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:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

The input is overlay and right-linear. Switching to innermost
rewriting.

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))

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(a__f) = {2}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [a__f](x1, x2, x3) = [1] x2 + [4]
                                   
                 [a] = [0]         
                                   
              [a__b] = [0]         
                                   
                 [b] = [5]         
                                   
          [mark](x1) = [0]         
                                   
     [f](x1, x2, x3) = [1] x2 + [3]

The following symbols are considered usable

  {a__f, a__b, mark}

The order satisfies the following ordering constraints:

     [a__f(X1, X2, X3)] =  [1] X2 + [4]            
                        >  [1] X2 + [3]            
                        =  [f(X1, X2, X3)]         
                                                   
      [a__f(a(), X, X)] =  [1] X + [4]             
                        >= [4]                     
                        =  [a__f(X, a__b(), b())]  
                                                   
               [a__b()] =  [0]                     
                        >= [0]                     
                        =  [a()]                   
                                                   
               [a__b()] =  [0]                     
                        ?  [5]                     
                        =  [b()]                   
                                                   
            [mark(a())] =  [0]                     
                        >= [0]                     
                        =  [a()]                   
                                                   
            [mark(b())] =  [0]                     
                        >= [0]                     
                        =  [a__b()]                
                                                   
  [mark(f(X1, X2, X3))] =  [0]                     
                        ?  [4]                     
                        =  [a__f(X1, mark(X2), X3)]
                                                   

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(a__f) = {2}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [a__f](x1, x2, x3) = [1] x2 + [4]
                                   
                 [a] = [0]         
                                   
              [a__b] = [4]         
                                   
                 [b] = [5]         
                                   
          [mark](x1) = [0]         
                                   
     [f](x1, x2, x3) = [1] x2 + [3]

The following symbols are considered usable

  {a__f, a__b, mark}

The order satisfies the following ordering constraints:

     [a__f(X1, X2, X3)] =  [1] X2 + [4]            
                        >  [1] X2 + [3]            
                        =  [f(X1, X2, X3)]         
                                                   
      [a__f(a(), X, X)] =  [1] X + [4]             
                        ?  [8]                     
                        =  [a__f(X, a__b(), b())]  
                                                   
               [a__b()] =  [4]                     
                        >  [0]                     
                        =  [a()]                   
                                                   
               [a__b()] =  [4]                     
                        ?  [5]                     
                        =  [b()]                   
                                                   
            [mark(a())] =  [0]                     
                        >= [0]                     
                        =  [a()]                   
                                                   
            [mark(b())] =  [0]                     
                        ?  [4]                     
                        =  [a__b()]                
                                                   
  [mark(f(X1, X2, X3))] =  [0]                     
                        ?  [4]                     
                        =  [a__f(X1, mark(X2), X3)]
                                                   

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(a__f) = {2}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [a__f](x1, x2, x3) = [1] x2 + [0]
                                   
                 [a] = [0]         
                                   
              [a__b] = [4]         
                                   
                 [b] = [1]         
                                   
          [mark](x1) = [0]         
                                   
     [f](x1, x2, x3) = [1] x2 + [0]

The following symbols are considered usable

  {a__f, a__b, mark}

The order satisfies the following ordering constraints:

     [a__f(X1, X2, X3)] =  [1] X2 + [0]            
                        >= [1] X2 + [0]            
                        =  [f(X1, X2, X3)]         
                                                   
      [a__f(a(), X, X)] =  [1] X + [0]             
                        ?  [4]                     
                        =  [a__f(X, a__b(), b())]  
                                                   
               [a__b()] =  [4]                     
                        >  [0]                     
                        =  [a()]                   
                                                   
               [a__b()] =  [4]                     
                        >  [1]                     
                        =  [b()]                   
                                                   
            [mark(a())] =  [0]                     
                        >= [0]                     
                        =  [a()]                   
                                                   
            [mark(b())] =  [0]                     
                        ?  [4]                     
                        =  [a__b()]                
                                                   
  [mark(f(X1, X2, X3))] =  [0]                     
                        >= [0]                     
                        =  [a__f(X1, mark(X2), X3)]
                                                   

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(a__f) = {2}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [a__f](x1, x2, x3) = [1] x2 + [4]
                                   
                 [a] = [0]         
                                   
              [a__b] = [4]         
                                   
                 [b] = [1]         
                                   
          [mark](x1) = [4]         
                                   
     [f](x1, x2, x3) = [1] x2 + [3]

The following symbols are considered usable

  {a__f, a__b, mark}

The order satisfies the following ordering constraints:

     [a__f(X1, X2, X3)] =  [1] X2 + [4]            
                        >  [1] X2 + [3]            
                        =  [f(X1, X2, X3)]         
                                                   
