WORST_CASE(Omega(1),?) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: l3 0: l0 -> l1 : x^0'=x^post_1, [ x^0==x^post_1 ], cost: 1 1: l1 -> l2 : x^0'=x^post_2, [ 0<=x^0 && x^1_1==x^0 && x^post_2==-1+x^1_1 ], cost: 1 2: l2 -> l1 : x^0'=x^post_3, [ x^0==x^post_3 ], cost: 1 3: l3 -> l0 : x^0'=x^post_4, [ x^0==x^post_4 ], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 3: l3 -> l0 : x^0'=x^post_4, [ x^0==x^post_4 ], cost: 1 Simplified all rules, resulting in: Start location: l3 0: l0 -> l1 : [], cost: 1 1: l1 -> l2 : x^0'=-1+x^0, [ 0<=x^0 ], cost: 1 2: l2 -> l1 : [], cost: 1 3: l3 -> l0 : [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: l3 5: l1 -> l1 : x^0'=-1+x^0, [ 0<=x^0 ], cost: 2 4: l3 -> l1 : [], cost: 2 Accelerating simple loops of location 1. Accelerating the following rules: 5: l1 -> l1 : x^0'=-1+x^0, [ 0<=x^0 ], cost: 2 Accelerated rule 5 with backward acceleration, yielding the new rule 6. [accelerate] Nesting with 1 inner and 1 outer candidates Removing the simple loops: 5. Accelerated all simple loops using metering functions (where possible): Start location: l3 6: l1 -> l1 : x^0'=-1, [ 1+x^0>=0 ], cost: 2+2*x^0 4: l3 -> l1 : [], cost: 2 Chained accelerated rules (with incoming rules): Start location: l3 4: l3 -> l1 : [], cost: 2 7: l3 -> l1 : x^0'=-1, [ 1+x^0>=0 ], cost: 4+2*x^0 Removed unreachable locations (and leaf rules with constant cost): Start location: l3 7: l3 -> l1 : x^0'=-1, [ 1+x^0>=0 ], cost: 4+2*x^0 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: l3 7: l3 -> l1 : x^0'=-1, [ 1+x^0>=0 ], cost: 4+2*x^0 Computing asymptotic complexity for rule 7 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 ] WORST_CASE(Omega(1),?)