YES

Problem 1: 

(VAR v_NonEmpty:S x1:S)
(RULES
f(0(x1:S)) -> 0(x1:S)
f(s(x1:S)) -> s(s(f(p(s(x1:S)))))
p(s(x1:S)) -> x1:S
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 f(0(x1:S)) -> 0(x1:S)
 f(s(x1:S)) -> s(s(f(p(s(x1:S)))))
 p(s(x1:S)) -> x1:S
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 F(s(x1:S)) -> F(p(s(x1:S)))
 F(s(x1:S)) -> P(s(x1:S))
-> Rules:
 f(0(x1:S)) -> 0(x1:S)
 f(s(x1:S)) -> s(s(f(p(s(x1:S)))))
 p(s(x1:S)) -> x1:S

Problem 1: 

SCC Processor:
-> Pairs:
 F(s(x1:S)) -> F(p(s(x1:S)))
 F(s(x1:S)) -> P(s(x1:S))
-> Rules:
 f(0(x1:S)) -> 0(x1:S)
 f(s(x1:S)) -> s(s(f(p(s(x1:S)))))
 p(s(x1:S)) -> x1:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(s(x1:S)) -> F(p(s(x1:S)))
->->-> Rules:
 f(0(x1:S)) -> 0(x1:S)
 f(s(x1:S)) -> s(s(f(p(s(x1:S)))))
 p(s(x1:S)) -> x1:S

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 F(s(x1:S)) -> F(p(s(x1:S)))
-> Rules:
 f(0(x1:S)) -> 0(x1:S)
 f(s(x1:S)) -> s(s(f(p(s(x1:S)))))
 p(s(x1:S)) -> x1:S
-> Usable rules:
 p(s(x1:S)) -> x1:S
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[f](X) = 0
[p](X) = 1/2.X
[0](X) = 0
[fSNonEmpty] = 0
[s](X) = 2.X + 1/2
[F](X) = 2.X
[P](X) = 0

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 f(0(x1:S)) -> 0(x1:S)
 f(s(x1:S)) -> s(s(f(p(s(x1:S)))))
 p(s(x1:S)) -> x1:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.