YES

Problem:
 f(h(x)) -> f(i(x))
 g(i(x)) -> g(h(x))
 h(a()) -> b()
 i(a()) -> b()

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