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(a()) -> f(c(a()))
  , f(a()) -> f(d(a()))
  , f(c(X)) -> X
  , f(c(a())) -> f(d(b()))
  , f(c(b())) -> f(d(a()))
  , f(d(X)) -> X
  , e(g(X)) -> e(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: { e(g(X)) -> e(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) = [2] x1 + [0]
                          
        [a] = [0]         
                          
    [c](x1) = [1] x1 + [0]
                          
    [d](x1) = [1] x1 + [0]
                          
        [b] = [0]         
                          
    [e](x1) = [1] x1 + [3]
                          
    [g](x1) = [1] x1 + [3]
  
  This order satisfies the following ordering constraints:
  
       [f(a())] =  [0]        
                >= [0]        
                =  [f(c(a()))]
                              
       [f(a())] =  [0]        
                >= [0]        
                =  [f(d(a()))]
                              
      [f(c(X))] =  [2] X + [0]
                >= [1] X + [0]
                =  [X]        
                              
    [f(c(a()))] =  [0]        
                >= [0]        
                =  [f(d(b()))]
                              
    [f(c(b()))] =  [0]        
                >= [0]        
                =  [f(d(a()))]
                              
      [f(d(X))] =  [2] X + [0]
                >= [1] X + [0]
                =  [X]        
                              
      [e(g(X))] =  [1] X + [6]
                >  [1] X + [3]
                =  [e(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(a()) -> f(c(a()))
  , f(a()) -> f(d(a()))
  , f(c(X)) -> X
  , f(c(a())) -> f(d(b()))
  , f(c(b())) -> f(d(a()))
  , f(d(X)) -> X }
Weak Trs: { e(g(X)) -> e(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:
  { f(c(X)) -> X
  , f(d(X)) -> 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 + [1]
                          
        [a] = [0]         
                          
    [c](x1) = [1] x1 + [0]
                          
    [d](x1) = [1] x1 + [0]
                          
        [b] = [0]         
                          
    [e](x1) = [1] x1 + [3]
                          
    [g](x1) = [1] x1 + [3]
  
  This order satisfies the following ordering constraints:
  
       [f(a())] =  [1]        
                >= [1]        
                =  [f(c(a()))]
                              
       [f(a())] =  [1]        
                >= [1]        
                =  [f(d(a()))]
                              
      [f(c(X))] =  [1] X + [1]
                >  [1] X + [0]
                =  [X]        
                              
    [f(c(a()))] =  [1]        
                >= [1]        
                =  [f(d(b()))]
                              
    [f(c(b()))] =  [1]        
                >= [1]        
                =  [f(d(a()))]
                              
      [f(d(X))] =  [1] X + [1]
                >  [1] X + [0]
                =  [X]        
                              
      [e(g(X))] =  [1] X + [6]
                >  [1] X + [3]
                =  [e(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(a()) -> f(c(a()))
  , f(a()) -> f(d(a()))
  , f(c(a())) -> f(d(b()))
  , f(c(b())) -> f(d(a())) }
Weak Trs:
  { f(c(X)) -> X
  , f(d(X)) -> X
  , e(g(X)) -> e(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(a()) -> f(d(a()))
  , f(c(a())) -> f(d(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) and not(IDA(1)).
  
    [f](x1) = [1 2] x1 + [1]
              [1 2]      [0]
                            
        [a] = [0]           
              [2]           
                            
    [c](x1) = [1 2] x1 + [0]
              [0 0]      [0]
                            
    [d](x1) = [1 1] x1 + [0]
              [0 0]      [0]
                            
        [b] = [0]           
              [1]           
                            
    [e](x1) = [3 0] x1 + [1]
              [3 0]      [1]
                            
    [g](x1) = [1 0] x1 + [2]
              [0 0]      [3]
  
  This order satisfies the following ordering constraints:
  
       [f(a())] =  [5]          
                   [4]          
                >= [5]          
                   [4]          
                =  [f(c(a()))]  
                                
       [f(a())] =  [5]          
                   [4]          
                >  [3]          
                   [2]          
                =  [f(d(a()))]  
                                
      [f(c(X))] =  [1 2] X + [1]
                   [1 2]     [0]
                >  [1 0] X + [0]
                   [0 1]     [0]
                =  [X]          
                                
    [f(c(a()))] =  [5]          
                   [4]          
                >  [2]          
                   [1]          
                =  [f(d(b()))]  
                                
    [f(c(b()))] =  [3]          
                   [2]          
                >= [3]          
                   [2]          
                =  [f(d(a()))]  
                                
