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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,1)) x2 + (1,3)
            empty_A() = (1,1)
            cons_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,1)) x2 + (4,2)
    
      precedence:
      
        empty > f# = cons
      
      partial status:
    
        pi(f#) = [2]
        pi(empty) = []
        pi(cons) = [2]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1,x2) = (1,1)
            empty_A() = (3,1)
            cons_A(x1,x2) = (2,2)
    
      precedence:
      
        empty > cons > f#
      
      partial status:
    
        pi(f#) = []
        pi(empty) = []
        pi(cons) = []
    

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,1)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,2)
    
      precedence:
      
        f# = cons
      
      partial status:
    
        pi(f#) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,1)
    
      precedence:
      
        f# > cons
      
      partial status:
    
        pi(f#) = [1]
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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