YES

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

Proof:
 DP Processor:
  DPs:
   f#(g(x)) -> f#(x)
   f#(g(x)) -> f#(f(x))
   f'#(s(x),y,y) -> f'#(y,x,s(x))
  TRS:
   f(g(x)) -> g(f(f(x)))
   f(h(x)) -> h(g(x))
   f'(s(x),y,y) -> f'(y,x,s(x))
  Usable Rule Processor:
   DPs:
    f#(g(x)) -> f#(x)
    f#(g(x)) -> f#(f(x))
    f'#(s(x),y,y) -> f'#(y,x,s(x))
   TRS:
    f(g(x)) -> g(f(f(x)))
    f(h(x)) -> h(g(x))
   Matrix Interpretation Processor: dim=1
    
    usable rules:
     f(g(x)) -> g(f(f(x)))
     f(h(x)) -> h(g(x))
    interpretation:
     [f'#](x0, x1, x2) = 2x0 + 5/2x1 + 1,
     
     [f#](x0) = 1/2x0,
     
     [s](x0) = 2x0 + 2,
     
     [h](x0) = 3,
     
     [f](x0) = x0,
     
     [g](x0) = 2x0 + 3
    orientation:
     f#(g(x)) = x + 3/2 >= 1/2x = f#(x)
     
     f#(g(x)) = x + 3/2 >= 1/2x = f#(f(x))
     
     f'#(s(x),y,y) = 4x + 5/2y + 5 >= 5/2x + 2y + 1 = f'#(y,x,s(x))
     
     f(g(x)) = 2x + 3 >= 2x + 3 = g(f(f(x)))
     
     f(h(x)) = 3 >= 3 = h(g(x))
    problem:
     DPs:
      
     TRS:
      f(g(x)) -> g(f(f(x)))
      f(h(x)) -> h(g(x))
    Qed