YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          g#_A(x1) = x1
          s_A(x1) = x1 + 3
          f_A(x1) = x1 + 1
          |0|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            g#_A(x1) = 0
            s_A(x1) = x1 + 1
            f_A(x1) = x1 + 1
            |0|_A() = 2
            g#!_A(x1) = 0
            s!_A(x1) = 1
            f!_A(x1) = 3
            |0|!_A() = 0
  
    partial status:
  
      pi(g#) = [1]
      pi(s) = [1]
      pi(f) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.