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:

  lexicographic path order with precedence:
  
    precedence:
    
      |0| > f > g# > s
    
    argument filter:
  
      pi(g#) = [1]
      pi(s) = 1
      pi(|0|) = []
      pi(f) = []

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:

  lexicographic path order with precedence:
  
    precedence:
    
      s > f#
    
    argument filter:
  
      pi(f#) = [1]
      pi(s) = [1]

The next rules are strictly ordered:

  p1
  r1, r2, r3

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