YES

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

Proof:
 DP Processor:
  DPs:
   f#(g(h(x,y)),f(a(),a())) -> f#(y,a())
   f#(g(h(x,y)),f(a(),a())) -> f#(h(x,x),g(f(y,a())))
  TRS:
   f(g(h(x,y)),f(a(),a())) -> f(h(x,x),g(f(y,a())))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = 2x1 + 4,
    
    [f](x0, x1) = 2x1 + 0,
    
    [a] = 1,
    
    [g](x0) = 2,
    
    [h](x0, x1) = x0 + 6
   orientation:
    f#(g(h(x,y)),f(a(),a())) = 5 >= 4 = f#(y,a())
    
    f#(g(h(x,y)),f(a(),a())) = 5 >= 4 = f#(h(x,x),g(f(y,a())))
    
    f(g(h(x,y)),f(a(),a())) = 5 >= 4 = f(h(x,x),g(f(y,a())))
   problem:
    DPs:
     
    TRS:
     f(g(h(x,y)),f(a(),a())) -> f(h(x,x),g(f(y,a())))
   Qed