*** 1 Progress [(O(1),O(n^1))] *** Considered Problem: Strict DP Rules: Strict TRS Rules: 0() -> n__0() activate(X) -> X activate(n__0()) -> 0() activate(n__f(X)) -> f(activate(X)) activate(n__s(X)) -> s(activate(X)) f(X) -> n__f(X) f(0()) -> cons(0(),n__f(n__s(n__0()))) f(s(0())) -> f(p(s(0()))) p(s(X)) -> X s(X) -> n__s(X) Weak DP Rules: Weak TRS Rules: Signature: {0/0,activate/1,f/1,p/1,s/1} / {cons/2,n__0/0,n__f/1,n__s/1} Obligation: Innermost basic terms: {0,activate,f,p,s}/{cons,n__0,n__f,n__s} Applied Processor: InnermostRuleRemoval Proof: Arguments of following rules are not normal-forms. f(0()) -> cons(0(),n__f(n__s(n__0()))) f(s(0())) -> f(p(s(0()))) p(s(X)) -> X All above mentioned rules can be savely removed. *** 1.1 Progress [(O(1),O(n^1))] *** Considered Problem: Strict DP Rules: Strict TRS Rules: 0() -> n__0() activate(X) -> X activate(n__0()) -> 0() activate(n__f(X)) -> f(activate(X)) activate(n__s(X)) -> s(activate(X)) f(X) -> n__f(X) s(X) -> n__s(X) Weak DP Rules: Weak TRS Rules: Signature: {0/0,activate/1,f/1,p/1,s/1} / {cons/2,n__0/0,n__f/1,n__s/1} Obligation: Innermost basic terms: {0,activate,f,p,s}/{cons,n__0,n__f,n__s} Applied Processor: Bounds {initialAutomaton = minimal, enrichment = match} Proof: The problem is match-bounded by 2. The enriched problem is compatible with follwoing automaton. 0_0() -> 1 0_1() -> 1 0_1() -> 3 activate_0(2) -> 1 activate_1(2) -> 3 cons_0(2,2) -> 1 cons_0(2,2) -> 2 cons_0(2,2) -> 3 f_0(2) -> 1 f_1(3) -> 1 f_1(3) -> 3 n__0_0() -> 1 n__0_0() -> 2 n__0_0() -> 3 n__0_1() -> 1 n__0_2() -> 1 n__0_2() -> 3 n__f_0(2) -> 1 n__f_0(2) -> 2 n__f_0(2) -> 3 n__f_1(2) -> 1 n__f_2(3) -> 1 n__f_2(3) -> 3 n__s_0(2) -> 1 n__s_0(2) -> 2 n__s_0(2) -> 3 n__s_1(2) -> 1 n__s_2(3) -> 1 n__s_2(3) -> 3 p_0(2) -> 1 s_0(2) -> 1 s_1(3) -> 1 s_1(3) -> 3 2 -> 1 2 -> 3 *** 1.1.1 Progress [(O(1),O(1))] *** Considered Problem: Strict DP Rules: Strict TRS Rules: Weak DP Rules: Weak TRS Rules: 0() -> n__0() activate(X) -> X activate(n__0()) -> 0() activate(n__f(X)) -> f(activate(X)) activate(n__s(X)) -> s(activate(X)) f(X) -> n__f(X) s(X) -> n__s(X) Signature: {0/0,activate/1,f/1,p/1,s/1} / {cons/2,n__0/0,n__f/1,n__s/1} Obligation: Innermost basic terms: {0,activate,f,p,s}/{cons,n__0,n__f,n__s} Applied Processor: EmptyProcessor Proof: The problem is already closed. The intended complexity is O(1).