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, [ 1<=x^0 && x^post_1==z^0+x^0 && z^post_1==-1+z^0 && y^0==y^post_1 ], cost: 1
      2: l0 -> l2 : x^0'=x^post_3, y^0'=y^post_3, z^0'=z^post_3, [ 1<=x^0 && x^post_3==y^0+x^0 && y^post_3==-1+y^0 && 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, z^0'=-1+z^0, [ 1<=x^0 ], cost: 1
      2: l0 -> l2 : x^0'=y^0+x^0, y^0'=-1+y^0, [ 1<=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, z^0'=-1+z^0, [ 1<=x^0 ], cost: 2
      8: l0 -> l0 : x^0'=y^0+x^0, y^0'=-1+y^0, [ 1<=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, z^0'=-1+z^0, [ 1<=x^0 ], cost: 2
      8: l0 -> l0 : x^0'=y^0+x^0, y^0'=-1+y^0, [ 1<=x^0 ], cost: 2

[test] deduced pseudo-invariant 5+z^0<=0, also trying -5-z^0<=-1
[test] deduced pseudo-invariant 1+z^0<=0, also trying -1-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.
   Accelerated rule 7 with backward acceleration, yielding the new rule 11.
[test] deduced pseudo-invariant 5+y^0<=0, also trying -5-y^0<=-1
[test] deduced pseudo-invariant 5+2*y^0<=0, also trying -5-2*y^0<=-1
[test] deduced pseudo-invariant 1+y^0<=0, also trying -1-y^0<=-1
   Accelerated rule 8 with backward acceleration, yielding the new rule 12.
   Accelerated rule 8 with backward acceleration, yielding the new rule 13.
   Accelerated rule 8 with backward acceleration, yielding the new rule 14.
   Accelerated rule 8 with backward acceleration, yielding the new rule 15.
[accelerate] Nesting with 7 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, z^0'=-1+z^0, [ 1<=x^0 ], cost: 2
      8: l0 -> l0 : x^0'=y^0+x^0, y^0'=-1+y^0, [ 1<=x^0 ], cost: 2
      9: l0 -> l0 : x^0'=z^0*k-1/2*k^2+1/2*k+x^0, z^0'=z^0-k, [ 5+z^0<=0 && k>=0 && 1<=-1/2+z^0*(-1+k)-1/2*(-1+k)^2+1/2*k+x^0 ], cost: 2*k
     10: l0 -> l0 : x^0'=5/2-1/2*(5+z^0)^2+1/2*z^0+x^0+(5+z^0)*z^0, z^0'=-5, [ 1+z^0<=0 && 5+z^0>=0 && 1<=2-1/2*(4+z^0)^2+1/2*z^0+(4+z^0)*z^0+x^0 ], cost: 10+2*z^0
     11: l0 -> l0 : x^0'=1/2-1/2*(1+z^0)^2+1/2*z^0+z^0*(1+z^0)+x^0, z^0'=-1, [ 1<=x^0 && 1+z^0>=0 ], cost: 2+2*z^0
     12: l0 -> l0 : x^0'=1/2*k_4-1/2*k_4^2+x^0+y^0*k_4, y^0'=y^0-k_4, [ 5+y^0<=0 && k_4>=0 && 1<=-1/2-1/2*(-1+k_4)^2+1/2*k_4+y^0*(-1+k_4)+x^0 ], cost: 2*k_4
     13: l0 -> l0 : x^0'=5/2+1/2*y^0-1/2*(5+y^0)^2+x^0+y^0*(5+y^0), y^0'=-5, [ 5+2*y^0<=0 && 5+y^0>=0 && 1<=2+1/2*y^0+y^0*(4+y^0)+x^0-1/2*(4+y^0)^2 ], cost: 10+2*y^0
     14: l0 -> l0 : x^0'=3/2-1/2*(3+y^0)^2+y^0*(3+y^0)+1/2*y^0+x^0, y^0'=-3, [ 1+y^0<=0 && 3+y^0>=0 && 1<=1+1/2*y^0-1/2*(2+y^0)^2+y^0*(2+y^0)+x^0 ], cost: 6+2*y^0
     15: l0 -> l0 : x^0'=1/2+1/2*y^0+y^0*(1+y^0)-1/2*(1+y^0)^2+x^0, y^0'=-1, [ 1<=x^0 && -5-2*y^0<=-1 && 1+y^0>=0 ], cost: 2+2*y^0
      6: l4 -> l0 : [], cost: 2

