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(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
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(X1, X2, X3) -> f(X1, X2, X3)
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b() }

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, x2, x3) = [2] x1 + [1] x2 + [2] x3 + [2]
                                                       
                   [a] = [2]                           
                                                       
                [a__b] = [2]                           
                                                       
                   [b] = [1]                           
                                                       
            [mark](x1) = [2] x1 + [2]                  
                                                       
       [f](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X1, X2, X3)] =  [2] X1 + [1] X2 + [2] X3 + [2]
                          >  [1] X1 + [1] X2 + [1] X3 + [1]
                          =  [f(X1, X2, X3)]               
                                                           
        [a__f(a(), X, X)] =  [3] X + [6]                   
                          >= [2] X + [6]                   
                          =  [a__f(X, a__b(), b())]        
                                                           
                 [a__b()] =  [2]                           
                          >= [2]                           
                          =  [a()]                         
                                                           
                 [a__b()] =  [2]                           
                          >  [1]                           
                          =  [b()]                         
                                                           
              [mark(a())] =  [6]                           
                          >  [2]                           
                          =  [a()]                         
                                                           
              [mark(b())] =  [4]                           
                          >  [2]                           
                          =  [a__b()]                      
                                                           
    [mark(f(X1, X2, X3))] =  [2] X1 + [2] X2 + [2] X3 + [4]
                          >= [2] X1 + [2] X2 + [2] X3 + [4]
                          =  [a__f(X1, mark(X2), X3)]      
                                                           

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(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b() }
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(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }

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, x2, x3) = [2] x1 + [1] x2 + [1] x3 + [3]
                                                       
                   [a] = [0]                           
                                                       
                [a__b] = [0]                           
                                                       
                   [b] = [0]                           
                                                       
            [mark](x1) = [2] x1 + [0]                  
                                                       
       [f](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [2]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X1, X2, X3)] =  [2] X1 + [1] X2 + [1] X3 + [3]
                          >  [1] X1 + [1] X2 + [1] X3 + [2]
                          =  [f(X1, X2, X3)]               
                                                           
        [a__f(a(), X, X)] =  [2] X + [3]                   
                          >= [2] X + [3]                   
                          =  [a__f(X, a__b(), b())]        
                                                           
                 [a__b()] =  [0]                           
                          >= [0]                           
                          =  [a()]                         
                                                           
                 [a__b()] =  [0]                           
                          >= [0]                           
                          =  [b()]                         
                                                           
              [mark(a())] =  [0]                           
                          >= [0]                           
                          =  [a()]                         
                                                           
              [mark(b())] =  [0]                           
                          >= [0]                           
                          =  [a__b()]                      
                                                           
    [mark(f(X1, X2, X3))] =  [2] X1 + [2] X2 + [2] X3 + [4]
                          >  [2] X1 + [2] X2 + [1] X3 + [3]
                          =  [a__f(X1, mark(X2), X3)]      
                                                           

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(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a() }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
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(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> 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, x2, x3) = [2] x1 + [1] x2 + [1] x3 + [0]
                                                       
                   [a] = [2]                           
                                                       
                [a__b] = [3]                           
                                                       
                   [b] = [0]                           
                                                       
            [mark](x1) = [2] x1 + [3]                  
                                                       
       [f](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X1, X2, X3)] =  [2] X1 + [1] X2 + [1] X3 + [0]
                          >= [1] X1 + [1] X2 + [1] X3 + [0]
                          =  [f(X1, X2, X3)]               
                                                           
        [a__f(a(), X, X)] =  [2] X + [4]                   
                          >  [2] X + [3]                   
                          =  [a__f(X, a__b(), b())]        
                                                           
                 [a__b()] =  [3]                           
                          >  [2]                           
                          =  [a()]                         
                                                           
                 [a__b()] =  [3]                           
                          >  [0]                           
                          =  [b()]                         
                                                           
              [mark(a())] =  [7]                           
                          >  [2]                           
                          =  [a()]                         
                                                           
              [mark(b())] =  [3]                           
                          >= [3]                           
                          =  [a__b()]                      
                                                           
    [mark(f(X1, X2, X3))] =  [2] X1 + [2] X2 + [2] X3 + [3]
                          >= [2] X1 + [2] X2 + [1] X3 + [3]
                          =  [a__f(X1, mark(X2), X3)]      
                                                           

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(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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