YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S)
(RULES
plus(x:S,0) -> x:S
times(x:S,plus(y:S,1)) -> plus(times(x:S,plus(y:S,times(1,0))),x:S)
times(x:S,0) -> 0
times(x:S,1) -> x:S
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 plus(x:S,0) -> x:S
 times(x:S,plus(y:S,1)) -> plus(times(x:S,plus(y:S,times(1,0))),x:S)
 times(x:S,0) -> 0
 times(x:S,1) -> 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:
 TIMES(x:S,plus(y:S,1)) -> PLUS(times(x:S,plus(y:S,times(1,0))),x:S)
 TIMES(x:S,plus(y:S,1)) -> PLUS(y:S,times(1,0))
 TIMES(x:S,plus(y:S,1)) -> TIMES(1,0)
 TIMES(x:S,plus(y:S,1)) -> TIMES(x:S,plus(y:S,times(1,0)))
-> Rules:
 plus(x:S,0) -> x:S
 times(x:S,plus(y:S,1)) -> plus(times(x:S,plus(y:S,times(1,0))),x:S)
 times(x:S,0) -> 0
 times(x:S,1) -> x:S

Problem 1: 

SCC Processor:
-> Pairs:
 TIMES(x:S,plus(y:S,1)) -> PLUS(times(x:S,plus(y:S,times(1,0))),x:S)
 TIMES(x:S,plus(y:S,1)) -> PLUS(y:S,times(1,0))
 TIMES(x:S,plus(y:S,1)) -> TIMES(1,0)
 TIMES(x:S,plus(y:S,1)) -> TIMES(x:S,plus(y:S,times(1,0)))
-> Rules:
 plus(x:S,0) -> x:S
 times(x:S,plus(y:S,1)) -> plus(times(x:S,plus(y:S,times(1,0))),x:S)
 times(x:S,0) -> 0
 times(x:S,1) -> x:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 TIMES(x:S,plus(y:S,1)) -> TIMES(x:S,plus(y:S,times(1,0)))
->->-> Rules:
 plus(x:S,0) -> x:S
 times(x:S,plus(y:S,1)) -> plus(times(x:S,plus(y:S,times(1,0))),x:S)
 times(x:S,0) -> 0
 times(x:S,1) -> x:S

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 TIMES(x:S,plus(y:S,1)) -> TIMES(x:S,plus(y:S,times(1,0)))
-> Rules:
 plus(x:S,0) -> x:S
 times(x:S,plus(y:S,1)) -> plus(times(x:S,plus(y:S,times(1,0))),x:S)
 times(x:S,0) -> 0
 times(x:S,1) -> x:S
-> Usable rules:
 plus(x:S,0) -> x:S
 times(x:S,plus(y:S,1)) -> plus(times(x:S,plus(y:S,times(1,0))),x:S)
 times(x:S,0) -> 0
 times(x:S,1) -> x:S
->Interpretation type:
 Simple mixed
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[plus](X1,X2) = X1 + 2.X2
[times](X1,X2) = 2.X1.X2
[0] = 0
[1] = 2
[fSNonEmpty] = 0
[PLUS](X1,X2) = 0
[TIMES](X1,X2) = 2.X2

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 plus(x:S,0) -> x:S
 times(x:S,plus(y:S,1)) -> plus(times(x:S,plus(y:S,times(1,0))),x:S)
 times(x:S,0) -> 0
 times(x:S,1) -> x:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.