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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = ((0,0),(1,0)) x1 + (1,3)
          f_A(x1) = ((1,0),(1,1)) x1 + (4,6)
          g#_A(x1) = ((0,0),(1,0)) x1 + (1,1)
          g_A(x1) = ((1,0),(1,1)) x1 + (3,4)
  
    precedence:
    
      f# > f = g# = g
    
    partial status:
  
      pi(f#) = []
      pi(f) = [1]
      pi(g#) = []
      pi(g) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  (no SCCs)