YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(a(),y) -> f#(y,g(y))
p2: f#(a(),y) -> g#(y)

and R consists of:

r1: f(a(),y) -> f(y,g(y))
r2: g(a()) -> b()
r3: g(b()) -> b()

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: f(a(),y) -> f(y,g(y))
r2: g(a()) -> b()
r3: g(b()) -> b()

The set of usable rules consists of

  r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      f#_A(x1,x2) = x1 + x2
      a_A() = 1
      g_A(x1) = 0
      b_A() = 0

The next rules are strictly ordered:

  p1

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