WORST_CASE(Omega(n^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 && 1<=y^0 && x^0==x^post_1 && y^0==y^post_1 ], cost: 1
      1: l1 -> l0 : x^0'=x^post_2, y^0'=y^post_2, [ x^post_2==-1+x^0 && y^0==y^post_2 ], cost: 1
      2: l2 -> l1 : x^0'=x^post_3, y^0'=y^post_3, [ y^post_3==-1+y^0 && x^0==x^post_3 ], cost: 1
      3: l2 -> l0 : x^0'=x^post_4, y^0'=y^post_4, [ x^0==x^post_4 && y^0==y^post_4 ], cost: 1
      4: l3 -> l2 : x^0'=x^post_5, y^0'=y^post_5, [ x^0==x^post_5 && y^0==y^post_5 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      4: l3 -> l2 : x^0'=x^post_5, y^0'=y^post_5, [ x^0==x^post_5 && y^0==y^post_5 ], cost: 1

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

### Simplification by acceleration and chaining ###

Eliminated locations (on tree-shaped paths):
   Start location: l3
      0: l0 -> l1 : [ 1<=x^0 && 1<=y^0 ], cost: 1
      1: l1 -> l0 : x^0'=-1+x^0, [], cost: 1
      5: l3 -> l1 : y^0'=-1+y^0, [], cost: 2
      6: l3 -> l0 : [], cost: 2

Eliminated location l0 (as a last resort):
   Start location: l3
      7: l1 -> l1 : x^0'=-1+x^0, [ 1<=-1+x^0 && 1<=y^0 ], cost: 2
      5: l3 -> l1 : y^0'=-1+y^0, [], cost: 2
      8: l3 -> l1 : [ 1<=x^0 && 1<=y^0 ], cost: 3

Accelerating simple loops of location 1.
   Accelerating the following rules:
      7: l1 -> l1 : x^0'=-1+x^0, [ 1<=-1+x^0 && 1<=y^0 ], cost: 2

   Accelerated rule 7 with metering function -1+x^0, yielding the new rule 9.
   Removing the simple loops: 7.

Accelerated all simple loops using metering functions (where possible):
   Start location: l3
      9: l1 -> l1 : x^0'=1, [ 1<=-1+x^0 && 1<=y^0 ], cost: -2+2*x^0
      5: l3 -> l1 : y^0'=-1+y^0, [], cost: 2
      8: l3 -> l1 : [ 1<=x^0 && 1<=y^0 ], cost: 3

Chained accelerated rules (with incoming rules):
   Start location: l3
      5: l3 -> l1 : y^0'=-1+y^0, [], cost: 2
      8: l3 -> l1 : [ 1<=x^0 && 1<=y^0 ], cost: 3
     10: l3 -> l1 : x^0'=1, y^0'=-1+y^0, [ 1<=-1+x^0 && 1<=-1+y^0 ], cost: 2*x^0
     11: l3 -> l1 : x^0'=1, [ 1<=y^0 && 1<=-1+x^0 ], cost: 1+2*x^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l3
     10: l3 -> l1 : x^0'=1, y^0'=-1+y^0, [ 1<=-1+x^0 && 1<=-1+y^0 ], cost: 2*x^0
     11: l3 -> l1 : x^0'=1, [ 1<=y^0 && 1<=-1+x^0 ], cost: 1+2*x^0

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l3
     10: l3 -> l1 : x^0'=1, y^0'=-1+y^0, [ 1<=-1+x^0 && 1<=-1+y^0 ], cost: 2*x^0
     11: l3 -> l1 : x^0'=1, [ 1<=y^0 && 1<=-1+x^0 ], cost: 1+2*x^0

Computing asymptotic complexity for rule 10
   Solved the limit problem by the following transformations:
      Created initial limit problem:
      -1+y^0 (+/+!), -1+x^0 (+/+!), 2*x^0 (+) [not solved]

      removing all constraints (solved by SMT)
      resulting limit problem: [solved]

      applying transformation rule (C) using substitution {x^0==n,y^0==2}
      resulting limit problem:
      [solved]

   Solution:
      x^0 / n
      y^0 / 2
   Resulting cost 2*n has complexity: Poly(n^1)

Found new complexity Poly(n^1).

Obtained the following overall complexity (w.r.t. the length of the input n):
   Complexity:  Poly(n^1)
   Cpx degree:  1
   Solved cost: 2*n
   Rule cost:   2*x^0
   Rule guard:  [ 1<=-1+x^0 && 1<=-1+y^0 ]

WORST_CASE(Omega(n^1),?)