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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = x1 + x2 + (2,2)
            a_A() = (3,2)
            g_A(x1) = ((0,0),(1,0)) x1 + (2,1)
            b_A() = (1,1)
    
      precedence:
      
        a > f# = g > b
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(a) = []
        pi(g) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = (3,1)
            a_A() = (4,2)
            g_A(x1) = (2,4)
            b_A() = (1,3)
    
      precedence:
      
        b > f# = a = g
      
      partial status:
    
        pi(f#) = []
        pi(a) = []
        pi(g) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p1

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