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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = x1
        g_A(x1) = ((0,1),(0,0)) x1 + (1,0)
        f_A(x1) = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = ((0,1),(0,0)) x1
        g_A(x1) = (1,1)
        f_A(x1) = (1,1)
    

The next rules are strictly ordered:

  p1

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