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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        f > g# > h
      
      argument filter:
    
        pi(g#) = 1
        pi(f) = [2]
        pi(h) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        f > g# > h
      
      argument filter:
    
        pi(g#) = [1]
        pi(f) = 2
        pi(h) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        g# > f > h
      
      argument filter:
    
        pi(g#) = [1]
        pi(f) = []
        pi(h) = [1]
    

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: 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}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        h > g#
      
      argument filter:
    
        pi(g#) = [2]
        pi(h) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        h > g#
      
      argument filter:
    
        pi(g#) = [2]
        pi(h) = [1, 2]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        h > g#
      
      argument filter:
    
        pi(g#) = [2]
        pi(h) = [1, 2]
    

The next rules are strictly ordered:

  p1

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