YES

We show the termination of the TRS R:

  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)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(a()) -> f#(c(a()))
p2: f#(c(a())) -> f#(d(b()))
p3: f#(a()) -> f#(d(a()))
p4: f#(c(b())) -> f#(d(a()))
p5: e#(g(X)) -> e#(X)

and R consists of:

r1: f(a()) -> f(c(a()))
r2: f(c(X)) -> X
r3: f(c(a())) -> f(d(b()))
r4: f(a()) -> f(d(a()))
r5: f(d(X)) -> X
r6: f(c(b())) -> f(d(a()))
r7: e(g(X)) -> e(X)

The estimated dependency graph contains the following SCCs:

  {p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: e#(g(X)) -> e#(X)

and R consists of:

r1: f(a()) -> f(c(a()))
r2: f(c(X)) -> X
r3: f(c(a())) -> f(d(b()))
r4: f(a()) -> f(d(a()))
r5: f(d(X)) -> X
r6: f(c(b())) -> f(d(a()))
r7: e(g(X)) -> e(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        e#_A(x1) = ((1,1),(1,1)) x1
        g_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        e#_A(x1) = x1
        g_A(x1) = x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        e#_A(x1) = ((0,1),(1,1)) x1
        g_A(x1) = ((0,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.