WORST_CASE(?,O(n^1)) * Step 1: Bounds WORST_CASE(?,O(n^1)) + Considered Problem: - Strict TRS: fold#3(Cons(x4,x2)) -> plus#2(x4,fold#3(x2)) fold#3(Nil()) -> 0() main(x1) -> fold#3(x1) plus#2(0(),x12) -> x12 plus#2(S(x4),x2) -> S(plus#2(x4,x2)) - Signature: {fold#3/1,main/1,plus#2/2} / {0/0,Cons/2,Nil/0,S/1} - Obligation: innermost runtime complexity wrt. defined symbols {fold#3,main,plus#2} and constructors {0,Cons,Nil,S} + Applied Processor: Bounds {initialAutomaton = minimal, enrichment = match} + Details: The problem is match-bounded by 1. The enriched problem is compatible with follwoing automaton. 0_0() -> 1 0_0() -> 2 0_0() -> 4 0_1() -> 1 0_1() -> 3 Cons_0(2,2) -> 1 Cons_0(2,2) -> 2 Cons_0(2,2) -> 4 Nil_0() -> 1 Nil_0() -> 2 Nil_0() -> 4 S_0(2) -> 1 S_0(2) -> 2 S_0(2) -> 4 S_1(1) -> 1 S_1(3) -> 1 S_1(3) -> 3 S_1(4) -> 1 S_1(4) -> 4 fold#3_0(2) -> 1 fold#3_1(2) -> 1 fold#3_1(2) -> 3 main_0(2) -> 1 plus#2_0(2,2) -> 1 plus#2_1(2,2) -> 4 plus#2_1(2,3) -> 1 plus#2_1(2,3) -> 3 2 -> 1 2 -> 4 3 -> 1 * Step 2: EmptyProcessor WORST_CASE(?,O(1)) + Considered Problem: - Weak TRS: fold#3(Cons(x4,x2)) -> plus#2(x4,fold#3(x2)) fold#3(Nil()) -> 0() main(x1) -> fold#3(x1) plus#2(0(),x12) -> x12 plus#2(S(x4),x2) -> S(plus#2(x4,x2)) - Signature: {fold#3/1,main/1,plus#2/2} / {0/0,Cons/2,Nil/0,S/1} - Obligation: innermost runtime complexity wrt. defined symbols {fold#3,main,plus#2} and constructors {0,Cons,Nil,S} + Applied Processor: EmptyProcessor + Details: The problem is already closed. The intended complexity is O(1). WORST_CASE(?,O(n^1))