Chained accelerated rules (with incoming rules):
   Start location: l4
      6: l4 -> l0 : [], cost: 2
     16: l4 -> l0 : x^0'=z^0+x^0, z^0'=-1+z^0, [ 1<=x^0 ], cost: 4
     17: l4 -> l0 : x^0'=y^0+x^0, y^0'=-1+y^0, [ 1<=x^0 ], cost: 4
     18: l4 -> l0 : x^0'=z^0*k-1/2*k^2+1/2*k+x^0, z^0'=z^0-k, [ 5+z^0<=0 && k>=0 && 1<=-1/2+z^0*(-1+k)-1/2*(-1+k)^2+1/2*k+x^0 ], cost: 2+2*k
     19: l4 -> l0 : x^0'=5/2-1/2*(5+z^0)^2+1/2*z^0+x^0+(5+z^0)*z^0, z^0'=-5, [ 1+z^0<=0 && 5+z^0>=0 && 1<=2-1/2*(4+z^0)^2+1/2*z^0+(4+z^0)*z^0+x^0 ], cost: 12+2*z^0
     20: l4 -> l0 : x^0'=1/2-1/2*(1+z^0)^2+1/2*z^0+z^0*(1+z^0)+x^0, z^0'=-1, [ 1<=x^0 && 1+z^0>=0 ], cost: 4+2*z^0
     21: l4 -> l0 : x^0'=1/2*k_4-1/2*k_4^2+x^0+y^0*k_4, y^0'=y^0-k_4, [ 5+y^0<=0 && k_4>=0 && 1<=-1/2-1/2*(-1+k_4)^2+1/2*k_4+y^0*(-1+k_4)+x^0 ], cost: 2+2*k_4
     22: l4 -> l0 : x^0'=5/2+1/2*y^0-1/2*(5+y^0)^2+x^0+y^0*(5+y^0), y^0'=-5, [ 5+2*y^0<=0 && 5+y^0>=0 && 1<=2+1/2*y^0+y^0*(4+y^0)+x^0-1/2*(4+y^0)^2 ], cost: 12+2*y^0
     23: l4 -> l0 : x^0'=3/2-1/2*(3+y^0)^2+y^0*(3+y^0)+1/2*y^0+x^0, y^0'=-3, [ 1+y^0<=0 && 3+y^0>=0 && 1<=1+1/2*y^0-1/2*(2+y^0)^2+y^0*(2+y^0)+x^0 ], cost: 8+2*y^0
     24: l4 -> l0 : x^0'=1/2+1/2*y^0+y^0*(1+y^0)-1/2*(1+y^0)^2+x^0, y^0'=-1, [ 1<=x^0 && -5-2*y^0<=-1 && 1+y^0>=0 ], cost: 4+2*y^0

