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

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            f#_A(0) = 0
            f#_A(1) = 0
            g_A(0) = 1
            g_A(1) = 1
            f_A(0) = 0
            f_A(1) = 0
            a_A(0,0) = 0
            a_A(0,1) = 0
            a_A(1,0) = 0
            a_A(1,1) = 1
            f#!_A(0) = 0
            f#!_A(1) = 0
            g!_A(0) = 1
            g!_A(1) = 1
            f!_A(0) = 0
            f!_A(1) = 0
            a!_A(0,0) = 0
            a!_A(0,1) = 0
            a!_A(1,0) = 0
            a!_A(1,1) = 1
  
    partial status:
  
      pi(f#) = [1]
      pi(g) = [1]
      pi(f) = []
      pi(a) = []

The following rules are removed:

  p1

Then no dependency pair remains.