YES

We show the termination of the TRS R:

  f(g(x),g(y)) -> f(p(f(g(x),s(y))),g(s(p(x))))
  p(|0|()) -> g(|0|())
  g(s(p(x))) -> p(x)

-- SCC decomposition.

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

p1: f#(g(x),g(y)) -> f#(p(f(g(x),s(y))),g(s(p(x))))
p2: f#(g(x),g(y)) -> p#(f(g(x),s(y)))
p3: f#(g(x),g(y)) -> f#(g(x),s(y))
p4: f#(g(x),g(y)) -> g#(s(p(x)))
p5: f#(g(x),g(y)) -> p#(x)
p6: p#(|0|()) -> g#(|0|())

and R consists of:

r1: f(g(x),g(y)) -> f(p(f(g(x),s(y))),g(s(p(x))))
r2: p(|0|()) -> g(|0|())
r3: g(s(p(x))) -> p(x)

The estimated dependency graph contains the following SCCs:

  (no SCCs)