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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          h#_A(x1) = x1 + (1,0)
          g_A(x1) = ((0,1),(0,0)) x1 + (3,1)
          f_A(x1) = ((0,1),(0,0)) x1 + (1,1)
          a_A() = (6,4)
          h_A(x1) = (4,1)
  
    precedence:
    
      h# = a > g = f > h
    
    partial status:
  
      pi(h#) = [1]
      pi(g) = []
      pi(f) = []
      pi(a) = []
      pi(h) = []

The next rules are strictly ordered:

  p1

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