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__c() -> a__f(g(c()))
  , a__c() -> c()
  , a__f(X) -> f(X)
  , a__f(g(X)) -> g(X)
  , mark(g(X)) -> g(X)
  , mark(c()) -> a__c()
  , mark(f(X)) -> a__f(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__c() -> a__f(g(c()))
  , a__c() -> c()
  , a__f(X) -> f(X)
  , a__f(g(X)) -> g(X)
  , mark(g(X)) -> g(X)
  , mark(c()) -> a__c()
  , mark(f(X)) -> a__f(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__c) = {}, safe(a__f) = {1}, safe(g) = {1}, safe(c) = {},
   safe(mark) = {1}, safe(f) = {1}
  
  and precedence
  
   a__c > a__f, mark > a__c, mark > a__f .
  
  Following symbols are considered recursive:
  
   {a__f}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
            a__c() > a__f(; g(; c()))
                                     
            a__c() > c()             
                                     
         a__f(; X) > f(; X)          
                                     
    a__f(; g(; X)) > g(; X)          
                                     
    mark(; g(; X)) > g(; X)          
                                     
       mark(; c()) > a__c()          
                                     
    mark(; f(; X)) > a__f(; 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__c() -> a__f(g(c()))
  , a__c() -> c()
  , a__f(X) -> f(X)
  , a__f(g(X)) -> g(X)
  , mark(g(X)) -> g(X)
  , mark(c()) -> a__c()
  , mark(f(X)) -> a__f(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))