YES

We show the termination of the relative TRS R/S:

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

  S:
  g(x) -> g(g(x))

-- SCC decomposition.

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

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

and R consists of:

r1: f(g(f(x))) -> f(g(g(g(f(x)))))
r2: g(x) -> g(g(x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: f(g(f(x))) -> f(g(g(g(f(x)))))
r2: g(x) -> g(g(x))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = x1 + 13
        g_A(x1) = max{0, x1 - 2}
        f_A(x1) = x1 + 6
  
    precedence:
    
      f# = g = f
    
    partial status:
  
      pi(f#) = []
      pi(g) = []
      pi(f) = [1]

The next rules are strictly ordered:

  p1

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