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 reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      f# = s
    
    partial status:
  
      pi(f#) = []
      pi(s) = []

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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          h#_A(x1) = x1 + (2,1)
          cons_A(x1,x2) = ((1,0),(1,1)) x2 + (3,2)
          g_A(x1) = (2,4)
          |0|_A() = (1,1)
          s_A(x1) = (1,1)
  
    precedence:
    
      h# = cons = g = |0| = s
    
    partial status:
  
      pi(h#) = []
      pi(cons) = [2]
      pi(g) = []
      pi(|0|) = []
      pi(s) = []

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 reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          g#_A(x1) = ((1,0),(0,0)) x1 + (2,2)
          cons_A(x1,x2) = x1 + x2 + (2,2)
          |0|_A() = (1,1)
  
    precedence:
    
      g# = cons = |0|
    
    partial status:
  
      pi(g#) = []
      pi(cons) = [2]
      pi(|0|) = []

The next rules are strictly ordered:

  p1

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