YES

Problem:
 g(a()) -> g(b())
 b() -> f(a(),a())
 f(a(),a()) -> g(d())

Proof:
 DP Processor:
  DPs:
   g#(a()) -> b#()
   g#(a()) -> g#(b())
   b#() -> f#(a(),a())
   f#(a(),a()) -> g#(d())
  TRS:
   g(a()) -> g(b())
   b() -> f(a(),a())
   f(a(),a()) -> g(d())
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = -7x0 + -11x1 + 1,
    
    [b#] = 2,
    
    [g#](x0) = x0 + 0,
    
    [d] = 0,
    
    [f](x0, x1) = -2x0 + -2x1 + 1,
    
    [b] = 2,
    
    [g](x0) = x0 + 0,
    
    [a] = 3
   orientation:
    g#(a()) = 3 >= 2 = b#()
    
    g#(a()) = 3 >= 2 = g#(b())
    
    b#() = 2 >= 1 = f#(a(),a())
    
    f#(a(),a()) = 1 >= 0 = g#(d())
    
    g(a()) = 3 >= 2 = g(b())
    
    b() = 2 >= 1 = f(a(),a())
    
    f(a(),a()) = 1 >= 0 = g(d())
   problem:
    DPs:
     
    TRS:
     g(a()) -> g(b())
     b() -> f(a(),a())
     f(a(),a()) -> g(d())
   Qed