YES

Problem:
 minus(minus(x)) -> x
 minus(h(x)) -> h(minus(x))
 minus(f(x,y)) -> f(minus(y),minus(x))

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