YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [k](x0, x1, x2) = x0 + x1 + 2x2 + 4,
   
   [f](x0) = x0,
   
   [g](x0) = 2x0,
   
   [a] = 0,
   
   [h](x0) = 4x0
  orientation:
   f(a()) = 0 >= 0 = g(h(a()))
   
   h(g(x)) = 8x >= 8x = g(h(f(x)))
   
   k(x,h(x),a()) = 5x + 4 >= 4x = h(x)
   
   k(f(x),y,x) = 3x + y + 4 >= x = f(x)
  problem:
   f(a()) -> g(h(a()))
   h(g(x)) -> g(h(f(x)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]  
    [f](x0) = [0 0 1]x0
              [0 0 0]  ,
    
              [1 0 0]     [0]
    [g](x0) = [0 1 1]x0 + [1]
              [0 0 0]     [0],
    
          [1]
    [a] = [0]
          [1],
    
              [1 1 0]  
    [h](x0) = [0 1 0]x0
              [0 0 0]  
   orientation:
             [1]    [1]            
    f(a()) = [1] >= [1] = g(h(a()))
             [0]    [0]            
    
              [1 1 1]    [1]    [1 0 1]    [0]             
    h(g(x)) = [0 1 1]x + [1] >= [0 0 1]x + [1] = g(h(f(x)))
              [0 0 0]    [0]    [0 0 0]    [0]             
   problem:
    f(a()) -> g(h(a()))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]     [1]
     [f](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [0],
     
               [1 0 0]  
     [g](x0) = [0 0 0]x0
               [0 0 0]  ,
     
           [0]
     [a] = [0]
           [0],
     
               [1 0 0]  
     [h](x0) = [0 0 0]x0
               [0 0 0]  
    orientation:
              [1]    [0]            
     f(a()) = [0] >= [0] = g(h(a()))
              [0]    [0]            
    problem:
     
    Qed