YES

We show the termination of the TRS R:

  f(g(x)) -> g(g(f(x)))
  f(g(x)) -> g(g(g(x)))

-- SCC decomposition.

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)) -> g(g(f(x)))
r2: f(g(x)) -> g(g(g(x)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- 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)) -> g(g(f(x)))
r2: f(g(x)) -> g(g(g(x)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      f#_A(x1) = x1
      g_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1
  r1, r2

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