YES

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

Proof:
 DP Processor:
  DPs:
   f#(a()) -> f#(c(a()))
   f#(c(a())) -> f#(d(b()))
   f#(a()) -> f#(d(a()))
   f#(c(b())) -> f#(d(a()))
   e#(g(X)) -> e#(X)
  TRS:
   f(a()) -> f(c(a()))
   f(c(X)) -> X
   f(c(a())) -> f(d(b()))
   f(a()) -> f(d(a()))
   f(d(X)) -> X
   f(c(b())) -> f(d(a()))
   e(g(X)) -> e(X)
  Usable Rule Processor:
   DPs:
    f#(a()) -> f#(c(a()))
    f#(c(a())) -> f#(d(b()))
    f#(a()) -> f#(d(a()))
    f#(c(b())) -> f#(d(a()))
    e#(g(X)) -> e#(X)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [e#](x0) = -8x0 + 0,
     
     [f#](x0) = x0 + 0,
     
     [g](x0) = 2x0 + 10,
     
     [d](x0) = 0,
     
     [b] = 1,
     
     [c](x0) = -1x0 + 1,
     
     [a] = 2
    orientation:
     f#(a()) = 2 >= 1 = f#(c(a()))
     
     f#(c(a())) = 1 >= 0 = f#(d(b()))
     
     f#(a()) = 2 >= 0 = f#(d(a()))
     
     f#(c(b())) = 1 >= 0 = f#(d(a()))
     
     e#(g(X)) = -6X + 2 >= -8X + 0 = e#(X)
    problem:
     DPs:
      
     TRS:
      
    Qed