YES

We show the termination of R/S, where R is

  f(x,0) -> s(x)
  g(x) -> h(x,gen)
  h(0,x) -> f(x,x)
  a -> b

and S is:

  gen -> s(gen)

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  g#(x) -> h#(x,gen)
  h#(0,x) -> f#(x,x)

and Q consists of the rules:

  f(x,0) -> s(x)
  g(x) -> h(x,gen)
  h(0,x) -> f(x,x)
  a -> b
  gen -> s(gen)

The weakly monotone algebra (N^2, >_lex) with

  g#_A((x1,y1)) = (x1 + 2, 0)
  h#_A((x1,y1),(x2,y2)) = (x1 + x2, y1 + y2)
  gen_A = (1, 1)
  0_A = (1, 1)
  f#_A((x1,y1),(x2,y2)) = (0, 0)
  f_A((x1,y1),(x2,y2)) = (x1 + 1, 2)
  s_A((x1,y1)) = (0, 3)
  g_A((x1,y1)) = (x1 + 4, 0)
  h_A((x1,y1),(x2,y2)) = (x2 + 2, 1)
  a_A = (1, 0)
  b_A = (0, 1)

strictly orients the following dependency pairs:

  g#(x) -> h#(x,gen)
  h#(0,x) -> f#(x,x)

We remove them from the set of dependency pairs.

No dependency pair remains.