YES

We show the termination of the TRS R:

  f(|1|()) -> f(g(|1|()))
  f(f(x)) -> f(x)
  g(|0|()) -> g(f(|0|()))
  g(g(x)) -> g(x)

-- SCC decomposition.

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

p1: f#(|1|()) -> f#(g(|1|()))
p2: f#(|1|()) -> g#(|1|())
p3: g#(|0|()) -> g#(f(|0|()))
p4: g#(|0|()) -> f#(|0|())

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)