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

Strict Trs:
  { comp_f_g#1(plus_x(x3), comp_f_g(x1, x2), 0()) ->
    plus_x#1(x3, comp_f_g#1(x1, x2, 0()))
  , comp_f_g#1(plus_x(x3), id(), 0()) -> plus_x#1(x3, 0())
  , plus_x#1(0(), x6) -> x6
  , plus_x#1(S(x8), x10) -> S(plus_x#1(x8, x10))
  , map#2(Nil()) -> Nil()
  , map#2(Cons(x16, x6)) -> Cons(plus_x(x16), map#2(x6))
  , foldr_f#3(Nil(), 0()) -> 0()
  , foldr_f#3(Cons(x16, x5), x24) ->
    comp_f_g#1(x16, foldr#3(x5), x24)
  , foldr#3(Nil()) -> id()
  , foldr#3(Cons(x32, x6)) -> comp_f_g(x32, foldr#3(x6))
  , main(x3) -> foldr_f#3(map#2(x3), 0()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The problem is match-bounded by 2. The enriched problem is
compatible with the following automaton.
{ comp_f_g#1_0(2, 2, 2) -> 1
, comp_f_g#1_0(2, 2, 2) -> 10
, comp_f_g#1_1(2, 2, 4) -> 1
, comp_f_g#1_1(2, 2, 4) -> 3
, comp_f_g#1_1(2, 2, 4) -> 10
, comp_f_g#1_1(2, 8, 2) -> 1
, comp_f_g#1_1(2, 8, 2) -> 10
, comp_f_g#1_1(2, 8, 4) -> 1
, comp_f_g#1_1(2, 8, 4) -> 3
, comp_f_g#1_1(2, 8, 4) -> 10
, comp_f_g#1_1(6, 9, 10) -> 1
, comp_f_g#1_1(6, 9, 10) -> 10
, comp_f_g#1_2(6, 9, 1) -> 1
, comp_f_g#1_2(6, 9, 1) -> 10
, comp_f_g#1_2(6, 9, 4) -> 1
, comp_f_g#1_2(6, 9, 4) -> 10
, plus_x_0(2) -> 1
, plus_x_0(2) -> 2
, plus_x_0(2) -> 5
, plus_x_0(2) -> 10
, plus_x_1(2) -> 6
, comp_f_g_0(2, 2) -> 1
, comp_f_g_0(2, 2) -> 2
, comp_f_g_0(2, 2) -> 5
, comp_f_g_0(2, 2) -> 10
, comp_f_g_1(2, 8) -> 1
, comp_f_g_1(2, 8) -> 8
, comp_f_g_1(2, 8) -> 10
, comp_f_g_2(6, 9) -> 9
, 0_0() -> 1
, 0_0() -> 2
, 0_0() -> 5
, 0_0() -> 10
, 0_1() -> 1
, 0_1() -> 3
, 0_1() -> 4
, 0_1() -> 10
, 0_2() -> 1
, 0_2() -> 10
, plus_x#1_0(2, 2) -> 1
, plus_x#1_0(2, 2) -> 10
, plus_x#1_1(2, 1) -> 1
, plus_x#1_1(2, 1) -> 10
, plus_x#1_1(2, 2) -> 5
, plus_x#1_1(2, 3) -> 1
, plus_x#1_1(2, 3) -> 3
, plus_x#1_1(2, 3) -> 10
, plus_x#1_1(2, 4) -> 1
, plus_x#1_1(2, 4) -> 3
, plus_x#1_1(2, 4) -> 10
, plus_x#1_1(2, 10) -> 1
, plus_x#1_1(2, 10) -> 10
, plus_x#1_2(2, 1) -> 1
, plus_x#1_2(2, 1) -> 10
, plus_x#1_2(2, 10) -> 1
, plus_x#1_2(2, 10) -> 10
, id_0() -> 1
, id_0() -> 2
, id_0() -> 5
, id_0() -> 10
, id_1() -> 1
, id_1() -> 8
, id_1() -> 10
, id_2() -> 9
, map#2_0(2) -> 1
, map#2_0(2) -> 10
, map#2_1(2) -> 7
, Nil_0() -> 1
, Nil_0() -> 2
, Nil_0() -> 5
, Nil_0() -> 10
, Nil_1() -> 1
, Nil_1() -> 7
, Nil_1() -> 10
, Cons_0(2, 2) -> 1
, Cons_0(2, 2) -> 2
, Cons_0(2, 2) -> 5
, Cons_0(2, 2) -> 10
, Cons_1(6, 7) -> 1
, Cons_1(6, 7) -> 7
, Cons_1(6, 7) -> 10
, S_0(2) -> 1
, S_0(2) -> 2
, S_0(2) -> 5
, S_0(2) -> 10
, S_1(1) -> 1
, S_1(1) -> 10
, S_1(3) -> 1
, S_1(3) -> 3
, S_1(3) -> 10
, S_1(5) -> 1
, S_1(5) -> 5
, S_1(5) -> 10
, foldr_f#3_0(2, 2) -> 1
, foldr_f#3_0(2, 2) -> 10
, foldr_f#3_1(7, 4) -> 1
, foldr_f#3_1(7, 4) -> 10
, foldr#3_0(2) -> 1
, foldr#3_0(2) -> 10
, foldr#3_1(2) -> 8
, foldr#3_2(7) -> 9
, main_0(2) -> 1
, main_0(2) -> 10 }

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