YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S)
(RULES
+(x:S,0) -> x:S
+(x:S,s(y:S)) -> s(+(x:S,y:S))
sum(0) -> 0
sum(s(x:S)) -> +(sum(x:S),s(x:S))
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 +(x:S,0) -> x:S
 +(x:S,s(y:S)) -> s(+(x:S,y:S))
 sum(0) -> 0
 sum(s(x:S)) -> +(sum(x:S),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:
 +#(x:S,s(y:S)) -> +#(x:S,y:S)
 SUM(s(x:S)) -> +#(sum(x:S),s(x:S))
 SUM(s(x:S)) -> SUM(x:S)
-> Rules:
 +(x:S,0) -> x:S
 +(x:S,s(y:S)) -> s(+(x:S,y:S))
 sum(0) -> 0
 sum(s(x:S)) -> +(sum(x:S),s(x:S))

Problem 1: 

SCC Processor:
-> Pairs:
 +#(x:S,s(y:S)) -> +#(x:S,y:S)
 SUM(s(x:S)) -> +#(sum(x:S),s(x:S))
 SUM(s(x:S)) -> SUM(x:S)
-> Rules:
 +(x:S,0) -> x:S
 +(x:S,s(y:S)) -> s(+(x:S,y:S))
 sum(0) -> 0
 sum(s(x:S)) -> +(sum(x:S),s(x:S))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 +#(x:S,s(y:S)) -> +#(x:S,y:S)
->->-> Rules:
 +(x:S,0) -> x:S
 +(x:S,s(y:S)) -> s(+(x:S,y:S))
 sum(0) -> 0
 sum(s(x:S)) -> +(sum(x:S),s(x:S))
->->Cycle:
->->-> Pairs:
 SUM(s(x:S)) -> SUM(x:S)
->->-> Rules:
 +(x:S,0) -> x:S
 +(x:S,s(y:S)) -> s(+(x:S,y:S))
 sum(0) -> 0
 sum(s(x:S)) -> +(sum(x:S),s(x:S))


The problem is decomposed in 2 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 +#(x:S,s(y:S)) -> +#(x:S,y:S)
-> Rules:
 +(x:S,0) -> x:S
 +(x:S,s(y:S)) -> s(+(x:S,y:S))
 sum(0) -> 0
 sum(s(x:S)) -> +(sum(x:S),s(x:S))
->Projection:
 pi(+#) = 2

Problem 1.1: 

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

The problem is finite.

Problem 1.2: 

Subterm Processor:
-> Pairs:
 SUM(s(x:S)) -> SUM(x:S)
-> Rules:
 +(x:S,0) -> x:S
 +(x:S,s(y:S)) -> s(+(x:S,y:S))
 sum(0) -> 0
 sum(s(x:S)) -> +(sum(x:S),s(x:S))
->Projection:
 pi(SUM) = 1

Problem 1.2: 

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

The problem is finite.