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)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [e#](x0) = x0,
    
    [f#](x0) = 1x0 + 0,
    
    [e](x0) = 3x0 + 0,
    
    [g](x0) = 3x0 + 1,
    
    [d](x0) = -3x0 + 0,
    
    [b] = 3,
    
    [c](x0) = -3x0 + 1,
    
    [f](x0) = 4x0 + 0,
    
    [a] = 3
   orientation:
    f#(a()) = 4 >= 2 = f#(c(a()))
    
    f#(c(a())) = 2 >= 1 = f#(d(b()))
    
    f#(a()) = 4 >= 1 = f#(d(a()))
    
    f#(c(b())) = 2 >= 1 = f#(d(a()))
    
    e#(g(X)) = 3X + 1 >= X = e#(X)
    
    f(a()) = 7 >= 5 = f(c(a()))
    
    f(c(X)) = 1X + 5 >= X = X
    
    f(c(a())) = 5 >= 4 = f(d(b()))
    
    f(a()) = 7 >= 4 = f(d(a()))
    
    f(d(X)) = 1X + 4 >= X = X
    
    f(c(b())) = 5 >= 4 = f(d(a()))
    
    e(g(X)) = 6X + 4 >= 3X + 0 = e(X)
   problem:
    DPs:
     
    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)
   Qed