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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1) = x1
          g_A(x1) = x1 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            f#_A(x1) = x1
            g_A(x1) = x1 + 1
            f#!_A(x1) = x1 + 1
            g!_A(x1) = 0
  
    partial status:
  
      pi(f#) = [1]
      pi(g) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.