YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.