YES

We show the termination of the TRS R:

  f(x,empty()) -> x
  f(empty(),cons(a,k)) -> f(cons(a,k),k)
  f(cons(a,k),y) -> f(y,k)

-- SCC decomposition.

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

p1: f#(empty(),cons(a,k)) -> f#(cons(a,k),k)
p2: f#(cons(a,k),y) -> f#(y,k)

and R consists of:

r1: f(x,empty()) -> x
r2: f(empty(),cons(a,k)) -> f(cons(a,k),k)
r3: f(cons(a,k),y) -> f(y,k)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(empty(),cons(a,k)) -> f#(cons(a,k),k)
p2: f#(cons(a,k),y) -> f#(y,k)

and R consists of:

r1: f(x,empty()) -> x
r2: f(empty(),cons(a,k)) -> f(cons(a,k),k)
r3: f(cons(a,k),y) -> f(y,k)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{3, x1 - 4, x2 - 3}
        empty_A = 8
        cons_A(x1,x2) = max{x1 + 2, x2 + 2}
  
    precedence:
    
      f# = empty = cons
    
    partial status:
  
      pi(f#) = []
      pi(empty) = []
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: f(x,empty()) -> x
r2: f(empty(),cons(a,k)) -> f(cons(a,k),k)
r3: f(cons(a,k),y) -> f(y,k)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: f(x,empty()) -> x
r2: f(empty(),cons(a,k)) -> f(cons(a,k),k)
r3: f(cons(a,k),y) -> f(y,k)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{x1, x2 + 1}
        cons_A(x1,x2) = max{x1 + 1, x2 + 2}
  
    precedence:
    
      f# = cons
    
    partial status:
  
      pi(f#) = [1]
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

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