Term Rewriting System R:
[N, X, Y, X1, X2, Z]
terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

TERMS(N) -> SQR(N)
TERMS(N) -> S(N)
SQR(s(X)) -> S(nadd(sqr(activate(X)), dbl(activate(X))))
SQR(s(X)) -> SQR(activate(X))
SQR(s(X)) -> ACTIVATE(X)
SQR(s(X)) -> DBL(activate(X))
DBL(s(X)) -> S(ns(ndbl(activate(X))))
DBL(s(X)) -> ACTIVATE(X)
ADD(s(X), Y) -> S(nadd(activate(X), Y))
ADD(s(X), Y) -> ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nterms(X)) -> TERMS(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
ACTIVATE(ns(X)) -> S(X)
ACTIVATE(ndbl(X)) -> DBL(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

SQR(s(X)) -> DBL(activate(X))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
DBL(s(X)) -> ACTIVATE(X)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
ACTIVATE(nterms(X)) -> TERMS(X)
SQR(s(X)) -> ACTIVATE(X)
SQR(s(X)) -> SQR(activate(X))
TERMS(N) -> SQR(N)


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(X)) -> SQR(activate(X))
six new Dependency Pairs are created:

SQR(s(nterms(X''))) -> SQR(terms(X''))
SQR(s(nadd(X1', X2'))) -> SQR(add(X1', X2'))
SQR(s(ns(X''))) -> SQR(s(X''))
SQR(s(ndbl(X''))) -> SQR(dbl(X''))
SQR(s(nfirst(X1', X2'))) -> SQR(first(X1', X2'))
SQR(s(X'')) -> SQR(X'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Narrowing Transformation


Dependency Pairs:

SQR(s(X'')) -> SQR(X'')
SQR(s(nfirst(X1', X2'))) -> SQR(first(X1', X2'))
SQR(s(ndbl(X''))) -> SQR(dbl(X''))
SQR(s(ns(X''))) -> SQR(s(X''))
SQR(s(nadd(X1', X2'))) -> SQR(add(X1', X2'))
SQR(s(nterms(X''))) -> SQR(terms(X''))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
SQR(s(X)) -> ACTIVATE(X)
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(X)) -> DBL(activate(X))


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(X)) -> DBL(activate(X))
six new Dependency Pairs are created:

SQR(s(nterms(X''))) -> DBL(terms(X''))
SQR(s(nadd(X1', X2'))) -> DBL(add(X1', X2'))
SQR(s(ns(X''))) -> DBL(s(X''))
SQR(s(ndbl(X''))) -> DBL(dbl(X''))
SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))
SQR(s(X'')) -> DBL(X'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Narrowing Transformation


Dependency Pairs:

SQR(s(X'')) -> DBL(X'')
SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))
SQR(s(ndbl(X''))) -> DBL(dbl(X''))
SQR(s(ns(X''))) -> DBL(s(X''))
SQR(s(nadd(X1', X2'))) -> DBL(add(X1', X2'))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(nterms(X''))) -> DBL(terms(X''))
SQR(s(nfirst(X1', X2'))) -> SQR(first(X1', X2'))
SQR(s(ndbl(X''))) -> SQR(dbl(X''))
SQR(s(ns(X''))) -> SQR(s(X''))
SQR(s(nadd(X1', X2'))) -> SQR(add(X1', X2'))
SQR(s(nterms(X''))) -> SQR(terms(X''))
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
SQR(s(X)) -> ACTIVATE(X)
SQR(s(X'')) -> SQR(X'')


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(nterms(X''))) -> SQR(terms(X''))
two new Dependency Pairs are created:

SQR(s(nterms(X'''))) -> SQR(cons(recip(sqr(X''')), nterms(s(X'''))))
SQR(s(nterms(X'''))) -> SQR(nterms(X'''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 4
                 ↳Narrowing Transformation


Dependency Pairs:

SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))
SQR(s(ndbl(X''))) -> DBL(dbl(X''))
SQR(s(ns(X''))) -> DBL(s(X''))
SQR(s(nadd(X1', X2'))) -> DBL(add(X1', X2'))
SQR(s(nterms(X''))) -> DBL(terms(X''))
SQR(s(X'')) -> SQR(X'')
SQR(s(nfirst(X1', X2'))) -> SQR(first(X1', X2'))
SQR(s(ndbl(X''))) -> SQR(dbl(X''))
SQR(s(ns(X''))) -> SQR(s(X''))
SQR(s(nadd(X1', X2'))) -> SQR(add(X1', X2'))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
SQR(s(X)) -> ACTIVATE(X)
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(X'')) -> DBL(X'')


