YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S)
(RULES
f(s(x:S),s(y:S)) -> f(x:S,s(c(s(y:S))))
f(x:S,c(y:S)) -> f(x:S,s(f(y:S,y:S)))
) 
(STRATEGY INNERMOST)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 F(s(x:S),s(y:S)) -> F(x:S,s(c(s(y:S))))
 F(x:S,c(y:S)) -> F(x:S,s(f(y:S,y:S)))
 F(x:S,c(y:S)) -> F(y:S,y:S)
-> Rules:
 f(s(x:S),s(y:S)) -> f(x:S,s(c(s(y:S))))
 f(x:S,c(y:S)) -> f(x:S,s(f(y:S,y:S)))

Problem 1: 

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


The problem is decomposed in 2 subproblems.

Problem 1.1: 

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

Problem 1.1: 

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

The problem is finite.

Problem 1.2: 

Subterm Processor:
-> Pairs:
 F(x:S,c(y:S)) -> F(y:S,y:S)
-> Rules:
 f(s(x:S),s(y:S)) -> f(x:S,s(c(s(y:S))))
 f(x:S,c(y:S)) -> f(x:S,s(f(y:S,y:S)))
->Projection:
 pi(F) = 2

Problem 1.2: 

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

The problem is finite.