YES

We show the termination of the TRS R:

  rev(ls) -> r1(ls,empty())
  r1(empty(),a) -> a
  r1(cons(x,k),a) -> r1(k,cons(x,a))

-- SCC decomposition.

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

p1: rev#(ls) -> r1#(ls,empty())
p2: r1#(cons(x,k),a) -> r1#(k,cons(x,a))

and R consists of:

r1: rev(ls) -> r1(ls,empty())
r2: r1(empty(),a) -> a
r3: r1(cons(x,k),a) -> r1(k,cons(x,a))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: r1#(cons(x,k),a) -> r1#(k,cons(x,a))

and R consists of:

r1: rev(ls) -> r1(ls,empty())
r2: r1(empty(),a) -> a
r3: r1(cons(x,k),a) -> r1(k,cons(x,a))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          r1#_A(x1,x2) = max{x1 + 3, x2 - 1}
          cons_A(x1,x2) = max{2, x1, x2}
    
      precedence:
      
        r1# = cons
      
      partial status:
    
        pi(r1#) = [1]
        pi(cons) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          r1#_A(x1,x2) = max{0, x1 - 2}
          cons_A(x1,x2) = max{x1 + 1, x2}
    
      precedence:
      
        r1# = cons
      
      partial status:
    
        pi(r1#) = []
        pi(cons) = [2]
    

The next rules are strictly ordered:

  p1

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