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(nadd(X1', X2'))) -> SQR(add(X1', X2'))
three new Dependency Pairs are created:

SQR(s(nadd(0, X2''))) -> SQR(X2'')
SQR(s(nadd(s(X'), X2''))) -> SQR(s(nadd(activate(X'), X2'')))
SQR(s(nadd(X1'', X2''))) -> SQR(nadd(X1'', X2''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 5
                 ↳Narrowing Transformation


Dependency Pairs:

SQR(s(nadd(s(X'), X2''))) -> SQR(s(nadd(activate(X'), X2'')))
SQR(s(nadd(0, X2''))) -> SQR(X2'')
SQR(s(X'')) -> DBL(X'')
SQR(s(ndbl(X''))) -> DBL(dbl(X''))
SQR(s(ns(X''))) -> DBL(s(X''))
SQR(s(nadd(X1', X2'))) -> DBL(add(X1', X2'))
SQR(s(nterms(X''))) -> DBL(terms(X''))
SQR(s(X'')) -> SQR(X'')
SQR(s(nfirst(X1', X2'))) -> SQR(first(X1', X2'))
SQR(s(ndbl(X''))) -> SQR(dbl(X''))
SQR(s(ns(X''))) -> SQR(s(X''))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
SQR(s(X)) -> ACTIVATE(X)
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(ndbl(X''))) -> SQR(dbl(X''))
three new Dependency Pairs are created:

SQR(s(ndbl(0))) -> SQR(0)
SQR(s(ndbl(s(X')))) -> SQR(s(ns(ndbl(activate(X')))))
SQR(s(ndbl(X'''))) -> SQR(ndbl(X'''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 6
                 ↳Narrowing Transformation


Dependency Pairs:

SQR(s(ndbl(s(X')))) -> SQR(s(ns(ndbl(activate(X')))))
SQR(s(nadd(0, X2''))) -> SQR(X2'')
SQR(s(X'')) -> DBL(X'')
SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))
SQR(s(ndbl(X''))) -> DBL(dbl(X''))
SQR(s(ns(X''))) -> DBL(s(X''))
SQR(s(nadd(X1', X2'))) -> DBL(add(X1', X2'))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(nterms(X''))) -> DBL(terms(X''))
SQR(s(X'')) -> SQR(X'')
SQR(s(nfirst(X1', X2'))) -> SQR(first(X1', X2'))
SQR(s(ns(X''))) -> SQR(s(X''))
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
SQR(s(X)) -> ACTIVATE(X)
SQR(s(nadd(s(X'), X2''))) -> SQR(s(nadd(activate(X'), X2'')))


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(nfirst(X1', X2'))) -> SQR(first(X1', X2'))
three new Dependency Pairs are created:

SQR(s(nfirst(0, X2''))) -> SQR(nil)
SQR(s(nfirst(s(X'), cons(Y', Z')))) -> SQR(cons(Y', nfirst(activate(X'), activate(Z'))))
SQR(s(nfirst(X1'', X2''))) -> SQR(nfirst(X1'', X2''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 7
                 ↳Narrowing Transformation


Dependency Pairs:

SQR(s(nadd(s(X'), X2''))) -> SQR(s(nadd(activate(X'), X2'')))
SQR(s(nadd(0, X2''))) -> SQR(X2'')
SQR(s(X'')) -> DBL(X'')
SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))
SQR(s(ndbl(X''))) -> DBL(dbl(X''))
SQR(s(ns(X''))) -> DBL(s(X''))
SQR(s(nadd(X1', X2'))) -> DBL(add(X1', X2'))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(nterms(X''))) -> DBL(terms(X''))
SQR(s(X'')) -> SQR(X'')
SQR(s(ns(X''))) -> SQR(s(X''))
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
SQR(s(X)) -> ACTIVATE(X)
SQR(s(ndbl(s(X')))) -> SQR(s(ns(ndbl(activate(X')))))


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(nterms(X''))) -> DBL(terms(X''))
two new Dependency Pairs are created:

SQR(s(nterms(X'''))) -> DBL(cons(recip(sqr(X''')), nterms(s(X'''))))
SQR(s(nterms(X'''))) -> DBL(nterms(X'''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 8
                 ↳Narrowing Transformation


Dependency Pairs:

SQR(s(ndbl(s(X')))) -> SQR(s(ns(ndbl(activate(X')))))
SQR(s(nadd(0, X2''))) -> SQR(X2'')
SQR(s(X'')) -> DBL(X'')
SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))
SQR(s(ndbl(X''))) -> DBL(dbl(X''))
SQR(s(ns(X''))) -> DBL(s(X''))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(nadd(X1', X2'))) -> DBL(add(X1', X2'))
SQR(s(X'')) -> SQR(X'')
SQR(s(ns(X''))) -> SQR(s(X''))
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
SQR(s(X)) -> ACTIVATE(X)
SQR(s(nadd(s(X'), X2''))) -> SQR(s(nadd(activate(X'), X2'')))


