YES

Problem:
 ++(nil(),y) -> y
 ++(x,nil()) -> x
 ++(.(x,y),z) -> .(x,++(y,z))
 ++(++(x,y),z) -> ++(x,++(y,z))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [++](x0, x1) = x0 + x1,
   
   [nil] = 5,
   
   [.](x0, x1) = x0 + x1 + 2
  orientation:
   ++(nil(),y) = y + 5 >= y = y
   
   ++(x,nil()) = x + 5 >= x = x
   
   ++(.(x,y),z) = x + y + z + 2 >= x + y + z + 2 = .(x,++(y,z))
   
   ++(++(x,y),z) = x + y + z >= x + y + z = ++(x,++(y,z))
  problem:
   ++(.(x,y),z) -> .(x,++(y,z))
   ++(++(x,y),z) -> ++(x,++(y,z))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 0 1]       
    [++](x0, x1) = [0 1 0]x0 + x1
                   [0 1 0]       ,
    
                  [1 0 0]          [0]
    [.](x0, x1) = [0 1 0]x0 + x1 + [1]
                  [0 1 0]          [1]
   orientation:
                   [1 1 0]    [1 0 1]        [1]    [1 0 0]    [1 0 1]        [0]               
    ++(.(x,y),z) = [0 1 0]x + [0 1 0]y + z + [1] >= [0 1 0]x + [0 1 0]y + z + [1] = .(x,++(y,z))
                   [0 1 0]    [0 1 0]        [1]    [0 1 0]    [0 1 0]        [1]               
    
                    [1 1 1]    [1 0 1]         [1 0 1]    [1 0 1]                     
    ++(++(x,y),z) = [0 1 0]x + [0 1 0]y + z >= [0 1 0]x + [0 1 0]y + z = ++(x,++(y,z))
                    [0 1 0]    [0 1 0]         [0 1 0]    [0 1 0]                     
   problem:
    ++(++(x,y),z) -> ++(x,++(y,z))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [++](x0, x1) = 2x0 + x1 + 5
    orientation:
     ++(++(x,y),z) = 4x + 2y + z + 15 >= 2x + 2y + z + 10 = ++(x,++(y,z))
    problem:
     
    Qed