YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

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

and R consists of:

r1: g(s(x)) -> f(x)
r2: f(|0|()) -> s(|0|())
r3: f(s(x)) -> s(s(g(x)))
r4: g(|0|()) -> |0|()

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: g(s(x)) -> f(x)
r2: f(|0|()) -> s(|0|())
r3: f(s(x)) -> s(s(g(x)))
r4: g(|0|()) -> |0|()

The set of usable rules consists of

  (no rules)

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
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          g#_A(x1) = max{0, x1 - 13}
          s_A(x1) = max{12, x1 + 6}
          f#_A(x1) = max{0, x1 - 13}
          g#!_A(x1) = max{2, x1 - 13}
          s!_A(x1) = max{4, x1 - 9}
          f#!_A(x1) = max{1, x1 - 9}
  
    partial status:
  
      pi(g#) = []
      pi(s) = [1]
      pi(f#) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.