      [f(d(X))] =  [1 1] X + [1]
                   [1 1]     [0]
                >  [1 0] X + [0]
                   [0 1]     [0]
                =  [X]          
                                
      [e(g(X))] =  [3 0] X + [7]
                   [3 0]     [7]
                >  [3 0] X + [1]
                   [3 0]     [1]
                =  [e(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(a()) -> f(c(a()))
  , f(c(b())) -> f(d(a())) }
Weak Trs:
  { f(a()) -> f(d(a()))
  , f(c(X)) -> X
  , f(c(a())) -> f(d(b()))
  , f(d(X)) -> X
  , e(g(X)) -> e(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(a()) -> f(c(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 3] x1 + [0]
              [1 2]      [0]
                            
        [a] = [0]           
              [2]           
                            
    [c](x1) = [1 1] x1 + [0]
              [0 0]      [0]
                            
    [d](x1) = [1 1] x1 + [0]
              [0 0]      [0]
                            
        [b] = [2]           
              [0]           
                            
    [e](x1) = [3 0] x1 + [1]
              [3 0]      [1]
                            
    [g](x1) = [1 0] x1 + [2]
              [0 0]      [3]
  
  This order satisfies the following ordering constraints:
  
       [f(a())] =  [6]          
                   [4]          
                >  [4]          
                   [2]          
                =  [f(c(a()))]  
                                
       [f(a())] =  [6]          
                   [4]          
                >  [4]          
                   [2]          
                =  [f(d(a()))]  
                                
      [f(c(X))] =  [2 2] X + [0]
                   [1 1]     [0]
                >= [1 0] X + [0]
                   [0 1]     [0]
                =  [X]          
                                
    [f(c(a()))] =  [4]          
                   [2]          
                >= [4]          
                   [2]          
                =  [f(d(b()))]  
                                
    [f(c(b()))] =  [4]          
                   [2]          
                >= [4]          
                   [2]          
                =  [f(d(a()))]  
                                
      [f(d(X))] =  [2 2] X + [0]
                   [1 1]     [0]
                >= [1 0] X + [0]
                   [0 1]     [0]
                =  [X]          
                                
      [e(g(X))] =  [3 0] X + [7]
                   [3 0]     [7]
                >  [3 0] X + [1]
                   [3 0]     [1]
                =  [e(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(c(b())) -> f(d(a())) }
Weak Trs:
  { f(a()) -> f(c(a()))
  , f(a()) -> f(d(a()))
  , f(c(X)) -> X
  , f(c(a())) -> f(d(b()))
  , f(d(X)) -> X
  , e(g(X)) -> e(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(c(b())) -> f(d(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) = [1 0] x1 + [0]
              [1 0]      [0]
                            
        [a] = [1]           
              [0]           
                            
    [c](x1) = [1 2] x1 + [0]
              [0 0]      [0]
                            
    [d](x1) = [1 1] x1 + [0]
              [0 0]      [0]
                            
        [b] = [0]           
              [1]           
                            
    [e](x1) = [3 0] x1 + [3]
              [3 0]      [3]
                            
    [g](x1) = [1 0] x1 + [1]
              [0 0]      [3]
  
  This order satisfies the following ordering constraints:
  
       [f(a())] =  [1]          
                   [1]          
                >= [1]          
                   [1]          
                =  [f(c(a()))]  
                                
       [f(a())] =  [1]          
                   [1]          
                >= [1]          
                   [1]          
                =  [f(d(a()))]  
                                
      [f(c(X))] =  [1 2] X + [0]
                   [1 2]     [0]
                >= [1 0] X + [0]
                   [0 1]     [0]
                =  [X]          
                                
    [f(c(a()))] =  [1]          
                   [1]          
                >= [1]          
                   [1]          
                =  [f(d(b()))]  
                                
    [f(c(b()))] =  [2]          
                   [2]          
                >  [1]          
                   [1]          
                =  [f(d(a()))]  
                                
      [f(d(X))] =  [1 1] X + [0]
                   [1 1]     [0]
                >= [1 0] X + [0]
                   [0 1]     [0]
                =  [X]          
                                
      [e(g(X))] =  [3 0] X + [6]
                   [3 0]     [6]
                >  [3 0] X + [3]
                   [3 0]     [3]
                =  [e(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(a()) -> f(c(a()))
  , f(a()) -> f(d(a()))
  , f(c(X)) -> X
  , f(c(a())) -> f(d(b()))
  , f(c(b())) -> f(d(a()))
  , f(d(X)) -> X
  , e(g(X)) -> e(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))