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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = ((1,0),(1,0)) x1 + (1,0)
        f_A(x1) = ((1,1),(0,0)) x1 + (3,1)
        g#_A(x1) = ((1,0),(1,0)) x1
        g_A(x1) = ((1,1),(0,0)) x1 + (2,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = x1 + (4,0)
        f_A(x1) = ((0,1),(1,1)) x1 + (0,1)
        g#_A(x1) = ((0,1),(1,1)) x1 + (0,1)
        g_A(x1) = (2,3)
    

The next rules are strictly ordered:

  p1, p2

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