      [a__f(a(), X, X)] =  [1] X + [4]             
                        ?  [8]                     
                        =  [a__f(X, a__b(), b())]  
                                                   
               [a__b()] =  [4]                     
                        >  [0]                     
                        =  [a()]                   
                                                   
               [a__b()] =  [4]                     
                        >  [1]                     
                        =  [b()]                   
                                                   
            [mark(a())] =  [4]                     
                        >  [0]                     
                        =  [a()]                   
                                                   
            [mark(b())] =  [4]                     
                        >= [4]                     
                        =  [a__b()]                
                                                   
  [mark(f(X1, X2, X3))] =  [4]                     
                        ?  [8]                     
                        =  [a__f(X1, mark(X2), X3)]
                                                   

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(a__f) = {2}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [a__f](x1, x2, x3) = [1] x2 + [4]
                                   
                 [a] = [0]         
                                   
              [a__b] = [0]         
                                   
                 [b] = [0]         
                                   
          [mark](x1) = [4]         
                                   
     [f](x1, x2, x3) = [1] x2 + [3]

The following symbols are considered usable

  {a__f, a__b, mark}

The order satisfies the following ordering constraints:

     [a__f(X1, X2, X3)] =  [1] X2 + [4]            
                        >  [1] X2 + [3]            
                        =  [f(X1, X2, X3)]         
                                                   
      [a__f(a(), X, X)] =  [1] X + [4]             
                        >= [4]                     
                        =  [a__f(X, a__b(), b())]  
                                                   
               [a__b()] =  [0]                     
                        >= [0]                     
                        =  [a()]                   
                                                   
               [a__b()] =  [0]                     
                        >= [0]                     
                        =  [b()]                   
                                                   
            [mark(a())] =  [4]                     
                        >  [0]                     
                        =  [a()]                   
                                                   
            [mark(b())] =  [4]                     
                        >  [0]                     
                        =  [a__b()]                
                                                   
  [mark(f(X1, X2, X3))] =  [4]                     
                        ?  [8]                     
                        =  [a__f(X1, mark(X2), X3)]
                                                   

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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())
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__b() -> a()
  , 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:
----------
  The following argument positions are usable:
    Uargs(a__f) = {2}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [a__f](x1, x2, x3) = [1] x2 + [4]
                                     
                   [a] = [0]         
                                     
                [a__b] = [0]         
                                     
                   [b] = [0]         
                                     
            [mark](x1) = [2] x1 + [0]
                                     
       [f](x1, x2, x3) = [1] x2 + [4]
  
  The following symbols are considered usable
  
    {a__f, a__b, mark}
  
  The order satisfies the following ordering constraints:
  
       [a__f(X1, X2, X3)] =  [1] X2 + [4]            
                          >= [1] X2 + [4]            
                          =  [f(X1, X2, X3)]         
                                                     
        [a__f(a(), X, X)] =  [1] X + [4]             
                          >= [4]                     
                          =  [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] X2 + [8]            
                          >  [2] X2 + [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()) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , 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(a(), X, X) -> a__f(X, a__b(), 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:
----------
  The following argument positions are usable:
    Uargs(a__f) = {2}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [a__f](x1, x2, x3) = [4] x1 + [1] x2 + [4] x3 + [0]
                                                       
                   [a] = [2]                           
                                                       
                [a__b] = [2]                           
                                                       
                   [b] = [0]                           
                                                       
            [mark](x1) = [4] x1 + [2]                  
                                                       
       [f](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
  
  The following symbols are considered usable
  
    {a__f, a__b, mark}
  
  The order satisfies the following ordering constraints:
  
       [a__f(X1, X2, X3)] =  [4] X1 + [1] X2 + [4] X3 + [0]
                          >= [1] X1 + [1] X2 + [1] X3 + [0]
                          =  [f(X1, X2, X3)]               
                                                           
        [a__f(a(), X, X)] =  [5] X + [8]                   
                          >  [4] X + [2]                   
                          =  [a__f(X, a__b(), b())]        
                                                           
                 [a__b()] =  [2]                           
                          >= [2]                           
                          =  [a()]                         
                                                           
                 [a__b()] =  [2]                           
                          >  [0]                           
                          =  [b()]                         
                                                           
              [mark(a())] =  [10]                          
                          >  [2]                           
                          =  [a()]                         
                                                           
              [mark(b())] =  [2]                           
                          >= [2]                           
                          =  [a__b()]                      
                                                           
    [mark(f(X1, X2, X3))] =  [4] X1 + [4] X2 + [4] X3 + [2]
                          >= [4] X1 + [4] X2 + [4] X3 + [2]
                          =  [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))