YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

Subterm Processor:
-> Pairs:
 +#(-(x:S,y:S),z:S) -> +#(x:S,z:S)
-> Rules:
 +(-(x:S,y:S),z:S) -> -(+(x:S,z:S),y:S)
 -(+(x:S,y:S),y:S) -> x:S
->Projection:
 pi(+#) = 1

Problem 1: 

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

The problem is finite.