YES

Problem:
 f(f(X)) -> f(a(b(f(X))))
 f(a(g(X))) -> b(X)
 b(X) -> a(X)

Proof:
 DP Processor:
  DPs:
   f#(f(X)) -> b#(f(X))
   f#(f(X)) -> f#(a(b(f(X))))
   f#(a(g(X))) -> b#(X)
  TRS:
   f(f(X)) -> f(a(b(f(X))))
   f(a(g(X))) -> b(X)
   b(X) -> a(X)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [b#](x0) = 2x0 + 0,
    
    [f#](x0) = 3x0 + 0,
    
    [g](x0) = 4x0,
    
    [a](x0) = -3x0 + 0,
    
    [b](x0) = x0 + 2,
    
    [f](x0) = 1x0 + 3
   orientation:
    f#(f(X)) = 4X + 6 >= 3X + 5 = b#(f(X))
    
    f#(f(X)) = 4X + 6 >= 1X + 3 = f#(a(b(f(X))))
    
    f#(a(g(X))) = 4X + 3 >= 2X + 0 = b#(X)
    
    f(f(X)) = 2X + 4 >= -1X + 3 = f(a(b(f(X))))
    
    f(a(g(X))) = 2X + 3 >= X + 2 = b(X)
    
    b(X) = X + 2 >= -3X + 0 = a(X)
   problem:
    DPs:
     
    TRS:
     f(f(X)) -> f(a(b(f(X))))
     f(a(g(X))) -> b(X)
     b(X) -> a(X)
   Qed