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:

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

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: 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      h#_A(x1) = ((1,0,0,0),(0,1,0,0),(1,1,1,0),(0,0,0,1)) x1
      cons_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,0,1,1)) x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x2 + (1,2,2,2)
      g_A(x1) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,1,0)) x1 + (0,1,0,0)
      |0|_A() = (1,1,1,0)
      s_A(x1) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x1 + (1,2,7,11)

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:

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

The next rules are strictly ordered:

  p1

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