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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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)
HALF(s(s(X))) -> HALF(X)
ACTIVATE(nterms(X)) -> TERMS(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains five SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 6
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 6
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 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 6
             ↳FwdInst
             ...
               →DP Problem 8
                 ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 9
             ↳Forward Instantiation Transformation
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 9
             ↳FwdInst
             ...
               →DP Problem 10
                 ↳Polynomial Ordering
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 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 9
             ↳FwdInst
             ...
               →DP Problem 11
                 ↳Dependency Graph
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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
       →DP Problem 5
         ↳FwdInst


Dependency Pair:

HALF(s(s(X))) -> HALF(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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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

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

HALF(s(s(s(s(X''))))) -> HALF(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 12
             ↳Forward Instantiation Transformation
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳FwdInst


Dependency Pair:

HALF(s(s(s(s(X''))))) -> HALF(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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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

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

HALF(s(s(s(s(s(s(X''''))))))) -> HALF(s(s(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 12
             ↳FwdInst
             ...
               →DP Problem 13
                 ↳Polynomial Ordering
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳FwdInst


Dependency Pair:

HALF(s(s(s(s(s(s(X''''))))))) -> HALF(s(s(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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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:

HALF(s(s(s(s(s(s(X''''))))))) -> HALF(s(s(s(s(X'''')))))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(HALF(x1))=  1 + x1  
  POL(s(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 12
             ↳FwdInst
             ...
               →DP Problem 14
                 ↳Dependency Graph
       →DP Problem 4
         ↳FwdInst
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 4
         ↳FwdInst
           →DP Problem 15
             ↳Forward Instantiation Transformation
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 4
         ↳FwdInst
           →DP Problem 15
             ↳FwdInst
             ...
               →DP Problem 16
                 ↳Polynomial Ordering
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 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 4
         ↳FwdInst
           →DP Problem 15
             ↳FwdInst
             ...
               →DP Problem 17
                 ↳Dependency Graph
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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
         ↳FwdInst
       →DP Problem 5
         ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 5
         ↳FwdInst
           →DP Problem 18
             ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 5
         ↳FwdInst
           →DP Problem 18
             ↳FwdInst
             ...
               →DP Problem 19
                 ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 5
         ↳FwdInst
           →DP Problem 18
             ↳FwdInst
             ...
               →DP Problem 20
                 ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 5
         ↳FwdInst
           →DP Problem 18
             ↳FwdInst
             ...
               →DP Problem 21
                 ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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 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))=  1  
  POL(ACTIVATE(x1))=  x1  
  POL(n__first(x1, x2))=  x1 + 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 5
         ↳FwdInst
           →DP Problem 18
             ↳FwdInst
             ...
               →DP Problem 22
                 ↳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)
half(0) -> 0
half(s(0)) -> 0
half(s(s(X))) -> s(half(X))
half(dbl(X)) -> X
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