YES

We show the termination of the TRS R:

  f(f(X)) -> f(g(f(g(f(X)))))
  f(g(f(X))) -> f(g(X))

-- SCC decomposition.

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

p1: f#(f(X)) -> f#(g(f(g(f(X)))))
p2: f#(f(X)) -> f#(g(f(X)))
p3: f#(g(f(X))) -> f#(g(X))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p3}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1) = x1 + (1,0)
          g_A(x1) = ((1,1),(0,0)) x1 + (1,3)
          f_A(x1) = ((0,1),(1,0)) x1 + (0,2)
  
    precedence:
    
      g > f# = f
    
    partial status:
  
      pi(f#) = [1]
      pi(g) = []
      pi(f) = []

The next rules are strictly ordered:

  p1

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