YES

We show the termination of the TRS R:

  f(h(x)) -> f(i(x))
  g(i(x)) -> g(h(x))
  h(a()) -> b()
  i(a()) -> b()

-- SCC decomposition.

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

p1: f#(h(x)) -> f#(i(x))
p2: f#(h(x)) -> i#(x)
p3: g#(i(x)) -> g#(h(x))
p4: g#(i(x)) -> h#(x)

and R consists of:

r1: f(h(x)) -> f(i(x))
r2: g(i(x)) -> g(h(x))
r3: h(a()) -> b()
r4: i(a()) -> b()

The estimated dependency graph contains the following SCCs:

  {p1}
  {p3}


-- Reduction pair.

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

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

and R consists of:

r1: f(h(x)) -> f(i(x))
r2: g(i(x)) -> g(h(x))
r3: h(a()) -> b()
r4: i(a()) -> b()

The set of usable rules consists of

  r4

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1) = x1
      h_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (2,1,1)
      i_A(x1) = ((0,0,0),(0,0,0),(1,0,0)) x1 + (1,2,2)
      a_A() = (1,1,1)
      b_A() = (0,3,4)

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

r1: f(h(x)) -> f(i(x))
r2: g(i(x)) -> g(h(x))
r3: h(a()) -> b()
r4: i(a()) -> b()

The set of usable rules consists of

  r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      g#_A(x1) = x1
      i_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (2,1,1)
      h_A(x1) = ((0,0,0),(0,0,0),(1,0,0)) x1 + (1,2,2)
      a_A() = (1,1,1)
      b_A() = (0,3,4)

The next rules are strictly ordered:

  p1

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