YES

Problem 1: 

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

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 f(0) -> s(0)
 f(s(0)) -> s(0)
 f(s(s(x:S))) -> f(f(s(x: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(s(x:S))) -> F(f(s(x:S)))
 F(s(s(x:S))) -> F(s(x:S))
-> Rules:
 f(0) -> s(0)
 f(s(0)) -> s(0)
 f(s(s(x:S))) -> f(f(s(x:S)))

Problem 1: 

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

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 F(s(s(x:S))) -> F(f(s(x:S)))
 F(s(s(x:S))) -> F(s(x:S))
-> Rules:
 f(0) -> s(0)
 f(s(0)) -> s(0)
 f(s(s(x:S))) -> f(f(s(x:S)))
-> Usable rules:
 f(0) -> s(0)
 f(s(0)) -> s(0)
 f(s(s(x:S))) -> f(f(s(x:S)))
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[f](X) = 2
[0] = 0
[fSNonEmpty] = 0
[s](X) = 2.X + 2
[F](X) = 2.X

Problem 1: 

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

Problem 1: 

Subterm Processor:
-> Pairs:
 F(s(s(x:S))) -> F(s(x:S))
-> Rules:
 f(0) -> s(0)
 f(s(0)) -> s(0)
 f(s(s(x:S))) -> f(f(s(x:S)))
->Projection:
 pi(F) = 1

Problem 1: 

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

The problem is finite.