YES

Problem 1: 

(VAR v_NonEmpty:S M:S N:S X:S)
(RULES
U11(tt,M:S,N:S) -> U12(tt,activate(M:S),activate(N:S))
U12(tt,M:S,N:S) -> s(plus(activate(N:S),activate(M:S)))
activate(X:S) -> X:S
plus(N:S,0) -> N:S
plus(N:S,s(M:S)) -> U11(tt,M:S,N:S)
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 U11(tt,M:S,N:S) -> U12(tt,activate(M:S),activate(N:S))
 U12(tt,M:S,N:S) -> s(plus(activate(N:S),activate(M:S)))
 activate(X:S) -> X:S
 plus(N:S,0) -> N:S
 plus(N:S,s(M:S)) -> U11(tt,M:S,N:S)
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 U11#(tt,M:S,N:S) -> U12#(tt,activate(M:S),activate(N:S))
 U11#(tt,M:S,N:S) -> ACTIVATE(M:S)
 U11#(tt,M:S,N:S) -> ACTIVATE(N:S)
 U12#(tt,M:S,N:S) -> ACTIVATE(M:S)
 U12#(tt,M:S,N:S) -> ACTIVATE(N:S)
 U12#(tt,M:S,N:S) -> PLUS(activate(N:S),activate(M:S))
 PLUS(N:S,s(M:S)) -> U11#(tt,M:S,N:S)
-> Rules:
 U11(tt,M:S,N:S) -> U12(tt,activate(M:S),activate(N:S))
 U12(tt,M:S,N:S) -> s(plus(activate(N:S),activate(M:S)))
 activate(X:S) -> X:S
 plus(N:S,0) -> N:S
 plus(N:S,s(M:S)) -> U11(tt,M:S,N:S)

Problem 1: 

SCC Processor:
-> Pairs:
 U11#(tt,M:S,N:S) -> U12#(tt,activate(M:S),activate(N:S))
 U11#(tt,M:S,N:S) -> ACTIVATE(M:S)
 U11#(tt,M:S,N:S) -> ACTIVATE(N:S)
 U12#(tt,M:S,N:S) -> ACTIVATE(M:S)
 U12#(tt,M:S,N:S) -> ACTIVATE(N:S)
 U12#(tt,M:S,N:S) -> PLUS(activate(N:S),activate(M:S))
 PLUS(N:S,s(M:S)) -> U11#(tt,M:S,N:S)
-> Rules:
 U11(tt,M:S,N:S) -> U12(tt,activate(M:S),activate(N:S))
 U12(tt,M:S,N:S) -> s(plus(activate(N:S),activate(M:S)))
 activate(X:S) -> X:S
 plus(N:S,0) -> N:S
 plus(N:S,s(M:S)) -> U11(tt,M:S,N:S)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 U11#(tt,M:S,N:S) -> U12#(tt,activate(M:S),activate(N:S))
 U12#(tt,M:S,N:S) -> PLUS(activate(N:S),activate(M:S))
 PLUS(N:S,s(M:S)) -> U11#(tt,M:S,N:S)
->->-> Rules:
 U11(tt,M:S,N:S) -> U12(tt,activate(M:S),activate(N:S))
 U12(tt,M:S,N:S) -> s(plus(activate(N:S),activate(M:S)))
 activate(X:S) -> X:S
 plus(N:S,0) -> N:S
 plus(N:S,s(M:S)) -> U11(tt,M:S,N:S)

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 U11#(tt,M:S,N:S) -> U12#(tt,activate(M:S),activate(N:S))
 U12#(tt,M:S,N:S) -> PLUS(activate(N:S),activate(M:S))
 PLUS(N:S,s(M:S)) -> U11#(tt,M:S,N:S)
-> Rules:
 U11(tt,M:S,N:S) -> U12(tt,activate(M:S),activate(N:S))
 U12(tt,M:S,N:S) -> s(plus(activate(N:S),activate(M:S)))
 activate(X:S) -> X:S
 plus(N:S,0) -> N:S
 plus(N:S,s(M:S)) -> U11(tt,M:S,N:S)
-> Usable rules:
 activate(X:S) -> X:S
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[U11](X1,X2,X3) = 0
[U12](X1,X2,X3) = 0
[activate](X) = X
[plus](X1,X2) = 0
[0] = 0
[fSNonEmpty] = 0
[s](X) = 2.X + 2
[tt] = 1
[U11#](X1,X2,X3) = 2.X1 + 2.X2 + 2.X3 + 1
[U12#](X1,X2,X3) = 2.X2 + 2.X3 + 2
[ACTIVATE](X) = 0
[PLUS](X1,X2) = 2.X1 + X2 + 1

Problem 1: 

SCC Processor:
-> Pairs:
 U12#(tt,M:S,N:S) -> PLUS(activate(N:S),activate(M:S))
 PLUS(N:S,s(M:S)) -> U11#(tt,M:S,N:S)
-> Rules:
 U11(tt,M:S,N:S) -> U12(tt,activate(M:S),activate(N:S))
 U12(tt,M:S,N:S) -> s(plus(activate(N:S),activate(M:S)))
 activate(X:S) -> X:S
 plus(N:S,0) -> N:S
 plus(N:S,s(M:S)) -> U11(tt,M:S,N:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.