YES

We show the termination of the TRS R:

  f(a()) -> g(h(a()))
  h(g(x)) -> g(h(f(x)))
  k(x,h(x),a()) -> h(x)
  k(f(x),y,x) -> f(x)

-- SCC decomposition.

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

p1: f#(a()) -> h#(a())
p2: h#(g(x)) -> h#(f(x))
p3: h#(g(x)) -> f#(x)

and R consists of:

r1: f(a()) -> g(h(a()))
r2: h(g(x)) -> g(h(f(x)))
r3: k(x,h(x),a()) -> h(x)
r4: k(f(x),y,x) -> f(x)

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: h#(g(x)) -> h#(f(x))

and R consists of:

r1: f(a()) -> g(h(a()))
r2: h(g(x)) -> g(h(f(x)))
r3: k(x,h(x),a()) -> h(x)
r4: k(f(x),y,x) -> f(x)

The set of usable rules consists of

  r1

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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