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:
  { fst(0(), Z) -> nil()
  , fst(s(), cons(Y)) -> cons(Y)
  , from(X) -> cons(X)
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , len(nil()) -> 0()
  , len(cons(X)) -> s() }
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).
  
    [fst](x1, x2) = [2] x1 + [3] x2 + [0]
                                         
              [0] = [0]                  
                                         
            [nil] = [0]                  
                                         
              [s] = [0]                  
                                         
       [cons](x1) = [1] x1 + [0]         
                                         
       [from](x1) = [3] x1 + [3]         
                                         
    [add](x1, x2) = [2] x1 + [3] x2 + [0]
                                         
        [len](x1) = [2] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
          [fst(0(), Z)] =  [3] Z + [0]
                        >= [0]        
                        =  [nil()]    
                                      
    [fst(s(), cons(Y))] =  [3] Y + [0]
                        >= [1] Y + [0]
                        =  [cons(Y)]  
                                      
              [from(X)] =  [3] X + [3]
                        >  [1] X + [0]
                        =  [cons(X)]  
                                      
          [add(0(), X)] =  [3] X + [0]
                        >= [1] X + [0]
                        =  [X]        
                                      
          [add(s(), Y)] =  [3] Y + [0]
                        >= [0]        
                        =  [s()]      
                                      
           [len(nil())] =  [0]        
                        >= [0]        
                        =  [0()]      
                                      
         [len(cons(X))] =  [2] X + [0]
                        >= [0]        
                        =  [s()]      
                                      

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:
  { fst(0(), Z) -> nil()
  , fst(s(), cons(Y)) -> cons(Y)
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , len(nil()) -> 0()
  , len(cons(X)) -> s() }
Weak Trs: { 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:
  { add(0(), X) -> X
  , add(s(), Y) -> s() }

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).
  
    [fst](x1, x2) = [2] x1 + [3] x2 + [0]
                                         
              [0] = [0]                  
                                         
            [nil] = [0]                  
                                         
              [s] = [0]                  
                                         
       [cons](x1) = [1] x1 + [0]         
                                         
       [from](x1) = [3] x1 + [3]         
                                         
    [add](x1, x2) = [2] x1 + [3] x2 + [1]
                                         
        [len](x1) = [2] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
          [fst(0(), Z)] =  [3] Z + [0]
                        >= [0]        
                        =  [nil()]    
                                      
    [fst(s(), cons(Y))] =  [3] Y + [0]
                        >= [1] Y + [0]
                        =  [cons(Y)]  
                                      
              [from(X)] =  [3] X + [3]
                        >  [1] X + [0]
                        =  [cons(X)]  
                                      
          [add(0(), X)] =  [3] X + [1]
                        >  [1] X + [0]
                        =  [X]        
                                      
          [add(s(), Y)] =  [3] Y + [1]
                        >  [0]        
                        =  [s()]      
                                      
           [len(nil())] =  [0]        
                        >= [0]        
                        =  [0()]      
                                      
         [len(cons(X))] =  [2] X + [0]
                        >= [0]        
                        =  [s()]      
                                      

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:
  { fst(0(), Z) -> nil()
  , fst(s(), cons(Y)) -> cons(Y)
  , len(nil()) -> 0()
  , len(cons(X)) -> s() }
Weak Trs:
  { from(X) -> cons(X)
  , add(0(), X) -> X
  , add(s(), Y) -> s() }
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:
  { fst(s(), cons(Y)) -> cons(Y)
  , len(nil()) -> 0() }

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).
  
    [fst](x1, x2) = [1] x1 + [3] x2 + [0]
                                         
              [0] = [0]                  
                                         
            [nil] = [0]                  
                                         
              [s] = [1]                  
                                         
       [cons](x1) = [1] x1 + [0]         
                                         
       [from](x1) = [3] x1 + [3]         
                                         
    [add](x1, x2) = [3] x1 + [3] x2 + [3]
                                         
        [len](x1) = [2] x1 + [1]         
  
  This order satisfies the following ordering constraints:
  
          [fst(0(), Z)] =  [3] Z + [0]
                        >= [0]        
                        =  [nil()]    
                                      
    [fst(s(), cons(Y))] =  [3] Y + [1]
                        >  [1] Y + [0]
                        =  [cons(Y)]  
                                      
