YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(x,y) -> g#(x,y)
p2: g#(h(x),y) -> f#(x,y)
p3: g#(h(x),y) -> g#(x,y)

and R consists of:

r1: f(x,y) -> g(x,y)
r2: g(h(x),y) -> h(f(x,y))
r3: g(h(x),y) -> h(g(x,y))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: f#(x,y) -> g#(x,y)
p2: g#(h(x),y) -> g#(x,y)
p3: g#(h(x),y) -> f#(x,y)

and R consists of:

r1: f(x,y) -> g(x,y)
r2: g(h(x),y) -> h(f(x,y))
r3: g(h(x),y) -> h(g(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + x2 + (0,1,1,2)
      g#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(1,0,1,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,0,1)) x2
      h_A(x1) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(1,1,1,0)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: f(x,y) -> g(x,y)
r2: g(h(x),y) -> h(f(x,y))
r3: g(h(x),y) -> h(g(x,y))

The estimated dependency graph contains the following SCCs:

  (no SCCs)