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

Strict Trs:
  { a__g(X) -> a__h(X)
  , a__g(X) -> g(X)
  , a__h(X) -> h(X)
  , a__h(d()) -> a__g(c())
  , a__c() -> d()
  , a__c() -> c()
  , mark(d()) -> d()
  , mark(c()) -> a__c()
  , mark(g(X)) -> a__g(X)
  , mark(h(X)) -> a__h(X) }
Obligation:
  runtime complexity
Answer:
  YES(?,O(n^1))

The input is overlay and right-linear. Switching to innermost
rewriting.

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

Strict Trs:
  { a__g(X) -> a__h(X)
  , a__g(X) -> g(X)
  , a__h(X) -> h(X)
  , a__h(d()) -> a__g(c())
  , a__c() -> d()
  , a__c() -> c()
  , mark(d()) -> d()
  , mark(c()) -> a__c()
  , mark(g(X)) -> a__g(X)
  , mark(h(X)) -> a__h(X) }
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.
{ a__g_0(2) -> 1
, a__g_1(2) -> 1
, a__g_1(3) -> 1
, a__h_0(2) -> 1
, a__h_1(2) -> 1
, a__h_2(2) -> 1
, a__h_2(3) -> 1
, a__c_0() -> 1
, a__c_1() -> 1
, d_0() -> 2
, d_1() -> 1
, d_2() -> 1
, c_0() -> 2
, c_1() -> 1
, c_1() -> 3
, c_2() -> 1
, mark_0(2) -> 1
, g_0(2) -> 2
, g_1(2) -> 1
, g_2(2) -> 1
, g_2(3) -> 1
, h_0(2) -> 2
, h_1(2) -> 1
, h_2(2) -> 1
, h_3(2) -> 1
, h_3(3) -> 1 }

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