YES(?,O(n^3))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict Trs:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(X))))
  , a__c(X) -> d(X)
  , a__c(X) -> c(X)
  , a__h(X) -> a__c(d(X))
  , a__h(X) -> h(X)
  , mark(f(X)) -> a__f(mark(X))
  , mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(X)
  , mark(h(X)) -> a__h(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

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

 safe(a__f) = {1}, safe(f) = {1}, safe(a__c) = {1}, safe(g) = {1},
 safe(d) = {1}, safe(a__h) = {1}, safe(mark) = {}, safe(c) = {1},
 safe(h) = {1}

and precedence

 a__f > a__c, a__h > a__c, mark > a__f, mark > a__c, mark > a__h,
 a__f ~ a__h .

Following symbols are considered recursive:

 {a__f, a__c, a__h, mark}

The recursion depth is 3.

For your convenience, here are the satisfied ordering constraints:

       a__f(; X) > f(; X)                  
                                           
  a__f(; f(; X)) > a__c(; f(; g(; f(; X))))
                                           
       a__c(; X) > d(; X)                  
                                           
       a__c(; X) > c(; X)                  
                                           
       a__h(; X) > a__c(; d(; X))          
                                           
       a__h(; X) > h(; X)                  
                                           
   mark(f(; X);) > a__f(; mark(X;))        
                                           
   mark(g(; X);) > g(; X)                  
                                           
   mark(d(; X);) > d(; X)                  
                                           
   mark(c(; X);) > a__c(; X)               
                                           
   mark(h(; X);) > a__h(; mark(X;))        
                                           

Hurray, we answered YES(?,O(n^3))