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