YES

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

Proof:
 DP Processor:
  DPs:
   a#() -> g#(c())
   f#(g(X),b()) -> a#()
   f#(g(X),b()) -> f#(a(),X)
  TRS:
   a() -> g(c())
   g(a()) -> b()
   f(g(X),b()) -> f(a(),X)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = -3x0 + -2x1 + 0,
    
    [g#](x0) = x0 + 0,
    
    [a#] = 1,
    
    [f](x0, x1) = x0 + 2x1 + 0,
    
    [b] = 4,
    
    [g](x0) = 3x0 + 0,
    
    [c] = 0,
    
    [a] = 4
   orientation:
    a#() = 1 >= 0 = g#(c())
    
    f#(g(X),b()) = X + 2 >= 1 = a#()
    
    f#(g(X),b()) = X + 2 >= -2X + 1 = f#(a(),X)
    
    a() = 4 >= 3 = g(c())
    
    g(a()) = 7 >= 4 = b()
    
    f(g(X),b()) = 3X + 6 >= 2X + 4 = f(a(),X)
   problem:
    DPs:
     
    TRS:
     a() -> g(c())
     g(a()) -> b()
     f(g(X),b()) -> f(a(),X)
   Qed