* Step 1: Bounds WORST_CASE(?,O(n^1)) + Considered Problem: - Strict TRS: activate(X) -> X activate(n__f(X)) -> f(X) f(X) -> n__f(X) f(0()) -> cons(0(),n__f(s(0()))) f(s(0())) -> f(p(s(0()))) p(s(X)) -> X - Signature: {activate/1,f/1,p/1} / {0/0,cons/2,n__f/1,s/1} - Obligation: innermost runtime complexity wrt. defined symbols {activate,f,p} and constructors {0,cons,n__f,s} + Applied Processor: Bounds {initialAutomaton = minimal, enrichment = match} + Details: The problem is match-bounded by 2. The enriched problem is compatible with follwoing automaton. 0_0() -> 1 0_0() -> 2 0_1() -> 1 0_1() -> 2 0_1() -> 3 0_1() -> 6 0_2() -> 7 0_2() -> 10 activate_0(2) -> 1 cons_0(2,2) -> 1 cons_0(2,2) -> 2 cons_1(3,4) -> 1 cons_2(7,8) -> 1 f_0(2) -> 1 f_1(2) -> 1 n__f_0(2) -> 1 n__f_0(2) -> 2 n__f_1(2) -> 1 n__f_1(5) -> 4 n__f_2(2) -> 1 n__f_2(9) -> 8 p_0(2) -> 1 p_1(5) -> 1 p_1(5) -> 2 s_0(2) -> 1 s_0(2) -> 2 s_1(6) -> 5 s_2(10) -> 9 2 -> 1 6 -> 1 6 -> 2 * Step 2: EmptyProcessor WORST_CASE(?,O(1)) + Considered Problem: - Weak TRS: activate(X) -> X activate(n__f(X)) -> f(X) f(X) -> n__f(X) f(0()) -> cons(0(),n__f(s(0()))) f(s(0())) -> f(p(s(0()))) p(s(X)) -> X - Signature: {activate/1,f/1,p/1} / {0/0,cons/2,n__f/1,s/1} - Obligation: innermost runtime complexity wrt. defined symbols {activate,f,p} and constructors {0,cons,n__f,s} + Applied Processor: EmptyProcessor + Details: The problem is already closed. The intended complexity is O(1). WORST_CASE(?,O(n^1))