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:
  { first(0(), X) -> nil()
  , first(s(X), cons(Y)) -> cons(Y)
  , from(X) -> cons(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:
  { first(0(), X) -> nil()
  , first(s(X), cons(Y)) -> cons(Y) }

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).
  
    [first](x1, x2) = [3] x1 + [1] x2 + [1]
                                           
                [0] = [1]                  
                                           
              [nil] = [3]                  
                                           
            [s](x1) = [1] x1 + [1]         
                                           
         [cons](x1) = [1] x1 + [3]         
                                           
         [from](x1) = [3] x1 + [3]         
  
  This order satisfies the following ordering constraints:
  
           [first(0(), X)] =  [1] X + [4]        
                           >  [3]                
                           =  [nil()]            
                                                 
    [first(s(X), cons(Y))] =  [3] X + [1] Y + [7]
                           >  [1] Y + [3]        
                           =  [cons(Y)]          
                                                 
                 [from(X)] =  [3] X + [3]        
                           >= [1] X + [3]        
                           =  [cons(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: { from(X) -> cons(X) }
Weak Trs:
  { first(0(), X) -> nil()
  , first(s(X), cons(Y)) -> cons(Y) }
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: { from(X) -> cons(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).
  
    [first](x1, x2) = [1] x1 + [3] x2 + [1]
                                           
                [0] = [3]                  
                                           
              [nil] = [3]                  
                                           
            [s](x1) = [1] x1 + [3]         
                                           
         [cons](x1) = [1] x1 + [1]         
                                           
         [from](x1) = [3] x1 + [3]         
  
  This order satisfies the following ordering constraints:
  
           [first(0(), X)] = [3] X + [4]        
                           > [3]                
                           = [nil()]            
                                                
    [first(s(X), cons(Y))] = [1] X + [3] Y + [7]
                           > [1] Y + [1]        
                           = [cons(Y)]          
                                                
                 [from(X)] = [3] X + [3]        
                           > [1] X + [1]        
                           = [cons(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:
  { first(0(), X) -> nil()
  , first(s(X), cons(Y)) -> cons(Y)
  , from(X) -> cons(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))