YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 +#(s(x:S),s(y:S)) -> +#(s(x:S),+(y:S,0))
 +#(s(x:S),s(y:S)) -> +#(y:S,0)
-> Rules:
 +(0,y:S) -> y:S
 +(s(x:S),0) -> s(x:S)
 +(s(x:S),s(y:S)) -> s(+(s(x:S),+(y:S,0)))

Problem 1: 

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

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 +#(s(x:S),s(y:S)) -> +#(s(x:S),+(y:S,0))
-> Rules:
 +(0,y:S) -> y:S
 +(s(x:S),0) -> s(x:S)
 +(s(x:S),s(y:S)) -> s(+(s(x:S),+(y:S,0)))
-> Usable rules:
 +(0,y:S) -> y:S
 +(s(x:S),0) -> s(x:S)
 +(s(x:S),s(y:S)) -> s(+(s(x:S),+(y:S,0)))
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[+](X1,X2) = X1 + 2.X2 + 1
[0] = 0
[fSNonEmpty] = 0
[s](X) = X + 2
[+#](X1,X2) = 2.X2

Problem 1: 

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

The problem is finite.