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__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(1),O(n^1))

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

Trs:
  { a__g(X) -> g(X)
  , a__h(X) -> h(X)
  , 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) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(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__g) = {}, safe(a__h) = {}, safe(a__c) = {}, safe(d) = {},
   safe(c) = {}, safe(mark) = {}, safe(g) = {1}, safe(h) = {1}
  
  and precedence
  
   a__c > a__g, a__c > a__h, mark > a__g, mark > a__h, mark > a__c,
   a__g ~ a__h .
  
  Following symbols are considered recursive:
  
   {}
  
  The recursion depth is 0.
  
  For your convenience, here are the satisfied ordering constraints:
  
         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;)  
                               

We return to the main proof.

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

Strict Trs:
  { a__g(X) -> a__h(X)
  , a__h(d()) -> a__g(c()) }
Weak Trs:
  { a__g(X) -> g(X)
  , a__h(X) -> h(X)
  , 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(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { a__g(X) -> a__h(X)
  , a__h(d()) -> a__g(c()) }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [a__g](x1) = [2] x1 + [1]
                             
    [a__h](x1) = [2] x1 + [0]
                             
        [a__c] = [2]         
                             
           [d] = [2]         
                             
           [c] = [1]         
                             
    [mark](x1) = [2] x1 + [0]
                             
       [g](x1) = [1] x1 + [1]
                             
       [h](x1) = [1] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
       [a__g(X)] =  [2] X + [1]
                 >  [2] X + [0]
                 =  [a__h(X)]  
                               
       [a__g(X)] =  [2] X + [1]
                 >= [1] X + [1]
                 =  [g(X)]     
                               
       [a__h(X)] =  [2] X + [0]
                 >= [1] X + [0]
                 =  [h(X)]     
                               
     [a__h(d())] =  [4]        
                 >  [3]        
                 =  [a__g(c())]
                               
        [a__c()] =  [2]        
                 >= [2]        
                 =  [d()]      
                               
        [a__c()] =  [2]        
                 >  [1]        
                 =  [c()]      
                               
     [mark(d())] =  [4]        
                 >  [2]        
                 =  [d()]      
                               
     [mark(c())] =  [2]        
                 >= [2]        
                 =  [a__c()]   
                               
    [mark(g(X))] =  [2] X + [2]
                 >  [2] X + [1]
                 =  [a__g(X)]  
                               
    [mark(h(X))] =  [2] X + [0]
                 >= [2] X + [0]
                 =  [a__h(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__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(1),O(1))

Empty rules are trivially bounded

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