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:
  { f(X) -> n__f(X)
  , f(n__f(n__a())) -> f(n__g(f(n__a())))
  , a() -> n__a()
  , g(X) -> n__g(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(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:
  { activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__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).
  
           [f](x1) = [1] x1 + [0]
                                 
        [n__f](x1) = [1] x1 + [0]
                                 
            [n__a] = [0]         
                                 
        [n__g](x1) = [1] x1 + [0]
                                 
               [a] = [0]         
                                 
           [g](x1) = [1] x1 + [0]
                                 
    [activate](x1) = [1] x1 + [1]
  
  This order satisfies the following ordering constraints:
  
                 [f(X)] =  [1] X + [0]         
                        >= [1] X + [0]         
                        =  [n__f(X)]           
                                               
      [f(n__f(n__a()))] =  [0]                 
                        >= [0]                 
                        =  [f(n__g(f(n__a())))]
                                               
                  [a()] =  [0]                 
                        >= [0]                 
                        =  [n__a()]            
                                               
                 [g(X)] =  [1] X + [0]         
                        >= [1] X + [0]         
                        =  [n__g(X)]           
                                               
          [activate(X)] =  [1] X + [1]         
                        >  [1] X + [0]         
                        =  [X]                 
                                               
    [activate(n__f(X))] =  [1] X + [1]         
                        >  [1] X + [0]         
                        =  [f(X)]              
                                               
     [activate(n__a())] =  [1]                 
                        >  [0]                 
                        =  [a()]               
                                               
    [activate(n__g(X))] =  [1] X + [1]         
                        >  [1] X + [0]         
                        =  [g(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:
  { f(X) -> n__f(X)
  , f(n__f(n__a())) -> f(n__g(f(n__a())))
  , a() -> n__a()
  , g(X) -> n__g(X) }
Weak Trs:
  { activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(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() -> n__a()
  , g(X) -> n__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).
  
           [f](x1) = [1] x1 + [0]
                                 
        [n__f](x1) = [1] x1 + [0]
                                 
            [n__a] = [0]         
                                 
        [n__g](x1) = [1] x1 + [0]
                                 
               [a] = [1]         
                                 
           [g](x1) = [2] x1 + [1]
                                 
    [activate](x1) = [2] x1 + [1]
  
  This order satisfies the following ordering constraints:
  
                 [f(X)] =  [1] X + [0]         
                        >= [1] X + [0]         
                        =  [n__f(X)]           
                                               
      [f(n__f(n__a()))] =  [0]                 
                        >= [0]                 
                        =  [f(n__g(f(n__a())))]
                                               
                  [a()] =  [1]                 
                        >  [0]                 
                        =  [n__a()]            
                                               
                 [g(X)] =  [2] X + [1]         
                        >  [1] X + [0]         
                        =  [n__g(X)]           
                                               
          [activate(X)] =  [2] X + [1]         
                        >  [1] X + [0]         
                        =  [X]                 
                                               
    [activate(n__f(X))] =  [2] X + [1]         
                        >  [1] X + [0]         
                        =  [f(X)]              
                                               
     [activate(n__a())] =  [1]                 
                        >= [1]                 
                        =  [a()]               
                                               
    [activate(n__g(X))] =  [2] X + [1]         
                        >= [2] X + [1]         
                        =  [g(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:
  { f(X) -> n__f(X)
  , f(n__f(n__a())) -> f(n__g(f(n__a()))) }
Weak Trs:
  { a() -> n__a()
  , g(X) -> n__g(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs:
  { f(X) -> n__f(X)
  , f(n__f(n__a())) -> f(n__g(f(n__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) and not(IDA(1)).
  
           [f](x1) = [2 1] x1 + [1]
                     [0 0]      [3]
                                   
        [n__f](x1) = [1 1] x1 + [0]
                     [0 0]      [3]
                                   
            [n__a] = [0]           
                     [0]           
                                   
        [n__g](x1) = [1 0] x1 + [0]
                     [0 0]      [0]
                                   
               [a] = [1]           
                     [2]           
                                   
           [g](x1) = [2 0] x1 + [1]
                     [0 0]      [2]
                                   
    [activate](x1) = [2 0] x1 + [1]
                     [0 1]      [2]
  
  This order satisfies the following ordering constraints:
  
                 [f(X)] =  [2 1] X + [1]       
                           [0 0]     [3]       
                        >  [1 1] X + [0]       
                           [0 0]     [3]       
                        =  [n__f(X)]           
                                               
      [f(n__f(n__a()))] =  [4]                 
                           [3]                 
                        >  [3]                 
                           [3]                 
                        =  [f(n__g(f(n__a())))]
                                               
                  [a()] =  [1]                 
                           [2]                 
                        >  [0]                 
                           [0]                 
                        =  [n__a()]            
                                               
                 [g(X)] =  [2 0] X + [1]       
                           [0 0]     [2]       
                        >  [1 0] X + [0]       
                           [0 0]     [0]       
                        =  [n__g(X)]           
                                               
          [activate(X)] =  [2 0] X + [1]       
                           [0 1]     [2]       
                        >  [1 0] X + [0]       
                           [0 1]     [0]       
                        =  [X]                 
                                               
    [activate(n__f(X))] =  [2 2] X + [1]       
                           [0 0]     [5]       
                        >= [2 1] X + [1]       
                           [0 0]     [3]       
                        =  [f(X)]              
                                               
     [activate(n__a())] =  [1]                 
                           [2]                 
                        >= [1]                 
                           [2]                 
                        =  [a()]               
                                               
    [activate(n__g(X))] =  [2 0] X + [1]       
                           [0 0]     [2]       
                        >= [2 0] X + [1]       
                           [0 0]     [2]       
                        =  [g(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:
  { f(X) -> n__f(X)
  , f(n__f(n__a())) -> f(n__g(f(n__a())))
  , a() -> n__a()
  , g(X) -> n__g(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(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))