cost: 2+2*k
     10: l3 -> l0 : x^0'=-1/2*y^0+1/2*y^0^2+x^0, y^0'=0, [ 1<=x^0 && -y^0>=0 ], cost: 2-2*y^0

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l3
      9: l3 -> l0 : x^0'=-y^0*k+x^0+1/2*k-1/2*k^2, y^0'=y^0+k, [ -y^0<=0 && k>=0 && 1<=-1/2-y^0*(-1+k)-1/2*(-1+k)^2+x^0+1/2*k ], cost: 2+2*k
     10: l3 -> l0 : x^0'=-1/2*y^0+1/2*y^0^2+x^0, y^0'=0, [ 1<=x^0 && -y^0>=0 ], cost: 2-2*y^0

Computing asymptotic complexity for rule 9
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 10
   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:  [ x^0==x^post_4 && y^0==y^post_4 ]

WORST_CASE(Omega(1),?)