WORST_CASE(Omega(1),?)

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

Initial linear ITS problem
   Start location: l4
      0: l0 -> l1 : x^0'=x^post_1, y^0'=y^post_1, z^0'=z^post_1, [ y^0<=x^0 && x^post_1==-z^0+x^0 && y^post_1==y^0-2*z^0 && z^post_1==-1+z^0 ], cost: 1
      2: l0 -> l2 : x^0'=x^post_3, y^0'=y^post_3, z^0'=z^post_3, [ y^0<=x^0 && x^post_3==1+x^0 && y^post_3==y^0+x^post_3 && z^0==z^post_3 ], cost: 1
      1: l1 -> l0 : x^0'=x^post_2, y^0'=y^post_2, z^0'=z^post_2, [ x^0==x^post_2 && y^0==y^post_2 && z^0==z^post_2 ], cost: 1
      3: l2 -> l0 : x^0'=x^post_4, y^0'=y^post_4, z^0'=z^post_4, [ x^0==x^post_4 && y^0==y^post_4 && z^0==z^post_4 ], cost: 1
      4: l3 -> l0 : x^0'=x^post_5, y^0'=y^post_5, z^0'=z^post_5, [ x^0==x^post_5 && y^0==y^post_5 && z^0==z^post_5 ], cost: 1
      5: l4 -> l3 : x^0'=x^post_6, y^0'=y^post_6, z^0'=z^post_6, [ x^0==x^post_6 && y^0==y^post_6 && z^0==z^post_6 ], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      5: l4 -> l3 : x^0'=x^post_6, y^0'=y^post_6, z^0'=z^post_6, [ x^0==x^post_6 && y^0==y^post_6 && z^0==z^post_6 ], cost: 1

Simplified all rules, resulting in:
   Start location: l4
      0: l0 -> l1 : x^0'=-z^0+x^0, y^0'=y^0-2*z^0, z^0'=-1+z^0, [ y^0<=x^0 ], cost: 1
      2: l0 -> l2 : x^0'=1+x^0, y^0'=1+y^0+x^0, [ y^0<=x^0 ], cost: 1
      1: l1 -> l0 : [], cost: 1
      3: l2 -> l0 : [], cost: 1
      4: l3 -> l0 : [], cost: 1
      5: l4 -> l3 : [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: l4
      7: l0 -> l0 : x^0'=-z^0+x^0, y^0'=y^0-2*z^0, z^0'=-1+z^0, [ y^0<=x^0 ], cost: 2
      8: l0 -> l0 : x^0'=1+x^0, y^0'=1+y^0+x^0, [ y^0<=x^0 ], cost: 2
      6: l4 -> l0 : [], cost: 2

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

[test] deduced pseudo-invariant z^0<=0, also trying -z^0<=-1
   Accelerated rule 7 with backward acceleration, yielding the new rule 9.
   Accelerated rule 7 with backward acceleration, yielding the new rule 10.
[test] deduced pseudo-invariant -x^0<=0, also trying x^0<=-1
   Accelerated rule 8 with backward acceleration, yielding the new rule 11.
   Accelerated rule 8 with backward acceleration, yielding the new rule 12.
[accelerate] Nesting with 4 inner and 2 outer candidates

