NO ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: l9 0: l0 -> l1 : i^0'=i^post_1, k^0'=k^post_1, [ i^post_1==1+i^0 && k^0==k^post_1 ], cost: 1 4: l1 -> l2 : i^0'=i^post_5, k^0'=k^post_5, [ i^0==i^post_5 && k^0==k^post_5 ], cost: 1 1: l2 -> l0 : i^0'=i^post_2, k^0'=k^post_2, [ i^0==i^post_2 && k^0==k^post_2 ], cost: 1 2: l2 -> l0 : i^0'=i^post_3, k^0'=k^post_3, [ i^0==i^post_3 && k^0==k^post_3 ], cost: 1 3: l2 -> l3 : i^0'=i^post_4, k^0'=k^post_4, [ i^0==i^post_4 && k^0==k^post_4 ], cost: 1 11: l3 -> l4 : i^0'=i^post_12, k^0'=k^post_12, [ 1+k^0<=0 && i^0==i^post_12 && k^0==k^post_12 ], cost: 1 12: l3 -> l7 : i^0'=i^post_13, k^0'=k^post_13, [ 0<=k^0 && i^0==i^post_13 && k^0==k^post_13 ], cost: 1 5: l4 -> l5 : i^0'=i^post_6, k^0'=k^post_6, [ i^0==i^post_6 && k^0==k^post_6 ], cost: 1 6: l6 -> l4 : i^0'=i^post_7, k^0'=k^post_7, [ i^0==i^post_7 && k^0==k^post_7 ], cost: 1 7: l6 -> l4 : i^0'=i^post_8, k^0'=k^post_8, [ i^0==i^post_8 && k^0==k^post_8 ], cost: 1 8: l6 -> l5 : i^0'=i^post_9, k^0'=k^post_9, [ i^0==i^post_9 && k^0==k^post_9 ], cost: 1 9: l7 -> l4 : i^0'=i^post_10, k^0'=k^post_10, [ i^0<=k^0 && i^0==i^post_10 && k^0==k^post_10 ], cost: 1 10: l7 -> l6 : i^0'=i^post_11, k^0'=k^post_11, [ 1+k^0<=i^0 && i^0==i^post_11 && k^0==k^post_11 ], cost: 1 13: l8 -> l1 : i^0'=i^post_14, k^0'=k^post_14, [ k^post_14==k^post_14 && i^post_14==0 ], cost: 1 14: l9 -> l8 : i^0'=i^post_15, k^0'=k^post_15, [ i^0==i^post_15 && k^0==k^post_15 ], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 14: l9 -> l8 : i^0'=i^post_15, k^0'=k^post_15, [ i^0==i^post_15 && k^0==k^post_15 ], cost: 1 Removed unreachable and leaf rules: Start location: l9 0: l0 -> l1 : i^0'=i^post_1, k^0'=k^post_1, [ i^post_1==1+i^0 && k^0==k^post_1 ], cost: 1 4: l1 -> l2 : i^0'=i^post_5, k^0'=k^post_5, [ i^0==i^post_5 && k^0==k^post_5 ], cost: 1 1: l2 -> l0 : i^0'=i^post_2, k^0'=k^post_2, [ i^0==i^post_2 && k^0==k^post_2 ], cost: 1 2: l2 -> l0 : i^0'=i^post_3, k^0'=k^post_3, [ i^0==i^post_3 && k^0==k^post_3 ], cost: 1 13: l8 -> l1 : i^0'=i^post_14, k^0'=k^post_14, [ k^post_14==k^post_14 && i^post_14==0 ], cost: 1 14: l9 -> l8 : i^0'=i^post_15, k^0'=k^post_15, [ i^0==i^post_15 && k^0==k^post_15 ], cost: 1 Simplified all rules, resulting in: Start location: l9 0: l0 -> l1 : i^0'=1+i^0, [], cost: 1 4: l1 -> l2 : [], cost: 1 2: l2 -> l0 : [], cost: 1 13: l8 -> l1 : i^0'=0, k^0'=k^post_14, [], cost: 1 14: l9 -> l8 : [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: l9 17: l1 -> l1 : i^0'=1+i^0, [], cost: 3 15: l9 -> l1 : i^0'=0, k^0'=k^post_14, [], cost: 2 Accelerating simple loops of location 1. Accelerating the following rules: 17: l1 -> l1 : i^0'=1+i^0, [], cost: 3 Accelerated rule 17 with non-termination, yielding the new rule 18. [accelerate] Nesting with 0 inner and 0 outer candidates Removing the simple loops: 17. Accelerated all simple loops using metering functions (where possible): Start location: l9 18: l1 -> [10] : [], cost: NONTERM 15: l9 -> l1 : i^0'=0, k^0'=k^post_14, [], cost: 2 Chained accelerated rules (with incoming rules): Start location: l9 15: l9 -> l1 : i^0'=0, k^0'=k^post_14, [], cost: 2 19: l9 -> [10] : [], cost: NONTERM Removed unreachable locations (and leaf rules with constant cost): Start location: l9 19: l9 -> [10] : [], cost: NONTERM ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: l9 19: l9 -> [10] : [], cost: NONTERM Computing asymptotic complexity for rule 19 Guard is satisfiable, yielding nontermination Resulting cost NONTERM has complexity: Nonterm Found new complexity Nonterm. Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Nonterm Cpx degree: Nonterm Solved cost: NONTERM Rule cost: NONTERM Rule guard: [] NO