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

Strict Trs:
  { a__f(X) -> f(X)
  , a__f(f(a())) -> c(f(g(f(a()))))
  , mark(f(X)) -> a__f(mark(X))
  , mark(a()) -> a()
  , mark(c(X)) -> c(X)
  , mark(g(X)) -> g(mark(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__f(X) -> f(X)
  , a__f(f(a())) -> c(f(g(f(a()))))
  , mark(f(X)) -> a__f(mark(X))
  , mark(a()) -> a()
  , mark(c(X)) -> c(X)
  , mark(g(X)) -> g(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping

 safe(a__f) = {1}, safe(f) = {1}, safe(a) = {}, safe(c) = {1},
 safe(g) = {1}, safe(mark) = {}

and precedence

 mark > a__f .

Following symbols are considered recursive:

 {mark}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

         a__f(; X) > f(; X)                 
                                            
  a__f(; f(; a())) > c(; f(; g(; f(; a()))))
                                            
     mark(f(; X);) > a__f(; mark(X;))       
                                            
        mark(a();) > a()                    
                                            
     mark(c(; X);) > c(; X)                 
                                            
     mark(g(; X);) > g(; mark(X;))          
                                            

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