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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,1),(1,1)) x2
        a_A() = (3,1)
        g_A(x1) = ((0,1),(0,0)) x1 + (1,1)
        b_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,1),(0,0)) x1 + x2
        a_A() = (1,1)
        g_A(x1) = (3,1)
        b_A() = (4,2)
    

The next rules are strictly ordered:

  p1

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