YES

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

Proof:
 DP Processor:
  DPs:
   g#(a()) -> h#(a(),b(),a())
   i#(x) -> f#(x,x)
   h#(x,x,y) -> g#(x)
  TRS:
   f(x,y) -> x
   g(a()) -> h(a(),b(),a())
   i(x) -> f(x,x)
   h(x,x,y) -> g(x)
  Usable Rule Processor:
   DPs:
    g#(a()) -> h#(a(),b(),a())
    i#(x) -> f#(x,x)
    h#(x,x,y) -> g#(x)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [i#](x0) = 8x0 + -8,
     
     [h#](x0, x1, x2) = -1x0 + 1x1 + -9x2 + 2,
     
     [g#](x0) = x0,
     
     [f#](x0, x1) = x0 + 4x1 + -12,
     
     [b] = 1,
     
     [a] = 4
    orientation:
     g#(a()) = 4 >= 3 = h#(a(),b(),a())
     
     i#(x) = 8x + -8 >= 4x + -12 = f#(x,x)
     
     h#(x,x,y) = 1x + -9y + 2 >= x = g#(x)
    problem:
     DPs:
      
     TRS:
      
    Qed