/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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NO

Problem 1: 

(VAR N X XS Y YS)
(RULES
from(X) -> cons(X,from(s(X)))
minus(s(X),s(Y)) -> minus(X,Y)
minus(X,0) -> 0
quot(0,s(Y)) -> 0
quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
sel(0,cons(X,XS)) -> X
sel(s(N),cons(X,XS)) -> sel(N,XS)
zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
zWquot(nil,XS) -> nil
zWquot(XS,nil) -> nil
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 FROM(X) -> FROM(s(X))
 MINUS(s(X),s(Y)) -> MINUS(X,Y)
 QUOT(s(X),s(Y)) -> MINUS(X,Y)
 QUOT(s(X),s(Y)) -> QUOT(minus(X,Y),s(Y))
 SEL(s(N),cons(X,XS)) -> SEL(N,XS)
 ZWQUOT(cons(X,XS),cons(Y,YS)) -> QUOT(X,Y)
 ZWQUOT(cons(X,XS),cons(Y,YS)) -> ZWQUOT(XS,YS)
-> Rules:
 from(X) -> cons(X,from(s(X)))
 minus(s(X),s(Y)) -> minus(X,Y)
 minus(X,0) -> 0
 quot(0,s(Y)) -> 0
 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
 sel(0,cons(X,XS)) -> X
 sel(s(N),cons(X,XS)) -> sel(N,XS)
 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
 zWquot(nil,XS) -> nil
 zWquot(XS,nil) -> nil

Problem 1: 

Infinite Processor:
-> Pairs:
 FROM(X) -> FROM(s(X))
 MINUS(s(X),s(Y)) -> MINUS(X,Y)
 QUOT(s(X),s(Y)) -> MINUS(X,Y)
 QUOT(s(X),s(Y)) -> QUOT(minus(X,Y),s(Y))
 SEL(s(N),cons(X,XS)) -> SEL(N,XS)
 ZWQUOT(cons(X,XS),cons(Y,YS)) -> QUOT(X,Y)
 ZWQUOT(cons(X,XS),cons(Y,YS)) -> ZWQUOT(XS,YS)
-> Rules:
 from(X) -> cons(X,from(s(X)))
 minus(s(X),s(Y)) -> minus(X,Y)
 minus(X,0) -> 0
 quot(0,s(Y)) -> 0
 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
 sel(0,cons(X,XS)) -> X
 sel(s(N),cons(X,XS)) -> sel(N,XS)
 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
 zWquot(nil,XS) -> nil
 zWquot(XS,nil) -> nil
-> Pairs in cycle:
 FROM(X) -> FROM(s(X))

The problem is infinite.