YES

We show the termination of the TRS R:

  f(f(x)) -> f(x)
  f(s(x)) -> f(x)
  g(s(|0|())) -> g(f(s(|0|())))

-- SCC decomposition.

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

p1: f#(s(x)) -> f#(x)
p2: g#(s(|0|())) -> g#(f(s(|0|())))
p3: g#(s(|0|())) -> f#(s(|0|()))

and R consists of:

r1: f(f(x)) -> f(x)
r2: f(s(x)) -> f(x)
r3: g(s(|0|())) -> g(f(s(|0|())))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}


-- Reduction pair.

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

p1: g#(s(|0|())) -> g#(f(s(|0|())))

and R consists of:

r1: f(f(x)) -> f(x)
r2: f(s(x)) -> f(x)
r3: g(s(|0|())) -> g(f(s(|0|())))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      g#_A(x1) = x1
      s_A(x1) = x1
      |0|_A() = 1
      f_A(x1) = 0

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

r1: f(f(x)) -> f(x)
r2: f(s(x)) -> f(x)
r3: g(s(|0|())) -> g(f(s(|0|())))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      f#_A(x1) = x1
      s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1
  r1, r2, r3

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