YES

We show the termination of the TRS R:

  f(a,empty()) -> g(a,empty())
  f(a,cons(x,k)) -> f(cons(x,a),k)
  g(empty(),d) -> d
  g(cons(x,k),d) -> g(k,cons(x,d))

-- SCC decomposition.

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

p1: f#(a,empty()) -> g#(a,empty())
p2: f#(a,cons(x,k)) -> f#(cons(x,a),k)
p3: g#(cons(x,k),d) -> g#(k,cons(x,d))

and R consists of:

r1: f(a,empty()) -> g(a,empty())
r2: f(a,cons(x,k)) -> f(cons(x,a),k)
r3: g(empty(),d) -> d
r4: g(cons(x,k),d) -> g(k,cons(x,d))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3}


-- Reduction pair.

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

p1: f#(a,cons(x,k)) -> f#(cons(x,a),k)

and R consists of:

r1: f(a,empty()) -> g(a,empty())
r2: f(a,cons(x,k)) -> f(cons(x,a),k)
r3: g(empty(),d) -> d
r4: g(cons(x,k),d) -> g(k,cons(x,d))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = ((0,0,0),(1,0,0),(0,0,0)) x2
      cons_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1 + ((1,0,0),(0,1,0),(1,1,1)) x2 + (1,1,1)

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(x,k),d) -> g#(k,cons(x,d))

and R consists of:

r1: f(a,empty()) -> g(a,empty())
r2: f(a,cons(x,k)) -> f(cons(x,a),k)
r3: g(empty(),d) -> d
r4: g(cons(x,k),d) -> g(k,cons(x,d))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      g#_A(x1,x2) = x1 + ((0,0,0),(1,0,0),(0,1,0)) x2
      cons_A(x1,x2) = ((0,0,0),(1,0,0),(1,1,0)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,2,1)

The next rules are strictly ordered:

  p1

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