YES

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

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