Accelerated all simple loops using metering functions (where possible):
   Start location: l4
      7: l0 -> l0 : x^0'=-z^0+x^0, y^0'=y^0-2*z^0, z^0'=-1+z^0, [ y^0<=x^0 ], cost: 2
      8: l0 -> l0 : x^0'=1+x^0, y^0'=1+y^0+x^0, [ y^0<=x^0 ], cost: 2
      9: l0 -> l0 : x^0'=-z^0*k+1/2*k^2-1/2*k+x^0, y^0'=-2*z^0*k+y^0+k^2-k, z^0'=z^0-k, [ z^0<=0 && k>=0 && 1+y^0+(-1+k)^2-2*z^0*(-1+k)-k<=1/2+1/2*(-1+k)^2-z^0*(-1+k)-1/2*k+x^0 ], cost: 2*k
     10: l0 -> l0 : x^0'=-1/2*z^0^2-1/2*z^0+x^0, y^0'=-z^0^2+y^0-z^0, z^0'=0, [ y^0<=x^0 && z^0>=0 ], cost: 2*z^0
     11: l0 -> l0 : x^0'=k_3+x^0, y^0'=1/2*k_3+1/2*k_3^2+k_3*x^0+y^0, [ -x^0<=0 && k_3>=0 && -1/2+1/2*k_3+y^0+(-1+k_3)*x^0+1/2*(-1+k_3)^2<=-1+k_3+x^0 ], cost: 2*k_3
     12: l0 -> l0 : x^0'=0, y^0'=y^0-1/2*x^0^2-1/2*x^0, [ y^0<=x^0 && -x^0>=0 ], cost: -2*x^0
      6: l4 -> l0 : [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l4
      6: l4 -> l0 : [], cost: 2
     13: l4 -> l0 : x^0'=-z^0+x^0, y^0'=y^0-2*z^0, z^0'=-1+z^0, [ y^0<=x^0 ], cost: 4
     14: l4 -> l0 : x^0'=1+x^0, y^0'=1+y^0+x^0, [ y^0<=x^0 ], cost: 4
     15: l4 -> l0 : x^0'=-z^0*k+1/2*k^2-1/2*k+x^0, y^0'=-2*z^0*k+y^0+k^2-k, z^0'=z^0-k, [ z^0<=0 && k>=0 && 1+y^0+(-1+k)^2-2*z^0*(-1+k)-k<=1/2+1/2*(-1+k)^2-z^0*(-1+k)-1/2*k+x^0 ], cost: 2+2*k
     16: l4 -> l0 : x^0'=-1/2*z^0^2-1/2*z^0+x^0, y^0'=-z^0^2+y^0-z^0, z^0'=0, [ y^0<=x^0 && z^0>=0 ], cost: 2+2*z^0
     17: l4 -> l0 : x^0'=k_3+x^0, y^0'=1/2*k_3+1/2*k_3^2+k_3*x^0+y^0, [ -x^0<=0 && k_3>=0 && -1/2+1/2*k_3+y^0+(-1+k_3)*x^0+1/2*(-1+k_3)^2<=-1+k_3+x^0 ], cost: 2+2*k_3
     18: l4 -> l0 : x^0'=0, y^0'=y^0-1/2*x^0^2-1/2*x^0, [ y^0<=x^0 && -x^0>=0 ], cost: 2-2*x^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l4
     15: l4 -> l0 : x^0'=-z^0*k+1/2*k^2-1/2*k+x^0, y^0'=-2*z^0*k+y^0+k^2-k, z^0'=z^0-k, [ z^0<=0 && k>=0 && 1+y^0+(-1+k)^2-2*z^0*(-1+k)-k<=1/2+1/2*(-1+k)^2-z^0*(-1+k)-1/2*k+x^0 ], cost: 2+2*k
     16: l4 -> l0 : x^0'=-1/2*z^0^2-1/2*z^0+x^0, y^0'=-z^0^2+y^0-z^0, z^0'=0, [ y^0<=x^0 && z^0>=0 ], cost: 2+2*z^0
     17: l4 -> l0 : x^0'=k_3+x^0, y^0'=1/2*k_3+1/2*k_3^2+k_3*x^0+y^0, [ -x^0<=0 && k_3>=0 && -1/2+1/2*k_3+y^0+(-1+k_3)*x^0+1/2*(-1+k_3)^2<=-1+k_3+x^0 ], cost: 2+2*k_3
     18: l4 -> l0 : x^0'=0, y^0'=y^0-1/2*x^0^2-1/2*x^0, [ y^0<=x^0 && -x^0>=0 ], cost: 2-2*x^0

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l4
     15: l4 -> l0 : x^0'=-z^0*k+1/2*k^2-1/2*k+x^0, y^0'=-2*z^0*k+y^0+k^2-k, z^0'=z^0-k, [ z^0<=0 && k>=0 && 1+y^0+(-1+k)^2-2*z^0*(-1+k)-k<=1/2+1/2*(-1+k)^2-z^0*(-1+k)-1/2*k+x^0 ], cost: 2+2*k
     16: l4 -> l0 : x^0'=-1/2*z^0^2-1/2*z^0+x^0, y^0'=-z^0^2+y^0-z^0, z^0'=0, [ y^0<=x^0 && z^0>=0 ], cost: 2+2*z^0
     17: l4 -> l0 : x^0'=k_3+x^0, y^0'=1/2*k_3+1/2*k_3^2+k_3*x^0+y^0, [ -x^0<=0 && k_3>=0 && -1/2+1/2*k_3+y^0+(-1+k_3)*x^0+1/2*(-1+k_3)^2<=-1+k_3+x^0 ], cost: 2+2*k_3
     18: l4 -> l0 : x^0'=0, y^0'=y^0-1/2*x^0^2-1/2*x^0, [ y^0<=x^0 && -x^0>=0 ], cost: 2-2*x^0

Computing asymptotic complexity for rule 15
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 17
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 16
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 18
   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_6 && y^0==y^post_6 && z^0==z^post_6 ]

WORST_CASE(Omega(1),?)