YES

Problem:
 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()

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
               [1 0 0]     [1]
   [len](x0) = [0 1 0]x0 + [0]
               [0 1 0]     [0],
   
                   [1 0 0]     [1 1 1]     [1]
   [add](x0, x1) = [0 1 0]x0 + [1 1 1]x1 + [0]
                   [0 1 0]     [1 1 1]     [0],
   
                [1 0 0]     [1]
   [from](x0) = [0 0 0]x0 + [0]
                [0 0 0]     [0],
   
                [1 0 0]  
   [cons](x0) = [0 0 0]x0
                [0 0 0]  ,
   
         [0]
   [s] = [0]
         [0],
   
           [0]
   [nil] = [1]
           [0],
   
                   [1 0 0]     [1 0 0]     [1]
   [fst](x0, x1) = [0 1 0]x0 + [0 0 0]x1 + [0]
                   [0 0 0]     [0 0 0]     [0],
   
         [0]
   [0] = [1]
         [1]
  orientation:
                [1 0 0]    [1]    [0]        
   fst(0(),Z) = [0 0 0]Z + [1] >= [1] = nil()
                [0 0 0]    [0]    [0]        
   
                      [1 0 0]    [1]    [1 0 0]           
   fst(s(),cons(Y)) = [0 0 0]Y + [0] >= [0 0 0]Y = cons(Y)
                      [0 0 0]    [0]    [0 0 0]           
   
             [1 0 0]    [1]    [1 0 0]           
   from(X) = [0 0 0]X + [0] >= [0 0 0]X = cons(X)
             [0 0 0]    [0]    [0 0 0]           
   
                [1 1 1]    [1]         
   add(0(),X) = [1 1 1]X + [1] >= X = X
                [1 1 1]    [1]         
   
                [1 1 1]    [1]    [0]      
   add(s(),Y) = [1 1 1]Y + [0] >= [0] = s()
                [1 1 1]    [0]    [0]      
   
                [1]    [0]      
   len(nil()) = [1] >= [1] = 0()
                [1]    [1]      
   
                  [1 0 0]    [1]    [0]      
   len(cons(X)) = [0 0 0]X + [0] >= [0] = s()
                  [0 0 0]    [0]    [0]      
  problem:
   
  Qed