YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),y) -> g#(y)
   f#(a(),y) -> f#(y,g(y))
  TRS:
   f(a(),y) -> f(y,g(y))
   g(a()) -> b()
   g(b()) -> b()
  TDG Processor:
   DPs:
    f#(a(),y) -> g#(y)
    f#(a(),y) -> f#(y,g(y))
   TRS:
    f(a(),y) -> f(y,g(y))
    g(a()) -> b()
    g(b()) -> b()
   graph:
    f#(a(),y) -> f#(y,g(y)) -> f#(a(),y) -> f#(y,g(y))
    f#(a(),y) -> f#(y,g(y)) -> f#(a(),y) -> g#(y)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(a(),y) -> f#(y,g(y))
    TRS:
     f(a(),y) -> f(y,g(y))
     g(a()) -> b()
     g(b()) -> b()
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [f#](x0, x1) = -7x0 + 1x1 + 0,
      
      [b] = 0,
      
      [g](x0) = -7x0 + 1,
      
      [f](x0, x1) = -7x0 + -1x1 + 0,
      
      [a] = 14
     orientation:
      f#(a(),y) = 1y + 7 >= -6y + 2 = f#(y,g(y))
      
      f(a(),y) = -1y + 7 >= -7y + 0 = f(y,g(y))
      
      g(a()) = 7 >= 0 = b()
      
      g(b()) = 1 >= 0 = b()
     problem:
      DPs:
       
      TRS:
       f(a(),y) -> f(y,g(y))
       g(a()) -> b()
       g(b()) -> b()
     Qed