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=1
  
  interpretation:
   [f](x0, x1) = 2x0 + 2x1 + 1,
   
   [h](x0) = 4x0 + 2,
   
   [minus](x0) = 4x0 + 1
  orientation:
   minus(minus(x)) = 16x + 5 >= x = x
   
   minus(h(x)) = 16x + 9 >= 16x + 6 = h(minus(x))
   
   minus(f(x,y)) = 8x + 8y + 5 >= 8x + 8y + 5 = f(minus(y),minus(x))
  problem:
   minus(f(x,y)) -> f(minus(y),minus(x))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                  [1 0 0]     [1 0 0]     [0]
    [f](x0, x1) = [0 0 1]x0 + [0 0 1]x1 + [0]
                  [0 1 0]     [0 1 0]     [1],
    
                  [1 1 1]  
    [minus](x0) = [0 1 1]x0
                  [0 1 1]  
   orientation:
                    [1 1 1]    [1 1 1]    [1]    [1 1 1]    [1 1 1]    [0]                       
    minus(f(x,y)) = [0 1 1]x + [0 1 1]y + [1] >= [0 1 1]x + [0 1 1]y + [0] = f(minus(y),minus(x))
                    [0 1 1]    [0 1 1]    [1]    [0 1 1]    [0 1 1]    [1]                       
   problem:
    
   Qed