YES

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

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