YES(O(1),O(n^1))

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

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

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)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  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;))          
                                              

We return to the main proof.

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

Weak 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(1),O(1))

Empty rules are trivially bounded

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