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)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [i#](x0) = 2x0 + 0,
    
    [h#](x0, x1, x2) = 4x1 + 1x2,
    
    [g#](x0) = 3x0,
    
    [f#](x0, x1) = x0 + 1x1,
    
    [i](x0) = 9x0 + 0,
    
    [h](x0, x1, x2) = x0 + 3x1 + 1,
    
    [b] = 1,
    
    [g](x0) = 2x0 + 0,
    
    [a] = 4,
    
    [f](x0, x1) = 2x0 + 8x1
   orientation:
    g#(a()) = 7 >= 5 = h#(a(),b(),a())
    
    i#(x) = 2x + 0 >= 1x = f#(x,x)
    
    h#(x,x,y) = 4x + 1y >= 3x = g#(x)
    
    f(x,y) = 2x + 8y >= x = x
    
    g(a()) = 6 >= 4 = h(a(),b(),a())
    
    i(x) = 9x + 0 >= 8x = f(x,x)
    
    h(x,x,y) = 3x + 1 >= 2x + 0 = g(x)
   problem:
    DPs:
     
    TRS:
     f(x,y) -> x
     g(a()) -> h(a(),b(),a())
     i(x) -> f(x,x)
     h(x,x,y) -> g(x)
   Qed