YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      g#_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x2
      f_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (1,2)
      h_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (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),z) -> g#(x,f(y,z))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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