YES

Problem 1: 

(VAR v_NonEmpty:S X:S)
(RULES
activate(n__f(X:S)) -> f(X:S)
activate(X:S) -> X:S
f(0) -> cons(0,n__f(s(0)))
f(s(0)) -> f(p(s(0)))
f(X:S) -> n__f(X:S)
p(s(X:S)) -> X:S
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 ACTIVATE(n__f(X:S)) -> F(X:S)
 F(s(0)) -> F(p(s(0)))
 F(s(0)) -> P(s(0))
-> Rules:
 activate(n__f(X:S)) -> f(X:S)
 activate(X:S) -> X:S
 f(0) -> cons(0,n__f(s(0)))
 f(s(0)) -> f(p(s(0)))
 f(X:S) -> n__f(X:S)
 p(s(X:S)) -> X:S

Problem 1: 

SCC Processor:
-> Pairs:
 ACTIVATE(n__f(X:S)) -> F(X:S)
 F(s(0)) -> F(p(s(0)))
 F(s(0)) -> P(s(0))
-> Rules:
 activate(n__f(X:S)) -> f(X:S)
 activate(X:S) -> X:S
 f(0) -> cons(0,n__f(s(0)))
 f(s(0)) -> f(p(s(0)))
 f(X:S) -> n__f(X:S)
 p(s(X:S)) -> X:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(s(0)) -> F(p(s(0)))
->->-> Rules:
 activate(n__f(X:S)) -> f(X:S)
 activate(X:S) -> X:S
 f(0) -> cons(0,n__f(s(0)))
 f(s(0)) -> f(p(s(0)))
 f(X:S) -> n__f(X:S)
 p(s(X:S)) -> X:S

Problem 1: 

Reduction Pair Processor:
-> Pairs:
 F(s(0)) -> F(p(s(0)))
-> Rules:
 activate(n__f(X:S)) -> f(X:S)
 activate(X:S) -> X:S
 f(0) -> cons(0,n__f(s(0)))
 f(s(0)) -> f(p(s(0)))
 f(X:S) -> n__f(X:S)
 p(s(X:S)) -> X:S
-> Usable rules:
 p(s(X:S)) -> X:S
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[p](X) = 1/2.X + 1/2
[0] = 0
[s](X) = 2.X + 2
[F](X) = 2.X

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 activate(n__f(X:S)) -> f(X:S)
 activate(X:S) -> X:S
 f(0) -> cons(0,n__f(s(0)))
 f(s(0)) -> f(p(s(0)))
 f(X:S) -> n__f(X:S)
 p(s(X:S)) -> X:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.