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(nadd(X1', X2'))) -> DBL(add(X1', X2'))
three new Dependency Pairs are created:

SQR(s(nadd(0, X2''))) -> DBL(X2'')
SQR(s(nadd(s(X'), X2''))) -> DBL(s(nadd(activate(X'), X2'')))
SQR(s(nadd(X1'', X2''))) -> DBL(nadd(X1'', X2''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 9
                 ↳Narrowing Transformation


Dependency Pairs:

SQR(s(nadd(s(X'), X2''))) -> DBL(s(nadd(activate(X'), X2'')))
SQR(s(nadd(0, X2''))) -> DBL(X2'')
SQR(s(nadd(s(X'), X2''))) -> SQR(s(nadd(activate(X'), X2'')))
SQR(s(nadd(0, X2''))) -> SQR(X2'')
SQR(s(X'')) -> DBL(X'')
SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))
SQR(s(ndbl(X''))) -> DBL(dbl(X''))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(ns(X''))) -> DBL(s(X''))
SQR(s(X'')) -> SQR(X'')
SQR(s(ns(X''))) -> SQR(s(X''))
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
SQR(s(X)) -> ACTIVATE(X)
SQR(s(ndbl(s(X')))) -> SQR(s(ns(ndbl(activate(X')))))


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(ndbl(X''))) -> DBL(dbl(X''))
three new Dependency Pairs are created:

SQR(s(ndbl(0))) -> DBL(0)
SQR(s(ndbl(s(X')))) -> DBL(s(ns(ndbl(activate(X')))))
SQR(s(ndbl(X'''))) -> DBL(ndbl(X'''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 10
                 ↳Narrowing Transformation


Dependency Pairs:

SQR(s(ndbl(s(X')))) -> DBL(s(ns(ndbl(activate(X')))))
SQR(s(nadd(0, X2''))) -> DBL(X2'')
SQR(s(ndbl(s(X')))) -> SQR(s(ns(ndbl(activate(X')))))
SQR(s(nadd(s(X'), X2''))) -> SQR(s(nadd(activate(X'), X2'')))
SQR(s(nadd(0, X2''))) -> SQR(X2'')
SQR(s(X'')) -> DBL(X'')
SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))
SQR(s(ns(X''))) -> DBL(s(X''))
SQR(s(X'')) -> SQR(X'')
SQR(s(ns(X''))) -> SQR(s(X''))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
SQR(s(X)) -> ACTIVATE(X)
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(nadd(s(X'), X2''))) -> DBL(s(nadd(activate(X'), X2'')))


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SQR(s(nfirst(X1', X2'))) -> DBL(first(X1', X2'))
three new Dependency Pairs are created:

SQR(s(nfirst(0, X2''))) -> DBL(nil)
SQR(s(nfirst(s(X'), cons(Y', Z')))) -> DBL(cons(Y', nfirst(activate(X'), activate(Z'))))
SQR(s(nfirst(X1'', X2''))) -> DBL(nfirst(X1'', X2''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 11
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

SQR(s(nadd(s(X'), X2''))) -> DBL(s(nadd(activate(X'), X2'')))
SQR(s(nadd(0, X2''))) -> DBL(X2'')
SQR(s(ndbl(s(X')))) -> SQR(s(ns(ndbl(activate(X')))))
SQR(s(nadd(s(X'), X2''))) -> SQR(s(nadd(activate(X'), X2'')))
SQR(s(nadd(0, X2''))) -> SQR(X2'')
SQR(s(X'')) -> DBL(X'')
SQR(s(ns(X''))) -> DBL(s(X''))
SQR(s(X'')) -> SQR(X'')
SQR(s(ns(X''))) -> SQR(s(X''))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(ndbl(X)) -> DBL(X)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
SQR(s(X)) -> ACTIVATE(X)
TERMS(N) -> SQR(N)
ACTIVATE(nterms(X)) -> TERMS(X)
DBL(s(X)) -> ACTIVATE(X)
SQR(s(ndbl(s(X')))) -> DBL(s(ns(ndbl(activate(X')))))


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(s(N)))
terms(X) -> nterms(X)
sqr(0) -> 0
sqr(s(X)) -> s(nadd(sqr(activate(X)), dbl(activate(X))))
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
s(X) -> ns(X)
activate(nterms(X)) -> terms(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X




Termination of R could not be shown.
Duration:
0:01 minutes