              [from(X)] =  [3] X + [3]
                        >  [1] X + [0]
                        =  [cons(X)]  
                                      
          [add(0(), X)] =  [3] X + [3]
                        >  [1] X + [0]
                        =  [X]        
                                      
          [add(s(), Y)] =  [3] Y + [6]
                        >  [1]        
                        =  [s()]      
                                      
           [len(nil())] =  [1]        
                        >  [0]        
                        =  [0()]      
                                      
         [len(cons(X))] =  [2] X + [1]
                        >= [1]        
                        =  [s()]      
                                      

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:
  { fst(0(), Z) -> nil()
  , len(cons(X)) -> s() }
Weak Trs:
  { fst(s(), cons(Y)) -> cons(Y)
  , from(X) -> cons(X)
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , len(nil()) -> 0() }
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: { len(cons(X)) -> s() }

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).
  
    [fst](x1, x2) = [2] x1 + [3] x2 + [0]
                                         
              [0] = [0]                  
                                         
            [nil] = [0]                  
                                         
              [s] = [0]                  
                                         
       [cons](x1) = [1] x1 + [0]         
                                         
       [from](x1) = [3] x1 + [3]         
                                         
    [add](x1, x2) = [3] x1 + [3] x2 + [3]
                                         
        [len](x1) = [2] x1 + [1]         
  
  This order satisfies the following ordering constraints:
  
          [fst(0(), Z)] =  [3] Z + [0]
                        >= [0]        
                        =  [nil()]    
                                      
    [fst(s(), cons(Y))] =  [3] Y + [0]
                        >= [1] Y + [0]
                        =  [cons(Y)]  
                                      
              [from(X)] =  [3] X + [3]
                        >  [1] X + [0]
                        =  [cons(X)]  
                                      
          [add(0(), X)] =  [3] X + [3]
                        >  [1] X + [0]
                        =  [X]        
                                      
          [add(s(), Y)] =  [3] Y + [3]
                        >  [0]        
                        =  [s()]      
                                      
           [len(nil())] =  [1]        
                        >  [0]        
                        =  [0()]      
                                      
         [len(cons(X))] =  [2] X + [1]
                        >  [0]        
                        =  [s()]      
                                      

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: { fst(0(), Z) -> nil() }
Weak Trs:
  { fst(s(), cons(Y)) -> cons(Y)
  , from(X) -> cons(X)
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , len(nil()) -> 0()
  , len(cons(X)) -> s() }
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: { fst(0(), Z) -> nil() }

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).
  
    [fst](x1, x2) = [1] x1 + [2] x2 + [0]
                                         
              [0] = [1]                  
                                         
            [nil] = [0]                  
                                         
              [s] = [0]                  
                                         
       [cons](x1) = [1] x1 + [2]         
                                         
       [from](x1) = [3] x1 + [3]         
                                         
    [add](x1, x2) = [3] x1 + [3] x2 + [3]
                                         
        [len](x1) = [2] x1 + [1]         
  
  This order satisfies the following ordering constraints:
  
          [fst(0(), Z)] =  [2] Z + [1]
                        >  [0]        
                        =  [nil()]    
                                      
    [fst(s(), cons(Y))] =  [2] Y + [4]
                        >  [1] Y + [2]
                        =  [cons(Y)]  
                                      
              [from(X)] =  [3] X + [3]
                        >  [1] X + [2]
                        =  [cons(X)]  
                                      
          [add(0(), X)] =  [3] X + [6]
                        >  [1] X + [0]
                        =  [X]        
                                      
          [add(s(), Y)] =  [3] Y + [3]
                        >  [0]        
                        =  [s()]      
                                      
           [len(nil())] =  [1]        
                        >= [1]        
                        =  [0()]      
                                      
         [len(cons(X))] =  [2] X + [5]
                        >  [0]        
                        =  [s()]      
                                      

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:
  { fst(0(), Z) -> nil()
  , fst(s(), cons(Y)) -> cons(Y)
  , from(X) -> cons(X)
  , add(0(), X) -> X
  , add(s(), Y) -> s()
  , len(nil()) -> 0()
  , len(cons(X)) -> s() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

Hurray, we answered YES(O(1),O(n^1))