Removed unreachable locations (and leaf rules with constant cost):
   Start location: l4
     18: l4 -> l0 : x^0'=z^0*k-1/2*k^2+1/2*k+x^0, z^0'=z^0-k, [ 5+z^0<=0 && k>=0 && 1<=-1/2+z^0*(-1+k)-1/2*(-1+k)^2+1/2*k+x^0 ], cost: 2+2*k
     19: l4 -> l0 : x^0'=5/2-1/2*(5+z^0)^2+1/2*z^0+x^0+(5+z^0)*z^0, z^0'=-5, [ 1+z^0<=0 && 5+z^0>=0 && 1<=2-1/2*(4+z^0)^2+1/2*z^0+(4+z^0)*z^0+x^0 ], cost: 12+2*z^0
     20: l4 -> l0 : x^0'=1/2-1/2*(1+z^0)^2+1/2*z^0+z^0*(1+z^0)+x^0, z^0'=-1, [ 1<=x^0 && 1+z^0>=0 ], cost: 4+2*z^0
     21: l4 -> l0 : x^0'=1/2*k_4-1/2*k_4^2+x^0+y^0*k_4, y^0'=y^0-k_4, [ 5+y^0<=0 && k_4>=0 && 1<=-1/2-1/2*(-1+k_4)^2+1/2*k_4+y^0*(-1+k_4)+x^0 ], cost: 2+2*k_4
     22: l4 -> l0 : x^0'=5/2+1/2*y^0-1/2*(5+y^0)^2+x^0+y^0*(5+y^0), y^0'=-5, [ 5+2*y^0<=0 && 5+y^0>=0 && 1<=2+1/2*y^0+y^0*(4+y^0)+x^0-1/2*(4+y^0)^2 ], cost: 12+2*y^0
     23: l4 -> l0 : x^0'=3/2-1/2*(3+y^0)^2+y^0*(3+y^0)+1/2*y^0+x^0, y^0'=-3, [ 1+y^0<=0 && 3+y^0>=0 && 1<=1+1/2*y^0-1/2*(2+y^0)^2+y^0*(2+y^0)+x^0 ], cost: 8+2*y^0
     24: l4 -> l0 : x^0'=1/2+1/2*y^0+y^0*(1+y^0)-1/2*(1+y^0)^2+x^0, y^0'=-1, [ 1<=x^0 && -5-2*y^0<=-1 && 1+y^0>=0 ], cost: 4+2*y^0

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: l4
     18: l4 -> l0 : x^0'=z^0*k-1/2*k^2+1/2*k+x^0, z^0'=z^0-k, [ 5+z^0<=0 && k>=0 && 1<=-1/2+z^0*(-1+k)-1/2*(-1+k)^2+1/2*k+x^0 ], cost: 2+2*k
     19: l4 -> l0 : x^0'=5/2-1/2*(5+z^0)^2+1/2*z^0+x^0+(5+z^0)*z^0, z^0'=-5, [ 1+z^0<=0 && 5+z^0>=0 && 1<=2-1/2*(4+z^0)^2+1/2*z^0+(4+z^0)*z^0+x^0 ], cost: 12+2*z^0
     20: l4 -> l0 : x^0'=1/2-1/2*(1+z^0)^2+1/2*z^0+z^0*(1+z^0)+x^0, z^0'=-1, [ 1<=x^0 && 1+z^0>=0 ], cost: 4+2*z^0
     21: l4 -> l0 : x^0'=1/2*k_4-1/2*k_4^2+x^0+y^0*k_4, y^0'=y^0-k_4, [ 5+y^0<=0 && k_4>=0 && 1<=-1/2-1/2*(-1+k_4)^2+1/2*k_4+y^0*(-1+k_4)+x^0 ], cost: 2+2*k_4
     22: l4 -> l0 : x^0'=5/2+1/2*y^0-1/2*(5+y^0)^2+x^0+y^0*(5+y^0), y^0'=-5, [ 5+2*y^0<=0 && 5+y^0>=0 && 1<=2+1/2*y^0+y^0*(4+y^0)+x^0-1/2*(4+y^0)^2 ], cost: 12+2*y^0
     23: l4 -> l0 : x^0'=3/2-1/2*(3+y^0)^2+y^0*(3+y^0)+1/2*y^0+x^0, y^0'=-3, [ 1+y^0<=0 && 3+y^0>=0 && 1<=1+1/2*y^0-1/2*(2+y^0)^2+y^0*(2+y^0)+x^0 ], cost: 8+2*y^0
     24: l4 -> l0 : x^0'=1/2+1/2*y^0+y^0*(1+y^0)-1/2*(1+y^0)^2+x^0, y^0'=-1, [ 1<=x^0 && -5-2*y^0<=-1 && 1+y^0>=0 ], cost: 4+2*y^0

Computing asymptotic complexity for rule 18
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 21
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 20
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 19
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 22
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 23
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 24
   Simplified the guard:
     24: l4 -> l0 : x^0'=1/2+1/2*y^0+y^0*(1+y^0)-1/2*(1+y^0)^2+x^0, y^0'=-1, [ 1<=x^0 && 1+y^0>=0 ], cost: 4+2*y^0
   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),?)