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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          a_A = 4
          g_A(x1) = 0
          b_A = 0
    
      precedence:
      
        f# = a > g = b
      
      partial status:
    
        pi(f#) = [2]
        pi(a) = []
        pi(g) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = x2 + 6
          a_A = 4
          g_A(x1) = 1
          b_A = 0
    
      precedence:
      
        f# = a = b > g
      
      partial status:
    
        pi(f#) = [2]
        pi(a) = []
        pi(g) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p1

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