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 combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = ((0,0),(1,0)) x1 + (3,1)
            f_A(x1) = x1 + (2,0)
            h#_A(x1,x2) = ((0,0),(1,0)) x1 + (3,2)
    
      precedence:
      
        f = h# > g#
      
      partial status:
    
        pi(g#) = []
        pi(f) = [1]
        pi(h#) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = (3,1)
            f_A(x1) = ((1,0),(1,1)) x1 + (1,3)
            h#_A(x1,x2) = (2,2)
    
      precedence:
      
        g# > f = h#
      
      partial status:
    
        pi(g#) = []
        pi(f) = [1]
        pi(h#) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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