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

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo


Dependency Pair:

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


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





The following dependency pair can be strictly oriented:

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


There are no usable rules 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))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 5
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo


Dependency Pair:


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Polo
       →DP Problem 4
Polo


Dependency Pair:

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


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





The following dependency pair can be strictly oriented:

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


There are no usable rules 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))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 6
Dependency Graph
       →DP Problem 3
Polo
       →DP Problem 4
Polo


Dependency Pair:


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polynomial Ordering
       →DP Problem 4
Polo


Dependency Pair:

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


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





The following dependency pair can be strictly oriented:

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


There are no usable rules 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))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 7
Dependency Graph
       →DP Problem 4
Polo


Dependency Pair:


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polynomial Ordering


Dependency Pairs:

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


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





The following dependency pairs can be strictly oriented:

ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nfirst(X1, X2)) -> FIRST(activate(X1), activate(X2))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(activate(x1))=  x1  
  POL(n__terms(x1))=  x1  
  POL(sqr(x1))=  0  
  POL(n__s(x1))=  x1  
  POL(terms(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(add(x1, x2))=  0  
  POL(first(x1, x2))=  1 + x1 + x2  
  POL(0)=  0  
  POL(cons(x1, x2))=  x2  
  POL(FIRST(x1, x2))=  x2  
  POL(dbl(x1))=  0  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(recip(x1))=  0  
  POL(n__first(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
           →DP Problem 8
Dependency Graph


Dependency Pairs:

ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nterms(X)) -> ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





Using the Dependency Graph the DP problem was split into 1 DP problems.


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


Dependency Pairs:

ACTIVATE(nterms(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> ACTIVATE(X)


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





The following dependency pair can be strictly oriented:

ACTIVATE(nterms(X)) -> ACTIVATE(X)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__terms(x1))=  1 + x1  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
           →DP Problem 8
DGraph
             ...
               →DP Problem 10
Polynomial Ordering


Dependency Pair:

ACTIVATE(ns(X)) -> ACTIVATE(X)


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





The following dependency pair can be strictly oriented:

ACTIVATE(ns(X)) -> ACTIVATE(X)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__s(x1))=  1 + x1  
  POL(ACTIVATE(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
           →DP Problem 8
DGraph
             ...
               →DP Problem 11
Dependency Graph


Dependency Pair:


Rules:


terms(N) -> cons(recip(sqr(N)), nterms(ns(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)
s(X) -> ns(X)
activate(nterms(X)) -> terms(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.

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