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

Strict Trs:
  { rev_l#2(x8, x10) -> Cons(x10, x8)
  , step_x_f#1(rev_l(), x5, step_x_f(x2, x3, x4), x1) ->
    step_x_f#1(x2, x3, x4, rev_l#2(x1, x5))
  , step_x_f#1(rev_l(), x5, fleft_op_e_xs_1(), x3) -> rev_l#2(x3, x5)
  , foldr#3(Cons(x16, x6)) -> step_x_f(rev_l(), x16, foldr#3(x6))
  , foldr#3(Nil()) -> fleft_op_e_xs_1()
  , main(Cons(x8, x9)) -> step_x_f#1(rev_l(), x8, foldr#3(x9), Nil())
  , main(Nil()) -> Nil() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The problem is match-bounded by 3. The enriched problem is
compatible with the following automaton.
{ rev_l#2_0(2, 2) -> 1
, rev_l#2_1(1, 2) -> 1
, rev_l#2_1(2, 2) -> 1
, rev_l#2_1(2, 2) -> 3
, rev_l#2_1(3, 2) -> 1
, rev_l#2_2(1, 2) -> 1
, rev_l#2_2(6, 2) -> 1
, Cons_0(2, 2) -> 2
, Cons_1(2, 2) -> 1
, Cons_2(2, 1) -> 1
, Cons_2(2, 2) -> 1
, Cons_2(2, 2) -> 3
, Cons_2(2, 3) -> 1
, Cons_3(2, 1) -> 1
, Cons_3(2, 6) -> 1
, step_x_f#1_0(2, 2, 2, 2) -> 1
, step_x_f#1_1(2, 2, 2, 1) -> 1
, step_x_f#1_1(2, 2, 2, 3) -> 1
, step_x_f#1_1(4, 2, 5, 6) -> 1
, step_x_f#1_2(4, 2, 5, 1) -> 1
, rev_l_0() -> 2
, rev_l_1() -> 4
, step_x_f_0(2, 2, 2) -> 2
, step_x_f_1(4, 2, 5) -> 1
, step_x_f_1(4, 2, 5) -> 5
, fleft_op_e_xs_1_0() -> 2
, fleft_op_e_xs_1_1() -> 1
, fleft_op_e_xs_1_1() -> 5
, foldr#3_0(2) -> 1
, foldr#3_1(2) -> 5
, Nil_0() -> 2
, Nil_1() -> 1
, Nil_1() -> 6
, main_0(2) -> 1 }

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