YES

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

Proof:
 DP Processor:
  DPs:
   g#(g(x)) -> h#(g(x))
   g#(g(x)) -> g#(h(g(x)))
   h#(h(x)) -> h#(f(h(x),x))
  TRS:
   g(h(g(x))) -> g(x)
   g(g(x)) -> g(h(g(x)))
   h(h(x)) -> h(f(h(x),x))
  Matrix Interpretation Processor: dim=1
   
   usable rules:
    g(h(g(x))) -> g(x)
    g(g(x)) -> g(h(g(x)))
    h(h(x)) -> h(f(h(x),x))
   interpretation:
    [h#](x0) = 2x0,
    
    [g#](x0) = x0 + 3,
    
    [f](x0, x1) = 0,
    
    [h](x0) = 1,
    
    [g](x0) = 2
   orientation:
    g#(g(x)) = 5 >= 4 = h#(g(x))
    
    g#(g(x)) = 5 >= 4 = g#(h(g(x)))
    
    h#(h(x)) = 2 >= 0 = h#(f(h(x),x))
    
    g(h(g(x))) = 2 >= 2 = g(x)
    
    g(g(x)) = 2 >= 2 = g(h(g(x)))
    
    h(h(x)) = 1 >= 1 = h(f(h(x),x))
   problem:
    DPs:
     
    TRS:
     g(h(g(x))) -> g(x)
     g(g(x)) -> g(h(g(x)))
     h(h(x)) -> h(f(h(x),x))
   Qed