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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1) = x1
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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