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))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f'#](x0, x1, x2) = 5x0 + 5x1,
    
    [f#](x0) = 5x0 + 2,
    
    [f'](x0, x1, x2) = 4x0 + 4x2,
    
    [s](x0) = x0 + 3,
    
    [h](x0) = 4,
    
    [f](x0) = x0,
    
    [g](x0) = x0 + 5
   orientation:
    f#(g(x)) = 5x + 27 >= 5x + 2 = f#(x)
    
    f#(g(x)) = 5x + 27 >= 5x + 2 = f#(f(x))
    
    f'#(s(x),y,y) = 5x + 5y + 15 >= 5x + 5y = f'#(y,x,s(x))
    
    f(g(x)) = x + 5 >= x + 5 = g(f(f(x)))
    
    f(h(x)) = 4 >= 4 = h(g(x))
    
    f'(s(x),y,y) = 4x + 4y + 12 >= 4x + 4y + 12 = f'(y,x,s(x))
   problem:
    DPs:
     
    TRS:
     f(g(x)) -> g(f(f(x)))
     f(h(x)) -> h(g(x))
     f'(s(x),y,y) -> f'(y,x,s(x))
   Qed