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 + 2,
   
   [++](x0, x1) = x0 + x1 + 1,
   
   [nil] = 6
  orientation:
   ++(nil(),y) = y + 7 >= y = y
   
   ++(x,nil()) = x + 7 >= x = x
   
   ++(.(x,y),z) = x + y + z + 3 >= x + y + z + 3 = .(x,++(y,z))
   
   ++(++(x,y),z) = x + y + z + 2 >= x + y + z + 2 = ++(x,++(y,z))
  problem:
   ++(.(x,y),z) -> .(x,++(y,z))
   ++(++(x,y),z) -> ++(x,++(y,z))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [.](x0, x1) = 2x0 + x1 + 4,
    
    [++](x0, x1) = 2x0 + x1 + 1
   orientation:
    ++(.(x,y),z) = 4x + 2y + z + 9 >= 2x + 2y + z + 5 = .(x,++(y,z))
    
    ++(++(x,y),z) = 4x + 2y + z + 3 >= 2x + 2y + z + 2 = ++(x,++(y,z))
   problem:
    
   Qed