NO ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: l4 0: l0 -> l2 : __disjvr_0^0'=__disjvr_0^post_1, x^0'=x^post_1, [ x^post_1==-1+x^0 && __disjvr_0^0==__disjvr_0^post_1 ], cost: 1 1: l2 -> l1 : __disjvr_0^0'=__disjvr_0^post_2, x^0'=x^post_2, [ __disjvr_0^post_2==__disjvr_0^0 && __disjvr_0^0==__disjvr_0^post_2 && x^0==x^post_2 ], cost: 1 2: l1 -> l0 : __disjvr_0^0'=__disjvr_0^post_3, x^0'=x^post_3, [ __disjvr_0^0==__disjvr_0^post_3 && x^0==x^post_3 ], cost: 1 3: l3 -> l0 : __disjvr_0^0'=__disjvr_0^post_4, x^0'=x^post_4, [ 1<=x^0 && __disjvr_0^0==__disjvr_0^post_4 && x^0==x^post_4 ], cost: 1 4: l4 -> l3 : __disjvr_0^0'=__disjvr_0^post_5, x^0'=x^post_5, [ __disjvr_0^0==__disjvr_0^post_5 && x^0==x^post_5 ], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 4: l4 -> l3 : __disjvr_0^0'=__disjvr_0^post_5, x^0'=x^post_5, [ __disjvr_0^0==__disjvr_0^post_5 && x^0==x^post_5 ], cost: 1 Simplified all rules, resulting in: Start location: l4 0: l0 -> l2 : x^0'=-1+x^0, [], cost: 1 1: l2 -> l1 : [], cost: 1 2: l1 -> l0 : [], cost: 1 3: l3 -> l0 : [ 1<=x^0 ], cost: 1 4: l4 -> l3 : [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: l4 7: l0 -> l0 : x^0'=-1+x^0, [], cost: 3 5: l4 -> l0 : [ 1<=x^0 ], cost: 2 Accelerating simple loops of location 0. Accelerating the following rules: 7: l0 -> l0 : x^0'=-1+x^0, [], cost: 3 Accelerated rule 7 with non-termination, yielding the new rule 8. [accelerate] Nesting with 0 inner and 0 outer candidates Removing the simple loops: 7. Accelerated all simple loops using metering functions (where possible): Start location: l4 8: l0 -> [5] : [], cost: NONTERM 5: l4 -> l0 : [ 1<=x^0 ], cost: 2 Chained accelerated rules (with incoming rules): Start location: l4 5: l4 -> l0 : [ 1<=x^0 ], cost: 2 9: l4 -> [5] : [ 1<=x^0 ], cost: NONTERM Removed unreachable locations (and leaf rules with constant cost): Start location: l4 9: l4 -> [5] : [ 1<=x^0 ], cost: NONTERM ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: l4 9: l4 -> [5] : [ 1<=x^0 ], cost: NONTERM Computing asymptotic complexity for rule 9 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: [ 1<=x^0 ] NO