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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      f#_A(x1) = ((1,0),(1,0)) x1
      f_A(x1) = ((1,1),(1,1)) x1 + (1,1)
      g#_A(x1) = ((0,1),(0,1)) x1
      g_A(x1) = ((0,1),(1,1)) x1 + (2,1)

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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)