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 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { 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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
        [a__c] = [0]         
                             
    [a__f](x1) = [2] x1 + [0]
                             
       [g](x1) = [1] x1 + [0]
                             
           [c] = [0]         
                             
    [mark](x1) = [2] x1 + [1]
                             
       [f](x1) = [1] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
        [a__c()] =  [0]           
                 >= [0]           
                 =  [a__f(g(c()))]
                                  
        [a__c()] =  [0]           
                 >= [0]           
                 =  [c()]         
                                  
       [a__f(X)] =  [2] X + [0]   
                 >= [1] X + [0]   
                 =  [f(X)]        
                                  
    [a__f(g(X))] =  [2] X + [0]   
                 >= [1] X + [0]   
                 =  [g(X)]        
                                  
    [mark(g(X))] =  [2] X + [1]   
                 >  [1] X + [0]   
                 =  [g(X)]        
                                  
     [mark(c())] =  [1]           
                 >  [0]           
                 =  [a__c()]      
                                  
    [mark(f(X))] =  [2] X + [1]   
                 >  [2] X + [0]   
                 =  [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(n^1)).

Strict Trs:
  { a__c() -> a__f(g(c()))
  , a__c() -> c()
  , a__f(X) -> f(X)
  , a__f(g(X)) -> g(X) }
Weak Trs:
  { 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 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { a__c() -> a__f(g(c()))
  , a__c() -> 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__c] = [1]         
                             
    [a__f](x1) = [2] x1 + [0]
                             
       [g](x1) = [1] x1 + [0]
                             
           [c] = [0]         
                             
    [mark](x1) = [2] x1 + [1]
                             
       [f](x1) = [1] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
        [a__c()] =  [1]           
                 >  [0]           
                 =  [a__f(g(c()))]
                                  
        [a__c()] =  [1]           
                 >  [0]           
                 =  [c()]         
                                  
       [a__f(X)] =  [2] X + [0]   
                 >= [1] X + [0]   
                 =  [f(X)]        
                                  
    [a__f(g(X))] =  [2] X + [0]   
                 >= [1] X + [0]   
                 =  [g(X)]        
                                  
    [mark(g(X))] =  [2] X + [1]   
                 >  [1] X + [0]   
                 =  [g(X)]        
                                  
     [mark(c())] =  [1]           
                 >= [1]           
                 =  [a__c()]      
                                  
    [mark(f(X))] =  [2] X + [1]   
                 >  [2] X + [0]   
                 =  [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(n^1)).

Strict Trs:
  { a__f(X) -> f(X)
  , a__f(g(X)) -> g(X) }
Weak Trs:
  { a__c() -> a__f(g(c()))
  , a__c() -> c()
  , 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 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { a__f(X) -> f(X)
  , a__f(g(X)) -> g(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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
        [a__c] = [3]         
                             
    [a__f](x1) = [1] x1 + [1]
                             
       [g](x1) = [1] x1 + [0]
                             
           [c] = [2]         
                             
    [mark](x1) = [2] x1 + [1]
                             
       [f](x1) = [1] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
        [a__c()] =  [3]           
                 >= [3]           
                 =  [a__f(g(c()))]
                                  
        [a__c()] =  [3]           
                 >  [2]           
                 =  [c()]         
                                  
       [a__f(X)] =  [1] X + [1]   
                 >  [1] X + [0]   
                 =  [f(X)]        
                                  
    [a__f(g(X))] =  [1] X + [1]   
                 >  [1] X + [0]   
                 =  [g(X)]        
                                  
    [mark(g(X))] =  [2] X + [1]   
                 >  [1] X + [0]   
                 =  [g(X)]        
                                  
     [mark(c())] =  [5]           
                 >  [3]           
                 =  [a__c()]      
                                  
    [mark(f(X))] =  [2] X + [1]   
                 >= [1] X + [1]   
                 =  [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))