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


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YES

Problem 1: 

(VAR N X Y Z)
(STRATEGY CONTEXTSENSITIVE
(add 1 2)
(dbl 1)
(first 1 2)
(sqr 1)
(terms 1)
(0)
(cons 1)
(nil)
(recip 1)
(s 1)
)
(RULES
add(0,X) -> X
add(s(X),Y) -> s(add(X,Y))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
first(0,X) -> nil
first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
sqr(0) -> 0
sqr(s(X)) -> s(add(sqr(X),dbl(X)))
terms(N) -> cons(recip(sqr(N)),terms(s(N)))
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
-> The context-sensitive term rewriting system is an orthogonal system. Therefore, innermost cs-termination implies cs-termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 ADD(s(X),Y) -> ADD(X,Y)
 DBL(s(X)) -> DBL(X)
 SQR(s(X)) -> ADD(sqr(X),dbl(X))
 SQR(s(X)) -> DBL(X)
 SQR(s(X)) -> SQR(X)
 TERMS(N) -> SQR(N)
-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
-> Unhiding Rules:
 Empty

Problem 1: 

SCC Processor:
-> Pairs:
 ADD(s(X),Y) -> ADD(X,Y)
 DBL(s(X)) -> DBL(X)
 SQR(s(X)) -> ADD(sqr(X),dbl(X))
 SQR(s(X)) -> DBL(X)
 SQR(s(X)) -> SQR(X)
 TERMS(N) -> SQR(N)
-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
-> Unhiding rules:
 Empty
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 DBL(s(X)) -> DBL(X)
->->-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
->->-> Unhiding rules:
 Empty
->->Cycle:
->->-> Pairs:
 ADD(s(X),Y) -> ADD(X,Y)
->->-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
->->-> Unhiding rules:
 Empty
->->Cycle:
->->-> Pairs:
 SQR(s(X)) -> SQR(X)
->->-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
->->-> Unhiding rules:
 Empty


The problem is decomposed in 3 subproblems.

Problem 1.1: 

SubNColl Processor:
-> Pairs:
 DBL(s(X)) -> DBL(X)
-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
-> Unhiding rules:
 Empty
->Projection:
 pi(DBL) = 1

Problem 1.1: 

Basic Processor:
-> Pairs:
 Empty
-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
-> Unhiding rules:
 Empty
-> Result:
 Set P is empty

The problem is finite.

Problem 1.2: 

SubNColl Processor:
-> Pairs:
 ADD(s(X),Y) -> ADD(X,Y)
-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
-> Unhiding rules:
 Empty
->Projection:
 pi(ADD) = 1

Problem 1.2: 

Basic Processor:
-> Pairs:
 Empty
-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
-> Unhiding rules:
 Empty
-> Result:
 Set P is empty

The problem is finite.

Problem 1.3: 

SubNColl Processor:
-> Pairs:
 SQR(s(X)) -> SQR(X)
-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
-> Unhiding rules:
 Empty
->Projection:
 pi(SQR) = 1

Problem 1.3: 

Basic Processor:
-> Pairs:
 Empty
-> Rules:
 add(0,X) -> X
 add(s(X),Y) -> s(add(X,Y))
 dbl(0) -> 0
 dbl(s(X)) -> s(s(dbl(X)))
 first(0,X) -> nil
 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
 sqr(0) -> 0
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 terms(N) -> cons(recip(sqr(N)),terms(s(N)))
-> Unhiding rules:
 Empty
-> Result:
 Set P is empty

The problem is finite.