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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1
      g_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + (1,0,0,1)

The next rules are strictly ordered:

  p1

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