We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict Trs:
  { plus#2(0(), x12) -> x12
  , plus#2(S(x4), x2) -> S(plus#2(x4, x2))
  , fold#3(Nil()) -> 0()
  , fold#3(Cons(x4, x2)) -> plus#2(x4, fold#3(x2))
  , main(x1) -> fold#3(x1) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The problem is match-bounded by 1. The enriched problem is
compatible with the following automaton.
{ plus#2_0(2, 2) -> 1
, plus#2_0(2, 3) -> 1
, plus#2_0(2, 5) -> 1
, plus#2_0(2, 6) -> 1
, plus#2_0(3, 2) -> 1
, plus#2_0(3, 3) -> 1
, plus#2_0(3, 5) -> 1
, plus#2_0(3, 6) -> 1
, plus#2_0(5, 2) -> 1
, plus#2_0(5, 3) -> 1
, plus#2_0(5, 5) -> 1
, plus#2_0(5, 6) -> 1
, plus#2_0(6, 2) -> 1
, plus#2_0(6, 3) -> 1
, plus#2_0(6, 5) -> 1
, plus#2_0(6, 6) -> 1
, plus#2_1(2, 2) -> 8
, plus#2_1(2, 3) -> 8
, plus#2_1(2, 5) -> 8
, plus#2_1(2, 6) -> 8
, plus#2_1(2, 9) -> 4
, plus#2_1(2, 9) -> 7
, plus#2_1(2, 9) -> 9
, plus#2_1(3, 2) -> 8
, plus#2_1(3, 3) -> 8
, plus#2_1(3, 5) -> 8
, plus#2_1(3, 6) -> 8
, plus#2_1(3, 9) -> 4
, plus#2_1(3, 9) -> 7
, plus#2_1(3, 9) -> 9
, plus#2_1(5, 2) -> 8
, plus#2_1(5, 3) -> 8
, plus#2_1(5, 5) -> 8
, plus#2_1(5, 6) -> 8
, plus#2_1(5, 9) -> 4
, plus#2_1(5, 9) -> 7
, plus#2_1(5, 9) -> 9
, plus#2_1(6, 2) -> 8
, plus#2_1(6, 3) -> 8
, plus#2_1(6, 5) -> 8
, plus#2_1(6, 6) -> 8
, plus#2_1(6, 9) -> 4
, plus#2_1(6, 9) -> 7
, plus#2_1(6, 9) -> 9
, 0_0() -> 1
, 0_0() -> 2
, 0_0() -> 8
, 0_1() -> 4
, 0_1() -> 7
, 0_1() -> 9
, S_0(2) -> 1
, S_0(2) -> 3
, S_0(2) -> 8
, S_0(3) -> 1
, S_0(3) -> 3
, S_0(3) -> 8
, S_0(5) -> 1
, S_0(5) -> 3
, S_0(5) -> 8
, S_0(6) -> 1
, S_0(6) -> 3
, S_0(6) -> 8
, S_1(4) -> 4
, S_1(8) -> 1
, S_1(8) -> 8
, S_1(9) -> 4
, S_1(9) -> 7
, S_1(9) -> 9
, fold#3_0(2) -> 4
, fold#3_0(3) -> 4
, fold#3_0(5) -> 4
, fold#3_0(6) -> 4
, fold#3_1(2) -> 4
, fold#3_1(2) -> 7
, fold#3_1(2) -> 9
, fold#3_1(3) -> 4
, fold#3_1(3) -> 7
, fold#3_1(3) -> 9
, fold#3_1(5) -> 4
, fold#3_1(5) -> 7
, fold#3_1(5) -> 9
, fold#3_1(6) -> 4
, fold#3_1(6) -> 7
, fold#3_1(6) -> 9
, Nil_0() -> 1
, Nil_0() -> 5
, Nil_0() -> 8
, Cons_0(2, 2) -> 1
, Cons_0(2, 2) -> 6
, Cons_0(2, 2) -> 8
, Cons_0(2, 3) -> 1
, Cons_0(2, 3) -> 6
, Cons_0(2, 3) -> 8
, Cons_0(2, 5) -> 1
, Cons_0(2, 5) -> 6
, Cons_0(2, 5) -> 8
, Cons_0(2, 6) -> 1
, Cons_0(2, 6) -> 6
, Cons_0(2, 6) -> 8
, Cons_0(3, 2) -> 1
, Cons_0(3, 2) -> 6
, Cons_0(3, 2) -> 8
, Cons_0(3, 3) -> 1
, Cons_0(3, 3) -> 6
, Cons_0(3, 3) -> 8
, Cons_0(3, 5) -> 1
, Cons_0(3, 5) -> 6
, Cons_0(3, 5) -> 8
, Cons_0(3, 6) -> 1
, Cons_0(3, 6) -> 6
, Cons_0(3, 6) -> 8
, Cons_0(5, 2) -> 1
, Cons_0(5, 2) -> 6
, Cons_0(5, 2) -> 8
, Cons_0(5, 3) -> 1
, Cons_0(5, 3) -> 6
, Cons_0(5, 3) -> 8
, Cons_0(5, 5) -> 1
, Cons_0(5, 5) -> 6
, Cons_0(5, 5) -> 8
, Cons_0(5, 6) -> 1
, Cons_0(5, 6) -> 6
, Cons_0(5, 6) -> 8
, Cons_0(6, 2) -> 1
, Cons_0(6, 2) -> 6
, Cons_0(6, 2) -> 8
, Cons_0(6, 3) -> 1
, Cons_0(6, 3) -> 6
, Cons_0(6, 3) -> 8
, Cons_0(6, 5) -> 1
, Cons_0(6, 5) -> 6
, Cons_0(6, 5) -> 8
, Cons_0(6, 6) -> 1
, Cons_0(6, 6) -> 6
, Cons_0(6, 6) -> 8
, main_0(2) -> 7
, main_0(3) -> 7
, main_0(5) -> 7
, main_0(6) -> 7 }

Hurray, we answered YES(?,O(n^1))