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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        g#_A(x1) = max{4, x1 + 2}
        s_A(x1) = max{6, x1 + 3}
        f_A(x1) = max{4, x1 - 6}
        |0|_A = 0
  
    base order 2:
  
      precedence:
      
        s > g# = f > |0|
      
  
    partial status:
  
      pi(g#) = [1]
      pi(s) = [1]
      pi(f) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.