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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1,x2) = max{x1 + 2, x2 + 2}
        f_A(x1) = x1
        h#_A(x1,x2) = max{x1 + 2, x2 + 2}
  
    precedence:
    
      g# = f = h#
    
    partial status:
  
      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)