YES

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

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