YES

Problem 1: 

(VAR v_NonEmpty:S a:S k:S x:S y:S)
(RULES
f(cons(a:S,k:S),y:S) -> f(y:S,k:S)
f(empty,cons(a:S,k:S)) -> f(cons(a:S,k:S),k:S)
f(x:S,empty) -> x:S
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 F(cons(a:S,k:S),y:S) -> F(y:S,k:S)
 F(empty,cons(a:S,k:S)) -> F(cons(a:S,k:S),k:S)
-> Rules:
 f(cons(a:S,k:S),y:S) -> f(y:S,k:S)
 f(empty,cons(a:S,k:S)) -> f(cons(a:S,k:S),k:S)
 f(x:S,empty) -> x:S

Problem 1: 

SCC Processor:
-> Pairs:
 F(cons(a:S,k:S),y:S) -> F(y:S,k:S)
 F(empty,cons(a:S,k:S)) -> F(cons(a:S,k:S),k:S)
-> Rules:
 f(cons(a:S,k:S),y:S) -> f(y:S,k:S)
 f(empty,cons(a:S,k:S)) -> f(cons(a:S,k:S),k:S)
 f(x:S,empty) -> x:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(cons(a:S,k:S),y:S) -> F(y:S,k:S)
 F(empty,cons(a:S,k:S)) -> F(cons(a:S,k:S),k:S)
->->-> Rules:
 f(cons(a:S,k:S),y:S) -> f(y:S,k:S)
 f(empty,cons(a:S,k:S)) -> f(cons(a:S,k:S),k:S)
 f(x:S,empty) -> x:S

Problem 1: 

Reduction Pair Processor:
-> Pairs:
 F(cons(a:S,k:S),y:S) -> F(y:S,k:S)
 F(empty,cons(a:S,k:S)) -> F(cons(a:S,k:S),k:S)
-> Rules:
 f(cons(a:S,k:S),y:S) -> f(y:S,k:S)
 f(empty,cons(a:S,k:S)) -> f(cons(a:S,k:S),k:S)
 f(x:S,empty) -> x:S
-> Usable rules:
 Empty
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[cons](X1,X2) = 2.X1 + 2.X2 + 2
[empty] = 0
[F](X1,X2) = X1 + 2.X2

Problem 1: 

SCC Processor:
-> Pairs:
 F(empty,cons(a:S,k:S)) -> F(cons(a:S,k:S),k:S)
-> Rules:
 f(cons(a:S,k:S),y:S) -> f(y:S,k:S)
 f(empty,cons(a:S,k:S)) -> f(cons(a:S,k:S),k:S)
 f(x:S,empty) -> x:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.