YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),x) -> g#(x)
   f#(a(),x) -> f#(g(x),x)
   h#(g(x)) -> h#(a())
   g#(h(x)) -> g#(x)
  TRS:
   f(a(),x) -> f(g(x),x)
   h(g(x)) -> h(a())
   g(h(x)) -> g(x)
   h(h(x)) -> x
  Usable Rule Processor:
   DPs:
    f#(a(),x) -> g#(x)
    f#(a(),x) -> f#(g(x),x)
    h#(g(x)) -> h#(a())
    g#(h(x)) -> g#(x)
   TRS:
    g(h(x)) -> g(x)
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     g(h(x)) -> g(x)
    interpretation:
     [h#](x0) = [0 1 1 0]x0,
     
     [g#](x0) = [0 0 0 1]x0 + [1],
     
     [f#](x0, x1) = [0 0 0 1]x0 + [0 0 0 1]x1 + [1],
     
               [0 0 1 0]     [0]
               [0 1 0 1]     [0]
     [h](x0) = [1 0 1 0]x0 + [1]
               [0 0 0 1]     [1],
     
               [0 1 0 1]     [0]
               [1 0 1 0]     [1]
     [g](x0) = [1 0 1 0]x0 + [1]
               [0 0 0 0]     [0],
     
           [0]
           [0]
     [a] = [0]
           [1]
    orientation:
     f#(a(),x) = [0 0 0 1]x + [2] >= [0 0 0 1]x + [1] = g#(x)
     
     f#(a(),x) = [0 0 0 1]x + [2] >= [0 0 0 1]x + [1] = f#(g(x),x)
     
     h#(g(x)) = [2 0 2 0]x + [2] >= [0] = h#(a())
     
     g#(h(x)) = [0 0 0 1]x + [2] >= [0 0 0 1]x + [1] = g#(x)
     
               [0 1 0 2]    [1]    [0 1 0 1]    [0]       
               [1 0 2 0]    [2]    [1 0 1 0]    [1]       
     g(h(x)) = [1 0 2 0]x + [2] >= [1 0 1 0]x + [1] = g(x)
               [0 0 0 0]    [0]    [0 0 0 0]    [0]       
    problem:
     DPs:
      
     TRS:
      g(h(x)) -> g(x)
    Qed