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, >) with

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

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.