YES

Problem:
 2nd(cons1(X,cons(Y,Z))) -> Y
 2nd(cons(X,X1)) -> 2nd(cons1(X,activate(X1)))
 from(X) -> cons(X,n__from(s(X)))
 from(X) -> n__from(X)
 activate(n__from(X)) -> from(X)
 activate(X) -> X

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
                   [1 0]  
   [n__from](x0) = [1 2]x0,
   
             [1 0]  
   [s](x0) = [0 0]x0,
   
                [3 2]     [0]
   [from](x0) = [3 2]x0 + [3],
   
                    [2 1]     [1]
   [activate](x0) = [0 3]x0 + [3],
   
               [1 2]     [2]
   [2nd](x0) = [1 2]x0 + [2],
   
                     [1 2]     [1 1]  
   [cons1](x0, x1) = [1 0]x0 + [0 0]x1,
   
                    [1 0]     [2 0]     [0]
   [cons](x0, x1) = [1 1]x0 + [0 2]x1 + [2]
  orientation:
                             [3 2]    [2 1]    [2 2]    [4]         
   2nd(cons1(X,cons(Y,Z))) = [3 2]X + [2 1]Y + [2 2]Z + [4] >= Y = Y
   
                     [3 2]    [2 4]     [6]    [3 2]    [2 4]     [6]                             
   2nd(cons(X,X1)) = [3 2]X + [2 4]X1 + [6] >= [3 2]X + [2 4]X1 + [6] = 2nd(cons1(X,activate(X1)))
   
             [3 2]    [0]    [3 0]    [0]                        
   from(X) = [3 2]X + [3] >= [3 1]X + [2] = cons(X,n__from(s(X)))
   
             [3 2]    [0]    [1 0]              
   from(X) = [3 2]X + [3] >= [1 2]X = n__from(X)
   
                          [3 2]    [1]    [3 2]    [0]          
   activate(n__from(X)) = [3 6]X + [3] >= [3 2]X + [3] = from(X)
   
                 [2 1]    [1]         
   activate(X) = [0 3]X + [3] >= X = X
  problem:
   2nd(cons(X,X1)) -> 2nd(cons1(X,activate(X1)))
   from(X) -> cons(X,n__from(s(X)))
   from(X) -> n__from(X)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [n__from](x0) = 4x0,
    
    [s](x0) = x0,
    
    [from](x0) = 6x0 + 3,
    
    [activate](x0) = x0 + 1,
    
    [2nd](x0) = 4x0,
    
    [cons1](x0, x1) = x0 + x1 + 2,
    
    [cons](x0, x1) = x0 + x1 + 3
   orientation:
    2nd(cons(X,X1)) = 4X + 4X1 + 12 >= 4X + 4X1 + 12 = 2nd(cons1(X,activate(X1)))
    
    from(X) = 6X + 3 >= 5X + 3 = cons(X,n__from(s(X)))
    
    from(X) = 6X + 3 >= 4X = n__from(X)
   problem:
    2nd(cons(X,X1)) -> 2nd(cons1(X,activate(X1)))
    from(X) -> cons(X,n__from(s(X)))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [n__from](x0) = 2x0 + 4,
     
     [s](x0) = 3x0,
     
     [from](x0) = 7x0 + 5,
     
     [activate](x0) = x0,
     
     [2nd](x0) = 2x0 + 4,
     
     [cons1](x0, x1) = x0 + x1,
     
     [cons](x0, x1) = x0 + x1
    orientation:
     2nd(cons(X,X1)) = 2X + 2X1 + 4 >= 2X + 2X1 + 4 = 2nd(cons1(X,activate(X1)))
     
     from(X) = 7X + 5 >= 7X + 4 = cons(X,n__from(s(X)))
    problem:
     2nd(cons(X,X1)) -> 2nd(cons1(X,activate(X1)))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                       [1 0 0]  
      [activate](x0) = [0 0 0]x0
                       [0 0 0]  ,
      
                  [1 0 1]  
      [2nd](x0) = [0 0 0]x0
                  [0 0 0]  ,
      
                        [1 0 0]     [1 0 0]  
      [cons1](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                        [0 0 0]     [0 0 0]  ,
      
                       [1 0 0]     [1 0 0]     [0]
      [cons](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                       [0 0 0]     [0 0 0]     [1]
     orientation:
                        [1 0 0]    [1 0 0]     [1]    [1 0 0]    [1 0 0]                               
      2nd(cons(X,X1)) = [0 0 0]X + [0 0 0]X1 + [0] >= [0 0 0]X + [0 0 0]X1 = 2nd(cons1(X,activate(X1)))
                        [0 0 0]    [0 0 0]     [0]    [0 0 0]    [0 0 0]                               
     problem:
      
     Qed