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

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

Strict Trs:
  { a__f(X) -> f(X)
  , a__f(f(a())) -> c(f(g(f(a()))))
  , mark(g(X)) -> g(mark(X)) }
Weak Trs:
  { mark(f(X)) -> a__f(mark(X))
  , mark(a()) -> a()
  , mark(c(X)) -> c(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)
  , 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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [a__f](x1) = [1] x1 + [3]
                             
       [f](x1) = [1] x1 + [2]
                             
           [a] = [2]         
                             
       [c](x1) = [1] x1 + [0]
                             
       [g](x1) = [1] x1 + [1]
                             
    [mark](x1) = [2] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
         [a__f(X)] =  [1] X + [3]      
                   >  [1] X + [2]      
                   =  [f(X)]           
                                       
    [a__f(f(a()))] =  [7]              
                   >= [7]              
                   =  [c(f(g(f(a()))))]
                                       
      [mark(f(X))] =  [2] X + [4]      
                   >  [2] X + [3]      
                   =  [a__f(mark(X))]  
                                       
       [mark(a())] =  [4]              
                   >  [2]              
                   =  [a()]            
                                       
      [mark(c(X))] =  [2] X + [0]      
                   >= [1] X + [0]      
                   =  [c(X)]           
                                       
      [mark(g(X))] =  [2] X + [2]      
                   >  [2] X + [1]      
                   =  [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(n^1)).

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

Trs: { a__f(f(a())) -> c(f(g(f(a())))) }

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__f](x1) = [1] x1 + [2]
                             
       [f](x1) = [1] x1 + [1]
                             
           [a] = [3]         
                             
       [c](x1) = [1] x1 + [0]
                             
       [g](x1) = [1] x1 + [0]
                             
    [mark](x1) = [2] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
         [a__f(X)] =  [1] X + [2]      
                   >  [1] X + [1]      
                   =  [f(X)]           
                                       
    [a__f(f(a()))] =  [6]              
                   >  [5]              
                   =  [c(f(g(f(a()))))]
                                       
      [mark(f(X))] =  [2] X + [2]      
                   >= [2] X + [2]      
                   =  [a__f(mark(X))]  
                                       
       [mark(a())] =  [6]              
                   >  [3]              
                   =  [a()]            
                                       
      [mark(c(X))] =  [2] X + [0]      
                   >= [1] X + [0]      
                   =  [c(X)]           
                                       
      [mark(g(X))] =  [2] X + [0]      
                   >= [2] X + [0]      
                   =  [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))