YES

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

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