YES

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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  f(f(g(g(x)))) -> f(a())
  g(a()) -> a()
  f(a()) -> a()
  f(f(f(x))) -> a()
  f(f(a())) -> a()
  
   T' = (P union S) with
  
   TRS P:
  
   TRS S:f(f(g(g(x)))) -> f(a())
         g(a()) -> a()
         f(a()) -> a()
         f(f(f(x))) -> a()
         f(f(a())) -> a()
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  f(f(a())) = a(), f(f(f(a()))) = a(), f(f(g(a()))) = f(a()), f(a()) = 
  a()
  
   CP(S,P union P^-1) = 
  
   CP(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f](x0) = x0 + 2,
    
    [g](x0) = 2x0 + 6,
    
    [a] = 0
   orientation:
    f(f(g(g(x)))) = 4x + 22 >= 2 = f(a())
    
    g(a()) = 6 >= 0 = a()
    
    f(a()) = 2 >= 0 = a()
    
    f(f(f(x))) = x + 6 >= 0 = a()
    
    f(f(a())) = 4 >= 0 = a()
   problem:
    
   Qed