YES

We show the termination of the TRS R:

  f(x,a()) -> x
  f(x,g(y)) -> f(g(x),y)

-- SCC decomposition.

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

p1: f#(x,g(y)) -> f#(g(x),y)

and R consists of:

r1: f(x,a()) -> x
r2: f(x,g(y)) -> f(g(x),y)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(x,g(y)) -> f#(g(x),y)

and R consists of:

r1: f(x,a()) -> x
r2: f(x,g(y)) -> f(g(x),y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      f#_A(x1,x2) = x2
      g_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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