YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

p1: f#(f(x)) -> g#(f(x))
p2: g#(g(x)) -> f#(x)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(f(x)) -> g#(f(x))
p2: g#(g(x)) -> f#(x)

and R consists of:

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

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1) = max{16, x1 + 1}
        f_A(x1) = max{18, x1 + 13}
        g#_A(x1) = max{17, x1 - 3}
        g_A(x1) = max{27, x1 + 7}
  
    base order 2:
  
      precedence:
      
        f# > g > f > g#
      
  
    partial status:
  
      pi(f#) = [1]
      pi(f) = [1]
      pi(g#) = []
      pi(g) = [1]

The following rules are removed:

  p2



-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)