YES

We show the termination of the TRS R:

  f(|0|()) -> |1|()
  f(s(x)) -> g(f(x))
  g(x) -> +(x,s(x))
  f(s(x)) -> +(f(x),s(f(x)))

-- SCC decomposition.

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

p1: f#(s(x)) -> g#(f(x))
p2: f#(s(x)) -> f#(x)
p3: f#(s(x)) -> f#(x)

and R consists of:

r1: f(|0|()) -> |1|()
r2: f(s(x)) -> g(f(x))
r3: g(x) -> +(x,s(x))
r4: f(s(x)) -> +(f(x),s(f(x)))

The estimated dependency graph contains the following SCCs:

  {p2, p3}


-- 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(|0|()) -> |1|()
r2: f(s(x)) -> g(f(x))
r3: g(x) -> +(x,s(x))
r4: f(s(x)) -> +(f(x),s(f(x)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = x1 + 2
        s_A(x1) = x1 + 2
  
    precedence:
    
      f# = s
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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