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, X) -> c(X)
  , f(X, c(X)) -> f(s(X), X)
  , f(s(X), X) -> f(X, a(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(X, 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).
  
    [f](x1, x2) = [1] x1 + [1] x2 + [1]
                                       
        [s](x1) = [1] x1 + [0]         
                                       
        [a](x1) = [1] x1 + [0]         
                                       
        [c](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
       [f(X, X)] =  [2] X + [1] 
                 >  [1] X + [0] 
                 =  [c(X)]      
                                
    [f(X, c(X))] =  [2] X + [1] 
                 >= [2] X + [1] 
                 =  [f(s(X), X)]
                                
    [f(s(X), X)] =  [2] X + [1] 
                 >= [2] X + [1] 
                 =  [f(X, a(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, c(X)) -> f(s(X), X)
  , f(s(X), X) -> f(X, a(X)) }
Weak Trs: { f(X, 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: { f(X, c(X)) -> f(s(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, x2) = [3] x1 + [1] x2 + [3]
                                       
        [s](x1) = [1] x1 + [0]         
                                       
        [a](x1) = [1] x1 + [0]         
                                       
        [c](x1) = [1] x1 + [2]         
  
  This order satisfies the following ordering constraints:
  
       [f(X, X)] =  [4] X + [3] 
                 >  [1] X + [2] 
                 =  [c(X)]      
                                
    [f(X, c(X))] =  [4] X + [5] 
                 >  [4] X + [3] 
                 =  [f(s(X), X)]
                                
    [f(s(X), X)] =  [4] X + [3] 
                 >= [4] X + [3] 
                 =  [f(X, a(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(s(X), X) -> f(X, a(X)) }
Weak Trs:
  { f(X, X) -> c(X)
  , f(X, c(X)) -> f(s(X), 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(s(X), X) -> f(X, a(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, x2) = [3] x1 + [1] x2 + [3]
                                       
        [s](x1) = [1] x1 + [1]         
                                       
        [a](x1) = [1] x1 + [0]         
                                       
        [c](x1) = [1] x1 + [3]         
  
  This order satisfies the following ordering constraints:
  
       [f(X, X)] =  [4] X + [3] 
                 >= [1] X + [3] 
                 =  [c(X)]      
                                
    [f(X, c(X))] =  [4] X + [6] 
                 >= [4] X + [6] 
                 =  [f(s(X), X)]
                                
    [f(s(X), X)] =  [4] X + [6] 
                 >  [4] X + [3] 
                 =  [f(X, a(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, X) -> c(X)
  , f(X, c(X)) -> f(s(X), X)
  , f(s(X), X) -> f(X, a(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))