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