YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: g#(f(x),y) -> h#(x,y)
p2: h#(x,y) -> g#(x,f(y))

and R consists of:

r1: g(f(x),y) -> f(h(x,y))
r2: h(x,y) -> g(x,f(y))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: g#(f(x),y) -> h#(x,y)
p2: h#(x,y) -> g#(x,f(y))

and R consists of:

r1: g(f(x),y) -> f(h(x,y))
r2: h(x,y) -> g(x,f(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      h# > f > g#
    
    argument filter:
  
      pi(g#) = 1
      pi(f) = [1]
      pi(h#) = 1

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: g(f(x),y) -> f(h(x,y))
r2: h(x,y) -> g(x,f(y))

The estimated dependency graph contains the following SCCs:

  (no SCCs)