YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S z:S)
(RULES
quot(0,s(y:S),s(z:S)) -> 0
quot(s(x:S),s(y:S),z:S) -> quot(x:S,y:S,z:S)
quot(x:S,0,s(z:S)) -> s(quot(x:S,s(z:S),s(z:S)))
) 
(STRATEGY INNERMOST)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 QUOT(s(x:S),s(y:S),z:S) -> QUOT(x:S,y:S,z:S)
 QUOT(x:S,0,s(z:S)) -> QUOT(x:S,s(z:S),s(z:S))
-> Rules:
 quot(0,s(y:S),s(z:S)) -> 0
 quot(s(x:S),s(y:S),z:S) -> quot(x:S,y:S,z:S)
 quot(x:S,0,s(z:S)) -> s(quot(x:S,s(z:S),s(z:S)))

Problem 1: 

SCC Processor:
-> Pairs:
 QUOT(s(x:S),s(y:S),z:S) -> QUOT(x:S,y:S,z:S)
 QUOT(x:S,0,s(z:S)) -> QUOT(x:S,s(z:S),s(z:S))
-> Rules:
 quot(0,s(y:S),s(z:S)) -> 0
 quot(s(x:S),s(y:S),z:S) -> quot(x:S,y:S,z:S)
 quot(x:S,0,s(z:S)) -> s(quot(x:S,s(z:S),s(z:S)))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 QUOT(s(x:S),s(y:S),z:S) -> QUOT(x:S,y:S,z:S)
 QUOT(x:S,0,s(z:S)) -> QUOT(x:S,s(z:S),s(z:S))
->->-> Rules:
 quot(0,s(y:S),s(z:S)) -> 0
 quot(s(x:S),s(y:S),z:S) -> quot(x:S,y:S,z:S)
 quot(x:S,0,s(z:S)) -> s(quot(x:S,s(z:S),s(z:S)))

Problem 1: 

Subterm Processor:
-> Pairs:
 QUOT(s(x:S),s(y:S),z:S) -> QUOT(x:S,y:S,z:S)
 QUOT(x:S,0,s(z:S)) -> QUOT(x:S,s(z:S),s(z:S))
-> Rules:
 quot(0,s(y:S),s(z:S)) -> 0
 quot(s(x:S),s(y:S),z:S) -> quot(x:S,y:S,z:S)
 quot(x:S,0,s(z:S)) -> s(quot(x:S,s(z:S),s(z:S)))
->Projection:
 pi(QUOT) = 1

Problem 1: 

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

The problem is finite.