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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst


Dependency Pair:

ADD(s(X), Y) -> ADD(X, Y)


Rules:


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


Strategy:

innermost




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

ADD(s(X), Y) -> ADD(X, Y)
one new Dependency Pair is created:

ADD(s(s(X'')), Y'') -> ADD(s(X''), Y'')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 5
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst


Dependency Pair:

ADD(s(s(X'')), Y'') -> ADD(s(X''), Y'')


Rules:


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


Strategy:

innermost




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

ADD(s(s(X'')), Y'') -> ADD(s(X''), Y'')
one new Dependency Pair is created:

ADD(s(s(s(X''''))), Y'''') -> ADD(s(s(X'''')), Y'''')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 6
Polynomial Ordering
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst


Dependency Pair:

ADD(s(s(s(X''''))), Y'''') -> ADD(s(s(X'''')), Y'''')


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

ADD(s(s(s(X''''))), Y'''') -> ADD(s(s(X'''')), Y'''')


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(ADD(x1, x2))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 7
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Forward Instantiation Transformation
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst


Dependency Pair:

DBL(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(add(sqr(X), dbl(X)))
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
activate(nterms(X)) -> terms(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X


Strategy:

innermost




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

DBL(s(X)) -> DBL(X)
one new Dependency Pair is created:

DBL(s(s(X''))) -> DBL(s(X''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 8
Forward Instantiation Transformation
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst


Dependency Pair:

DBL(s(s(X''))) -> DBL(s(X''))


Rules:


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


Strategy:

innermost




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

DBL(s(s(X''))) -> DBL(s(X''))
one new Dependency Pair is created:

DBL(s(s(s(X'''')))) -> DBL(s(s(X'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 8
FwdInst
             ...
               →DP Problem 9
Polynomial Ordering
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst


Dependency Pair:

DBL(s(s(s(X'''')))) -> DBL(s(s(X'''')))


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

DBL(s(s(s(X'''')))) -> DBL(s(s(X'''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(DBL(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 8
FwdInst
             ...
               →DP Problem 10
Dependency Graph
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Forward Instantiation Transformation
       →DP Problem 4
FwdInst


Dependency Pair:

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


Strategy:

innermost




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

SQR(s(X)) -> SQR(X)
one new Dependency Pair is created:

SQR(s(s(X''))) -> SQR(s(X''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 11
Forward Instantiation Transformation
       →DP Problem 4
FwdInst


Dependency Pair:

SQR(s(s(X''))) -> SQR(s(X''))


Rules:


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


Strategy:

innermost




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

SQR(s(s(X''))) -> SQR(s(X''))
one new Dependency Pair is created:

SQR(s(s(s(X'''')))) -> SQR(s(s(X'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 11
FwdInst
             ...
               →DP Problem 12
Polynomial Ordering
       →DP Problem 4
FwdInst


Dependency Pair:

SQR(s(s(s(X'''')))) -> SQR(s(s(X'''')))


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

SQR(s(s(s(X'''')))) -> SQR(s(s(X'''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(SQR(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 11
FwdInst
             ...
               →DP Problem 13
Dependency Graph
       →DP Problem 4
FwdInst


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Forward Instantiation Transformation


Dependency Pairs:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)


Rules:


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


Strategy:

innermost




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

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
one new Dependency Pair is created:

FIRST(s(X), cons(Y, nfirst(X1'', X2''))) -> ACTIVATE(nfirst(X1'', X2''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
           →DP Problem 14
Forward Instantiation Transformation


Dependency Pairs:

FIRST(s(X), cons(Y, nfirst(X1'', X2''))) -> ACTIVATE(nfirst(X1'', X2''))
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)


Rules:


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


Strategy:

innermost




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

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
one new Dependency Pair is created:

ACTIVATE(nfirst(s(X''), cons(Y'', nfirst(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', nfirst(X1'''', X2'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
           →DP Problem 14
FwdInst
             ...
               →DP Problem 15
Forward Instantiation Transformation


Dependency Pairs:

ACTIVATE(nfirst(s(X''), cons(Y'', nfirst(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', nfirst(X1'''', X2'''')))
FIRST(s(X), cons(Y, nfirst(X1'', X2''))) -> ACTIVATE(nfirst(X1'', X2''))


Rules:


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


Strategy:

innermost




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

FIRST(s(X), cons(Y, nfirst(X1'', X2''))) -> ACTIVATE(nfirst(X1'', X2''))
one new Dependency Pair is created:

FIRST(s(X), cons(Y, nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))) -> ACTIVATE(nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
           →DP Problem 14
FwdInst
             ...
               →DP Problem 16
Forward Instantiation Transformation


Dependency Pairs:

FIRST(s(X), cons(Y, nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))) -> ACTIVATE(nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))
ACTIVATE(nfirst(s(X''), cons(Y'', nfirst(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', nfirst(X1'''', X2'''')))


Rules:


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


Strategy:

innermost




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

ACTIVATE(nfirst(s(X''), cons(Y'', nfirst(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', nfirst(X1'''', X2'''')))
one new Dependency Pair is created:

ACTIVATE(nfirst(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
           →DP Problem 14
FwdInst
             ...
               →DP Problem 17
Polynomial Ordering


Dependency Pairs:

ACTIVATE(nfirst(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))
FIRST(s(X), cons(Y, nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))) -> ACTIVATE(nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

ACTIVATE(nfirst(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', nfirst(s(X''''''), cons(Y'''''', nfirst(X1'''''''', X2'''''''')))))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  x2  
  POL(FIRST(x1, x2))=  x2  
  POL(s(x1))=  0  
  POL(ACTIVATE(x1))=  x1  
  POL(n__first(x1, x2))=  1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
FwdInst
           →DP Problem 14
FwdInst
             ...
               →DP Problem 18
Dependency Graph


Dependency Pair:

FIRST(s(X), cons(Y, nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))) -> ACTIVATE(nfirst(s(X''''), cons(Y'''', nfirst(X1'''''', X2''''''))))


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes