YES

We show the termination of the TRS R:

  f(s(X)) -> f(X)
  g(cons(|0|(),Y)) -> g(Y)
  g(cons(s(X),Y)) -> s(X)
  h(cons(X,Y)) -> h(g(cons(X,Y)))

-- SCC decomposition.

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

p1: f#(s(X)) -> f#(X)
p2: g#(cons(|0|(),Y)) -> g#(Y)
p3: h#(cons(X,Y)) -> h#(g(cons(X,Y)))
p4: h#(cons(X,Y)) -> g#(cons(X,Y))

and R consists of:

r1: f(s(X)) -> f(X)
r2: g(cons(|0|(),Y)) -> g(Y)
r3: g(cons(s(X),Y)) -> s(X)
r4: h(cons(X,Y)) -> h(g(cons(X,Y)))

The estimated dependency graph contains the following SCCs:

  {p1}
  {p3}
  {p2}


-- Reduction pair.

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

p1: f#(s(X)) -> f#(X)

and R consists of:

r1: f(s(X)) -> f(X)
r2: g(cons(|0|(),Y)) -> g(Y)
r3: g(cons(s(X),Y)) -> s(X)
r4: h(cons(X,Y)) -> h(g(cons(X,Y)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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

-- Reduction pair.

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

p1: h#(cons(X,Y)) -> h#(g(cons(X,Y)))

and R consists of:

r1: f(s(X)) -> f(X)
r2: g(cons(|0|(),Y)) -> g(Y)
r3: g(cons(s(X),Y)) -> s(X)
r4: h(cons(X,Y)) -> h(g(cons(X,Y)))

The set of usable rules consists of

  r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > g > |0| > h# > cons
      
      argument filter:
    
        pi(h#) = 1
        pi(cons) = [1, 2]
        pi(g) = 1
        pi(|0|) = []
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > |0| > cons > g > h#
      
      argument filter:
    
        pi(h#) = [1]
        pi(cons) = [1]
        pi(g) = []
        pi(|0|) = []
        pi(s) = 1
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: g#(cons(|0|(),Y)) -> g#(Y)

and R consists of:

r1: f(s(X)) -> f(X)
r2: g(cons(|0|(),Y)) -> g(Y)
r3: g(cons(s(X),Y)) -> s(X)
r4: h(cons(X,Y)) -> h(g(cons(X,Y)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        g# > |0| > cons
      
      argument filter:
    
        pi(g#) = 1
        pi(cons) = [1, 2]
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        g# > |0| > cons
      
      argument filter:
    
        pi(g#) = 1
        pi(cons) = [1, 2]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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