YES

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

Proof:
 DP Processor:
  DPs:
   f#(a()) -> f#(b())
   g#(b()) -> g#(a())
   f#(x) -> g#(x)
  TRS:
   f(a()) -> f(b())
   g(b()) -> g(a())
   f(x) -> g(x)
  Usable Rule Processor:
   DPs:
    f#(a()) -> f#(b())
    g#(b()) -> g#(a())
    f#(x) -> g#(x)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 2
    usable rules:
     
    interpretation:
     [g#](x0) = [-& 0 ]x0 + [0],
     
     [f#](x0) = [0 1]x0 + [1],
     
           [0]
     [b] = [1],
     
           [3]
     [a] = [0]
    orientation:
     f#(a()) = 3 >= 2 = f#(b())
     
     g#(b()) = 1 >= 0 = g#(a())
     
     f#(x) = [0 1]x + [1] >= [-& 0 ]x + [0] = g#(x)
    problem:
     DPs:
      
     TRS:
      
    Qed