YES

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

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