YES

Problem:
 *(x,*(y,z)) -> *(*(x,y),z)
 *(x,x) -> x

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