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:

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