YES

We show the termination of the TRS R:

  f(g(x)) -> f(a(g(g(f(x))),g(f(x))))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(g(x)) -> f#(a(g(g(f(x))),g(f(x))))
p2: f#(g(x)) -> f#(x)

and R consists of:

r1: f(g(x)) -> f(a(g(g(f(x))),g(f(x))))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(g(x)) -> f#(x)

and R consists of:

r1: f(g(x)) -> f(a(g(g(f(x))),g(f(x))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{4, x1 + 3}
          g_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        f# = g
      
      partial status:
    
        pi(f#) = [1]
        pi(g) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{1, x1 - 1}
          g_A(x1) = x1
    
      precedence:
      
        f# = g
      
      partial status:
    
        pi(f#) = []
        pi(g) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.