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

  r2, r4

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        g#_A(x1) = max{13, x1 + 5}
        s_A(x1) = max{6, x1 + 4}
        f#_A(x1) = max{12, x1 + 8}
        f_A(x1) = max{8, x1}
        |0|_A = 0
        g_A(x1) = max{5, x1 - 4}
  
    base order 2:
  
      precedence:
      
        g# > f# = f > s > g > |0|
      
  
    partial status:
  
      pi(g#) = [1]
      pi(s) = [1]
      pi(f#) = []
      pi(f) = [1]
      pi(|0|) = []
      pi(g) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

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

  (no SCCs)